Answer:
n = 5
Explanation:
methane => 1 Carbon => CH₃
ethane => 2 Carbons => C₂H₆
propane => 3 Carbons => C₃H₈
butane => 4 Carbons => C₄H₁₀
pentane => 5 Carbons => C₅H₁₂
hexane => 6 Carbons => C₆H₁₄
heptane => 7 Carbons => C₇H₁₆ => CH₃(CH₂)₅CH₃
Octante => 8 Carbons => C₈H₁₈
Nonane => 9 Carbons => C₉H₂₀
Decane => 10 Carbons => C₁₀H₂₂
The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H2OH2O that can be produced by combining 70.170.1 g of each reactant?
Answer:
[tex]\boxed{\text{47.4 g}}[/tex]
Explanation:
We are given the mass of two reactants, so this is a limiting reactant problem.
We know that we will need mases, moles, and molar masses, so, let's assemble all the data in one place, with molar masses above the formulas and masses below them.
M_r: 17.03 32.00 18.02
4NH₃ + 5O₂ ⟶ 4NO + 6H₂O
m/g: 70.1 70.1
Step 1. Calculate the moles of each reactant
[tex]\text{Moles of CO } = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{17.03 g}} = \text{4.116 mol}\\\\\text{Moles of H$_{2}$O} = \text{70.1 g} \times \dfrac{\text{1 mol}}{\text{32.00 g}} = \text{2.191 mol}[/tex]
Step 2. Identify the limiting reactant
Calculate the moles of H₂O we can obtain from each reactant.
From NH₃:
The molar ratio of H₂O:NH₃ is 6:4.
[tex]\text{Moles of H$_{2}$O} = \text{4.116 mol NH$_{3}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{4 mol NH$_{3}$}} = \text{6.174 mol H$_{2}$O}[/tex]
From O₂:
The molar ratio of H₂O:O₂ is 6:5.
[tex]\text{Moles of H$_{2}$O} = \text{2.191 mol O$_{2}$} \times \dfrac{\text{6 mol H$_{2}$O}}{\text{5 mol O$_{2}$}} = \text{2.629 mol H$_{2}$O}[/tex]
O₂ is the limiting reactant because it gives the smaller amount of H₂O.
Step 3. Calculate the theoretical yield.
[tex]\text{Theor. yield } = \text{2.629 mol H$_{2}$O}\times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \textbf{47.4 g H$_{2}$O}\\\\\text{The maximum yield of H$_{2}$O is }\boxed{\textbf{47.4 g}}[/tex]
100 mL of a 0.300 M solution of AgNO3 reacts with 100 mL of a 0.300 M solution of HCl in a coffee-cup calorimeter and the temperature rises from 21.80 °C to 23.20 °C. Assuming the density and specific heat of the resulting solution is 1.00 g/mL and 4.18 J/g ∙ °C, respectfully, what is the ΔH°rxn?
Answer:
ΔH°rxn = 39013.33 J/mol = 39.013 kJ/mol.
Explanation:
We can calculate the amount of heat (Q) released from the solution using the relation:Q = m.c.ΔT,
Where, Q is the amount of heat released from the solution (Q = ??? J).
m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.
c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).
ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).
∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.
∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).
Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.
M = (no. of moles of AgNO₃)/(Volume of the solution (L)).
∴ no. of moles of AgNO₃ = (M)(Volume of the solution (L)) = (0.3 M)(0.1 L) = 0.03 mol.
∴ ΔH°rxn = Qrxn/(no. of moles of AgNO₃) = (1170.4 J)/(0.03 mol) = 39013.33 J/mol = 39.013 kJ/mol.
The heat of reaction (ΔH°rxn) for this chemical reaction in a coffee-cup calorimeter is -19.508 kJ/mol, indicating an exothermic reaction.
Explanation:In this question, we need to calculate the heat change (ΔH°rxn) of a chemical reaction in a coffee cup calorimeter. The reaction has a temperature rise from 21.80 °C to 23.20 °C. This increase in temperature represents an exothermic reaction, which means heat is released in the process.
We can calculate the heat absorbed by using the equation q = mcΔT, where m is the mass, c is the specific heat, and ΔT is the change in temperature. Given that the density of the solution is 1.00 g/mL and the volume is 200 mL (100 mL of AgNO3 + 100 mL of HCl), our m = 200 g. The specific heat c is given as 4.18 J/g ∙ °C. With ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.40 °C, we find that q = (200 g)(4.18 J/g ∙ °C)(1.40 °C) = 1170.48 J.
ΔH°rxn is given per mole of reaction, so let's convert the heat released to kJ (1.17048 kJ) and the reaction to moles using the 0.300 M concentration. In this reaction, 1 mol of AgNO3 reacts with 1 mol of HCl. So the molar quantity in 200 mL solution is 0.3 mol/L * 0.2 L = 0.06 mol. Then the ΔH°rxn = 1.17048 kJ / 0.06 mol = 19.508 kJ/mol (Note that this value is negative as it is an exothermic reaction).
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Explain why water is polar. Check all that apply. View Available Hint(s) Water is known as a polar molecule because Check all that apply. the hydrogen atom attracts electrons much more strongly than the oxygen atom. the oxygen atom has a greater attraction for electrons than the hydrogen atom does. the oxygen and hydrogen atoms have the same electronegativity. the electrons of the covalent bond are not shared equally between the hydrogen and oxygen atoms. water dipole moment is equal to zero.
Answer:
The oxygen atom has a greater attraction for electrons than the hydrogen atom does.
The electrons of the covalent bond are not shared equally between the hydrogen and oxygen atoms.
Its molecular geometry is bent.
Water is polar because the oxygen atom has a greater electronegativity than the hydrogen atoms, leading to unequal sharing of electrons and partial charges on the atoms.
Water is known as a polar molecule for various reasons. First, the oxygen atom in water has a greater attraction for electrons than the hydrogen atoms, which is due to its higher electronegativity. This results in the electrons of the covalent bond between hydrogen and oxygen not being shared equally, leading to a partial negative charge on the oxygen atom and a partial positive charge on the hydrogen atoms. The polarity of water is a critical factor in its ability to form hydrogen bonds, which are stronger than conventional dipole-dipole forces and contribute to water's unique properties such as high surface tension and the ability to dissolve many substances.
Which of the following is true for all exergonic reactions? The reaction releases energy. A net input of energy from the surroundings is required for the reactions to proceed. The reactions are rapid. The products have more total energy than the reactants. The reaction goes only in a forward direction: all reactants will be converted to products, but no products will be converted to reactants.
Answer:
The reaction releases energy
Explanation:
The products of an exergonic reaction have a lower energy state (Delta-G) compared to the reactants. Therefore there is a negative delta –G between products and reactants after the reactions. This means some energy is lost into the environment usually through light or heat.
Exergonic reactions are characterized by a net release of energy but they still require a small initial energy input to start, referred to as the 'activation energy'. The speed or direction of the reaction is not determined by whether it's exergonic.
Explanation:In the context of chemical reactions, the true statement for all exergonic reactions is that such reactions result in a net release of energy. However, even exergonic reactions, which are characterized by energy release, require a small initial input of energy to get started. This initial energy demand is referred to as the 'activation energy'. Also, it's important to note that the speed of the reaction or its directionality (whether it proceeds only in a forward direction) are not inherently determined by whether a reaction is exergonic. These aspects depend on other reaction conditions and catalysis.
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1.50 × 104 J of energy is transferred thermally into a huge tank filled with liquid water. The water temperature remains constant at 10.0 ∘C during the process.Part ABy how much does the entropy of the water change?
Final answer:
The change in entropy of the water is 53.0 J/K.
Explanation:
The change in entropy of the water can be calculated using the equation:
ΔS = Q / T
Where ΔS is the change in entropy, Q is the heat transfer, and T is the temperature in Kelvin.
In this case, the heat transfer is 1.50 × 10^4 J and the temperature is 10.0 °C, which is equal to 283.15 K.
So, ΔS = 1.50 × 10^4 J / 283.15 K = 53.0 J/K.
Suppose you are investigating the reaction: M(s) + 2 HCl(aq) → MCl2(aq) + H2(g). You weigh out a 0.295 gram piece of metal and combine it with 65 mL of 1.00 M HCl in a coffee-cup calorimeter. If the molar mass of the metal is 57.78 g/mol, and you measure that the reaction absorbed 104 J of heat, what is the enthalpy of this reaction in kJ per mole of limiting reactant? Enter your answer numerically to three significant figures in units of kJ/mol.
The enthalpy change of the reaction per mole of limiting reactant is -568 kJ/mol.
Explanation:The enthalpy change of a reaction can be calculated using the amount of reactant used. In this case, 0.0500 mol of HCl was used, and the reaction absorbed 104 J of heat. Since AH is an extensive property, it is proportional to the amount of acid neutralized. Therefore, the enthalpy change for 1 mol of HCl is -2.9 kJ. To find the enthalpy of the reaction per mole of limiting reactant, we can use the molar mass of the metal and the stoichiometric ratio between HCl and the metal in the balanced equation.
To calculate the enthalpy change per mole of limiting reactant, we need to determine the number of moles of the metal. Since the molar mass of the metal is 57.78 g/mol, we can calculate the number of moles by dividing the mass of the metal (0.295 g) by its molar mass: 0.295 g ÷ 57.78 g/mol = 0.00510 mol.
Next, we use the stoichiometric ratio in the balanced equation to relate the moles of HCl reacted to the moles of metal reacted. From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of metal. Therefore, the moles of HCl reacted is also 0.00510 mol. Now we can calculate the enthalpy change per mole of limiting reactant: -2.9 kJ ÷ 0.00510 mol = -568 kJ/mol (rounded to three significant figures).
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Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.50 g of sodium carbonate is mixed with one containing 5.00 g of silver nitrate. How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete?
Answer:
1) The no. of grams of Na₂CO₃ = 1.96 g.
2) The no. of grams of AgNO₃ = 0.0 g, because it is the limiting reactant that is consumed completely.
3) The no. of grams of Ag₂CO₃ = 3.998 g ≅ 4.0 g.
4) The no. of grams of NaNO₃ = 2.498 g ≅ 2.5 g.
Explanation:
For the balanced equation:Na₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃(s) + 2NaNO₃,
It is clear that 1 mol of Na₂CO₃ and 2 mol of AgNO₃ to produce 1 mol of Ag₂CO₃ and 2 mol of NaNO₃.
Firstly, we need to calculate the no. of moles of 3.50 g of Na₂CO₃ and 5.00 g of AgNO₃:no. of moles of Na₂CO₃ = mass/molar mass = (3.5 g)/(105.9888 g/mol) = 0.033 mol.
no. of moles of AgNO₃ = mass/molar mass = (5.0 g)/(169.87 g/mol) = 0.0294 mol.
1) the no. of grams of sodium carbonate, and silver nitrate:
We need to determine the limiting reactant:From stichiometry of the balanced equation:1 mol of Na₂CO₃ reacts completely with 2 mol of AgNO₃ with (1: 2 molar ratio).
∴ 0.0145 mol of Na₂CO₃ "the excess reactant, and the remaining is in excess (0.033 mol - 0.0145 mol = 0.0185 mol)" reacts completely with (0.0294 mol) of AgNO₃ "limiting reactant".
∴ The no. of grams of Na₂CO₃ = (no. of moles remaining)(molar mass) = (0.0185 mol)((105.9888 g/mol) = 1.96 g.
∴ The no. of grams of AgNO₃ = 0.0 g, because it is the limiting reactant that is consumed completely.
2) the no. of grams of silver carbonate:
Firstly, we need to find the no. of moles of silver carbonate:Using cross multiplication:
1 mol of Na₂CO₃ produces → 1 mol of Ag₂CO₃, from stichiometry.
∴ 0.0145 mol of Na₂CO₃ produces → 0.0145 mol of Ag₂CO₃.
∴ The no. of grams of Ag₂CO₃ = (no. of moles of Ag₂CO₃)(molar mass of Ag₂CO₃) = (0.0145 mol)(275.7453 g/mol) = 3.998 g ≅ 4.0 g.
3) the no. of grams of sodium nitrate:
Firstly, we need to find the no. of moles of sodium nitrate:Using cross multiplication:
2 mol of AgNO₃ produces → 2 mol of NaNO₃, from stichiometry.
∴ 0.0294 mol of AgNO₃ produces → 0.0294 mol of NaNO₃.
∴ The no. of grams of NaNO₃ = (no. of moles of NaNO₃)(molar mass of NaNO₃) = (0.0294 mol)(84.9947 g/mol) = 2.498 g ≅ 2.5 g.
After the reaction of 3.50 g sodium carbonate and 5.00 g silver nitrate, approximately 0.42 g of sodium carbonate, 0 g of silver nitrate, 7.8 g of silver carbonate, and 2.3 g of sodium nitrate will be present.
Explanation:The reaction between sodium carbonate (Na2CO3) and silver nitrate (AgNO3) forms solid silver carbonate (Ag2CO3) and a solution of sodium nitrate (NaNO3). The balanced chemical equation for this reaction is 2AgNO3 + Na2CO3 yields 2Ag2CO3 (s) + 2NaNO3 (aq). To determine the amount of each compound present after the reaction, we first need to calculate the moles of sodium carbonate and silver nitrate. We find that 3.50 g Na2CO3 amounts to 0.033 moles and 5.00 g AgNO3 equals 0.029 moles. Since AgNO3 is the limiting reagent, it will be completely consumed in the reaction, producing 0.029 moles of Ag2CO3 and 0.029 moles of NaNO3. This results 0.004 moles of Na2CO3 left unreacted. Converting these moles back into grams gives approximately 0.42 g Na2CO3, 0 g AgNO3, 7.8 g Ag2CO3, and 2.3 g NaNO3 present after the reaction.
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4. Propanol and isopropanol are isomers. This means that they have A) the same molecular formula but different chemical properties. B) different molecular formulas but the same chemical properties. C) the same molecular formula and the same chemical properties. D) the same molecular formula but represent different states of the compound
Answer:
A) the same molecular formula but different chemical properties.
Explanation:
Isomerism is the existence of a compound with the same molecular formula but different molecular structures due to the difference in the arrangement of atoms or spatial orientation of the atoms. In such a compound, they differ in physical and/or chemical properties. Such possible structures are known as isomers.
Propanol and isopropanol differs only in the arrangement of atoms but they both have the same molecular formula. Their physical and chemical properties differ.
Propanol and isopropanol are structural isomers, which means they have the same molecular formula but different chemical properties due to different arrangement of atoms within the molecules.
Explanation:Propanol and isopropanol are indeed isomers, specifically they are structural isomers. Structural isomers are compounds that share the same molecular formula but have a different spatial arrangement of their atoms, leading to different chemical properties. This means, option A) the same molecular formula but different chemical properties, is correct.
A classic example of this are the compounds n-butane and 2-methylpropane. Both compounds have the same molecular formula, C4H10, but n-butane contains an unbranched chain of carbon atoms, while 2-methylpropane has a branched chain. This difference in structure results in different chemical properties for each compound.
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Calculate the concentration of H3O⁺ in a solution that contains 5.5 × 10-5 M OH⁻ at 25°C. Identify the solution as acidic, basic, or neutral. A) 1.8 × 10-10 M, basic B) 1.8 × 10-10 M, acidic C) 5.5 × 10-10 M, neutral D) 9.2 × 10-1 M, acidic E) 9.2 × 10-1 M, basic
Answer : The correct option is, (A) [tex]1.8\times 10^{-10}M[/tex], basic.
Explanation : Given,
Concentration of [tex]OH^-[/tex] ion = [tex]5.5\times 10^{-5}M[/tex]
First we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (5.5\times 10^{-5})[/tex]
[tex]pOH=4.26[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.26=9.74[/tex]
Now we have to calculate the [tex]H_3O^+[/tex] concentration.
[tex]pH=-\log [H_3O^+][/tex]
[tex]9.74=-\log [H_3O^+][/tex]
[tex][H_3O^+]=1.8\times 10^{-10}M[/tex]
As we know that, when the pH value is less than 7 then the solution acidic in nature and when the pH value is more than 7 then the solution basic in nature.
From the pH value, 9.74 we conclude that the solution is basic in nature because the value of pH is greater than 7.
Therefore, the [tex]H_3O^+[/tex] concentration is, [tex]1.8\times 10^{-10}M[/tex], basic.
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) What is the maximum mass of aluminum chloride that can be formed when reacting 32.0 g of aluminum with 37.0 g of chlorine? Express your answer to three significant figures and include the appropriate units.
Answer: The mass of aluminium chloride that can be formed are 46.3 g
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
For Aluminium:Given mass of aluminium = 32 g
Molar mass of aluminium = 26.98 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of aluminium}=\frac{32g}{26.98g/mol}=1.186mol[/tex]
For Chlorine:Given mass of chlorine = 37 g
Molar mass of chlorine = 71 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of chlorine gas}=\frac{37g}{71g/mol}=0.521mol[/tex]
For the given chemical equation:
[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]
By Stoichiometry of the reaction:
3 moles of chlorine gas is reacting with 2 moles of aluminium.
So, 0.521 moles of chlorine gas will react with = [tex]\frac{2}{3}\times 0.521=0.347moles[/tex] of aluminium.
As, given amount of aluminium is more than the required amount. Thus, it is considered as an excess reagent.
So, chlorine gas is considered as a limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
3 moles of chlorine gas is producing 2 moles of aluminium chloride
So, 0.521 moles of chlorine gas will react with = [tex]\frac{2}{3}\times 0.521=0.347moles[/tex] of aluminium chloride.
Now, calculating the mass of aluminium chloride by using equation 1, we get:
Moles of aluminium chloride = 0.347 moles
Molar mass of aluminium chloride = 133.34 g/mol
Putting all the values in equation 1, we get:
[tex]0.347mol=\frac{\text{Mass of aluminium chloride}}{133.34g/mol}\\\\\text{Mass of aluminium chloride}=46.3g[/tex]
Hence, the mass of aluminium chloride that can be formed are 46.3 g
The quantity of the substance is given by the mass. The mass of aluminium chloride formed when reacting aluminium with chlorine will be 46.3 gm.
What is mass?Mass is a quantitative factor that determines the amount of substance or matter present in the sample.
The chemical reaction can be shown as,
[tex]\rm 2 Al + 3Cl_{2} \rightarrow 2AlCl_{3}[/tex]
Calculate the number of moles of Aluminium:
[tex]\begin{aligned}\rm Moles & = \dfrac {\rm Mass }{\rm Molar\; mass}\\\\& = \dfrac{32}{26.98}\\\\& = 1.186\;\rm moles\end{aligned}[/tex]
Calculate the number of moles of Chlorine:
[tex]\begin{aligned}\rm Moles & = \dfrac {\rm Mass }{\rm Molar\; mass}\\\\& = \dfrac{37}{71}\\\\& = 0.521 \;\rm moles\end{aligned}[/tex]
From the stoichiometry of the reaction above:
2 moles of aluminium reacts with 3 moles of chlorine
So, moles of aluminium will react with 0.521 moles chorine is,
[tex]\dfrac{2}{3} \times 0.521 = 0.347 \;\rm moles[/tex]
From this, it can infer that chlorine gas is a limiting reagent and aluminium is an excess reagent.
From the stoichiometry of the reaction,
3 moles of chlorine = 2 moles of aluminium chloride
So, 0.521 moles chorine will give,
[tex]\dfrac{2}{3} \times 0.521 = 0.347 \;\text{ moles of aluminium chloride.}[/tex]
Calculate the mass of the aluminium chloride as:
[tex]\begin{aligned} \rm mass &= \rm moles \times molar\; mass\\\\&= 0.347 \times 133.34\\\\&= 46.3 \;\rm g\end{aligned}[/tex]
Therefore, the mass of the aluminium chloride is 46.3 gm.
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A sample of SO3 is introduced into an evacuated sealed container and heated to 600 K. The following equilibrium is established: 2 SO3( g) ∆ 2 SO2( g) + O2( g) The total pressure in the system is 3.0 atm and the mole fraction of O2 is 0.12. Find Kp
Answer: The value of [tex]K_p[/tex] is 0.050.
Explanation:
According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.
[tex]p_x=x\times P[/tex]
As we know the mole fraction of [tex]O_2[/tex] is 0.12
The partial pressure of [tex]O_2=0.12\times 3.0atm=0.36atm[/tex]
The partial pressure of [tex]SO_2=2\times 0.36atm=0.72atm[/tex]Thus the partial pressure of [tex]SO_3[/tex] is = [3 - (0.36+0.720)] atm = 1.92 atm
[tex]p_{SO3}[/tex]= 1.92 atm
[tex]2SO_3(g)\rightleftharpoons 2SO_2(g)+O_2(g)[/tex]
[tex]K_p=\frac{p_{O_2}\times (p_{SO_}2)^2}{(p_{SO_3})^2}[/tex]
[tex]K_p=\frac{0.36\times (0.72)^2}{(1.92)^2}[/tex]
[tex]K_p=0.050[/tex]
The value of [tex]K_p[/tex] is 0.050.
The value of Kp is 0.050.
Raoult's law:As per this law, the vapor pressure of a component at a given temperature should be equivalent to the mole fraction of that component and then it should be multiplied by the vapor pressure of that component in the pure state.
[tex]p_x = x\times P[/tex]
Since mole fraction of oxygen is 0.12
Now the partial pressure should be = 0.12(3) = 0.36
The partial pressure of SO_2 is = 2(0.36) = 0.72
Now the partial pressure of SO_3 is [3 - (0.36+0.720)] atm = 1.92 atm
Now Kp is
[tex]= 0.36 \times (0.72)^3 \div (1.92)^3[/tex]
= 0.050
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In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white solid, but in these experiments it often looks gray, due to small amounts of Mg3N2 , a compound formed as some of the magnesium reacts with nitrogen. Write a balanced equation for each reaction.
Explanation:
When magnesium metal burns is heated i the air it forms magnesium oxide.The balanced chemical reaction is given as:
[tex]2Mg+O_2\rightarrow 2MgO[/tex]
2 moles of magnesium metal when reacts with 1 moles of oxygen it gives 2 moles of magnesium oxide which is white in color.
Some times along with formation of magnesium oxide small amount of magnesium nitride also produced due to which magnesium oxide appears grey in color .The balanced chemical reaction is given as:
[tex]3Mg+N_2\rightarrow Mg_3N_2[/tex]
3 moles of magnesium combines with 1 mol of nitrogen gas to to give 1 mol of magnesium nitride.
Final answer:
Magnesium reacts with oxygen to form magnesium oxide (2 Mg(s) + O₂(g) → 2 MgO(s)) and can also react with nitrogen to form magnesium nitride (3 Mg(s) + N₂(g) → Mg₃N₂(s)), causing the white magnesium oxide to appear gray.
Explanation:
Chemical Reactions of Magnesium
When magnesium metal is heated in air, it primarily reacts with oxygen to produce magnesium oxide (MgO), which is a white solid. The balanced chemical equation for this reaction is:
2 Mg(s) + O₂(g) → 2 MgO(s)
However, when there is nitrogen present in the air, magnesium can also react with nitrogen to form magnesium nitride (Mg₃N₂), which generally gives a gray appearance to the product. The balanced equation for the formation of magnesium nitride is:
3 Mg(s) + N₂(g) → Mg₃N₂(s)
Both reactions involve a combination with another element to form a compound, showcasing the highly reactive nature of magnesium at high temperatures.
Indicate what type, or types, of reaction each of the following represents: (a) H2 O(g) + C(s) ⟶ CO(g) + H2 (g) (b) 2KClO3 (s) ⟶ 2KCl(s) + 3O2 (g) (c) Al(OH)3 (aq) + 3HCl(aq) ⟶ AlCl3 (aq) + 3H2 O(l) (d) Pb(NO3 )2 (aq) + H2SO4 (aq) ⟶ PbSO4 (s) + 2HNO3 (aq)
The given equations (a)-(d) represent synthesis, decomposition, neutralization, and precipitation reactions, respectively.
Explanation:(a) The given equation represents a synthesis reaction, where hydrogen gas (H2) reacts with carbon solid (C) to form carbon monoxide gas (CO) and hydrogen gas (H2).
(b) The given equation represents a decomposition reaction, where potassium chlorate solid (KClO3) decomposes to form potassium chloride solid (KCl) and oxygen gas (O2).
(c) The given equation represents a neutralization reaction, where aluminum hydroxide (Al(OH)3) reacts with hydrochloric acid (HCl) to form aluminum chloride (AlCl3) and water (H2O).
(d) The given equation represents a precipitation reaction, where lead nitrate (Pb(NO3)2) reacts with sulfuric acid (H2SO4) to form lead sulfate solid (PbSO4) and nitric acid (HNO3) in aqueous form.
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The given equations (a)-(d) represent, a) Synthesis reaction, b) Decomposition reaction, c) neutralization reaction and d) precipitation reaction.
Reaction (a): H₂O(g) + C(s) ⟶ CO(g) + H₂(g) is a synthesis reaction, also known as a combination reaction, where two or more reactants combine to form a single product. In this case, hydrogen gas (H₂) and carbon monoxide (CO) are produced from the combination of water vapor (H₂O) and carbon (C).
Reaction (b): 2KClO₃(s) ⟶ 2KCl(s) + 3O₂(g) is a decomposition reaction, where a single compound breaks down into two or more simpler substances.
Reaction (c): Al(OH)₃(aq) + 3HCl(aq) ⟶ AlCl₃(aq) + 3H₂O(l) is an acid-base neutralization reaction, in which an acid (HCl) reacts with a base (Al(OH)₃) to produce a salt (AlCl₃) and water.
Reaction (d): Pb(NO₃)₂(aq) + H₂SO₃(aq) ⟶ PbSO₄(s) + 2HNO₃(aq) is a precipitation reaction, where two solutions react to form an insoluble solid (precipitate) and a soluble compound.
Please help 25 points :D
Phases of Matter Activity
Now it is your turn to show what you know about phases of matter and thermal energy transfer! Your task is to create a presentation to explain the transformation of a substance as it changes phases. You may choose to write a story or create a comic strip. For your story or comic, you will create a main character and detail the adventure as your character is exposed to thermal energy, causing it to undergo phase changes from a solid, to a liquid, to a gas. You may create your own comic strip using drawings, presentation software, or this comic strip template.
Your presentation must include the following:
title and introduction of your character, including what substance the character is made of
source of thermal energy your character encountered (conduction, convection, and/or radiation)
a detailed description and/or diagram of the particle transformation from solid to liquid phase
a detailed description and/or diagram of the particle transformation from liquid to the gas phase
You may get creative on this activity. If you are unsure if your idea or software for a presentation will work, contact your instructor for assistance. Be sure to review the grading rubric before you begin.
Answer:I can't art but I envision a comic of a swimming pool with chlorine water in it. The bottom of the pool is black. The chlorine is happy and excited to protect the people going to swim in it. Then the sun comes out, warms the black tile, the water is heated and the chlorine is boiled into gas form. Unable to control its movement through the atmosphere, the large amount of chlorine from the in-ground pool infiltrates the home of its lovely owners, and they die from chlorine gas inhalation, as well as half the neighborhood.
The end.
Here is a comic strip created to explain the transformation of a substance as it changes phases.
How to make a comic strip?Title: The Adventures of Mr. Ice Cube
Introduction: Mr. Ice Cube is a solid block of water. He lives in a freezer, where it is very cold.
Source of Thermal Energy: One day, Mr. Ice Cube is taken out of the freezer and placed in a hot cup of coffee. The hot coffee transfers thermal energy to Mr. Ice Cube, causing him to melt.
Particle Transformation from Solid to Liquid Phase: As Mr. Ice Cube melts, the particles in his solid structure start to move faster. They move so fast that they break free from the solid structure and become liquid particles.
Particle Transformation from Liquid to Gas Phase: As Mr. Ice Cube continues to heat up, the liquid particles start to move even faster. They move so fast that they escape from the liquid state and become gas particles.
Conclusion: Mr. Ice Cube has now transformed from a solid to a liquid to a gas. He is now a cloud of water vapor, floating in the air above the hot cup of coffee.
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ATF fluid should be A. red. B. yellow. C. green. D. blue.
ATF fluid should be A. Red
Answer:
According to Chemical Composition, the answer is A. red.
Explanation:
Which is a difference between molecular compounds and ionic compounds? Select the correct answer below: Question 5 options: Molecular compounds typically form between a metal and a nonmetal, while ionic compounds typically form between nonmetals. Molecular compounds result from the transfer of electrons between atoms to form ions, while ionic compounds result from the sharing of electrons between neutral atoms. Molecular compounds are formed of discrete, neutral molecules, while ionic compounds are formed of large repeating arrays of opposite charges. Molecular compounds have high melting points and high boiling points, while ionic
Answer:
Molecular compounds are formed of discrete, neutral molecules, while ionic compounds are formed of large repeating arrays of opposite charges.
Explanation:
in molecular a group of electron is shared while in ionic compounds electrons are transferred.
Molecular compounds are formed from the sharing of electrons between nonmetals to form discrete, neutral molecules. In contrast, ionic compounds result from the transfer of electrons between a metal and a nonmetal, forming a large repeating array of opposite charges.
Explanation:The main difference between molecular compounds and ionic compounds lies in the nature of the bonding between the atoms. Molecular compounds are formed of discrete, neutral molecules. These molecules are formed when atoms of two or more nonmetals share electrons to form covalent bonds. Examples include water (H2O) and carbon dioxide (CO2).
On the other hand, ionic compounds are formed of large repeating arrays of opposite charges. Ionic bonding typically occurs between a metal and a nonmetal atom, wherein electrons are transferred from the metal atom to the nonmetal atom. This transfer of electrons forms positively and negatively charged ions that attract one another to form the compound. Common examples of ionic compounds include sodium chloride (NaCl) and calcium chloride (CaCl2).
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What are the particles produced in a cathode tube? a. hydrogen b. atoms c. x-rays d. protons e. electrons
The particles produced in a cathode tube are primarily electrons, which strike the glass walls of the tube to produce X-rays.
Explanation:In a cathode tube, when a high voltage is applied, the particles that are produced are primarily electrons. These tubes were historic tools for scientists that led to the discovery of properties of electrons, and they played a significant role in the advancement of physics. Cathode rays were given their name because they emanate from the negatively charged electrode, or cathode, in the tube. When these electrons strike the glass walls of the tube, they can produce X-rays, but it's crucial to note that the X-rays are not the particles generated in the tube itself.
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In a cathode tube, the primary particles produced are electrons. These are emitted from the negatively charged cathode and comes from the ionization of the gas present in the tube, which can be various types of atoms, but not specifically hydrogen.
Explanation:In a cathode tube, or cathode ray tube, the primary particles produced are electrons. These particles are emitted from the negatively charged cathode, hence the name 'cathode rays'. This is a fundamental concept in physics, especially in the study of electricity and magnetism. The creation of these electrons comes from the ionization of the gas present in the tube, which can be various types of atoms, but not specifically hydrogen.
While the other particles listed (protons, atoms, and x-rays) can potentially be involved or produced in different types of tubes or under different conditions, the primary answer in the context of a standard cathode tube is electrons.
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When 131.0 mL of water at 26.0°C is mixed with 81.0 mL of water at 85.0°C, what is the final temperature? (Assume that no heat is lost to the surroundings; d of water is 1.00 g/mL.)
Answer:
63.52°C is the final temperature
Explanation:
1) 131.0 mL of water at 26.0°C
Mass of water = m
Volume of the water =131.0 mL
Density of the water = 1.00 g/mL
[tex]Density=1.00 g/mL=\frac{m}{131.0 mL}[/tex]
m = 131.0 g
Initial temperature of the water = [tex]T_i[/tex] = 26.0°C
Final temperature of the water = [tex]T_f[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i[/tex]
Heat absorbed 131.0 g of water = Q
[tex]Q=m\times c\times \Delta T[/tex]
2) 81.0 mL of water at 85.0°C
Mass of water = m'
Volume of the water =81.0 mL
Density of the water = 1.00 g/mL
[tex]Density=1.00 g/mL=\frac{m'}{81.0 mL}[/tex]
m' = 81.0 g
Initial temperature of the water = [tex]T_i'[/tex] = 85.0°C
Final temperature of the water = [tex]T_f'[/tex]
Change in temperature ,[tex]\Delta T'=T_f'-T_i'[/tex]
Heat lost by 81.0 g of water = Q'
[tex]Q'=m'\times c\times \Delta T'[/tex]
After mixing both liquids the final temperature will become equal fro both liquids.
[tex]T_f=T_f'[/tex]
Since, heat lost by the water at higher temperature will be equal to heat absorbed by the water at lower temperature.
Q=-Q' (Law of conservation of energy.)
Let the specific heat of water be c
[tex]m\times c\times \Delta T=m'\times c\times \Delta T'[/tex]
[tex]131.0 g\times c(T_f-26^oC)=-(81.0.0 g\times c(T_f-85^oC))[/tex]
[tex]T_f=63.52^oC[/tex]
63.52°C is the final temperature
According to conservation law, the amount of energy in a closed and isolated system remains constant and does not get reduced or added. The final temperature of the system is 63.52°C.
What are mass and temperature?First, calculate the mass of the 131 mL of water:
Given,
Mass (m)= ?Volume (V)= 131.0 mLDensity = 1.00 g/ml[tex]\begin{aligned}\rm Mass &= \rm Volume \times \rm density\\\\\rm m &= 131.0 \;\rm g\end{aligned}[/tex]
Given,
Initial temperature [tex]\rm (T_{i})[/tex] = 26.0°C
The final temperature of the water = [tex]\rm (T_{f})[/tex]
Change in the temperature is calculated as,
[tex]\rm \Delta T = T_{f} - T_{i}[/tex]
Heat absorbed by 131.0 g of water = Q
The formula to calculate heat absorbed is,
[tex]\rm Q = m \times c \times \Delta T[/tex]
Second, calculate the mass of the 81 mL of water:
Given,
Mass (m') = ?Volume (V')= 81.0 mLDensity = 1.00 g/ml[tex]\begin{aligned}\rm Mass' &= \rm Volume \times \rm density\\\\\rm m' &= 81.0 \;\rm g\end{aligned}[/tex]
Given,
Initial temperature [tex]\rm (T'_{i})[/tex] = 26.0°C
The final temperature of the water = [tex]\rm (T'_{f})[/tex]
Change in the temperature is calculated as,
[tex]\rm \Delta T' = T'_{f} - T'_{i}[/tex]
Heat lost by 81.0 g of water = Q'
The formula to calculate heat lost is,
[tex]\rm Q' = m' \times c \times \Delta T'[/tex]
Both the liquids are mixed and the final temperature will be equivalent and given as, [tex]\Delta \rm T_{f} = \Delta T'_{f}[/tex]
According to the law of conservation,
Heat lost by the water (Q') = Heat absorbed by the water (-Q)
[tex]\begin{aligned}\rm m \times c \times \Delta T &= \rm m' \times c \times \Delta T'\\\\131 \times\rm c(T_{f} -26) &= -(81 \times\rm c(T_{f}- 85))\\\\\rm T_{f} &= 63.52 \circC\end{aligned}[/tex]
Therefore, 63.52°C is the final temperature.
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A self-contained underwater breathing apparatus (SCUBA) uses canisters containing potassium superoxide. The superoxide consumes the CO2 exhaled by a person and replaces it with oxygen. 4 KO2(s) + 2 CO2(g) n 2 K2CO3(s) + 3 O2(g) What mass of KO2, in grams, is required to react with 8.90 L of CO2 at 22.0 °C and 767 mm Hg
Answer:
52.0004 grams of mass of potassium superoxide is required
Explanation:
Let moles carbon dioxide gas be n at 22.0 °C and 767 mm Hg occupying 8.90 L of volume.
Pressure of the gas,P = 767 mm Hg = 0.9971 atm
Temperature of the gas,T = 22.0 °C = 295.15 K
Using an ideal gas equation to calculate the number of moles.
[tex]PV=nRT[/tex]
[tex]n=\frac{0.9971 atm\times 8.90 L}{0.0821 atm L/mol K\times 295.15 K}[/tex]
n = 0.3662 mol
[tex]4KO_2(s)+2CO_2(g)\rightarrow 2K_2CO_3(s)+3O_2(g)[/tex]
According to reaction, 2 moles of carbon-dioxide reacts with 4 moles of potassium superoxide.
Then 0.3662 mol of of carbon-dioxide will react with:
[tex]\frac{4}{2}\times 0.3662 mol=0.7324 mol[/tex] of potassium superoxide.
Mass of 0.7324 mol potassium superoxide:
0.7324 mol × 71 g/mol = 52.0004 g
52.0004 grams of mass of potassium superoxide is required.
To find the mass of KO2 required to react with 8.90 L of CO2, we can use the stoichiometry of the balanced equation.
Explanation:To determine the mass of KO2 required to react with 8.90 L of CO2, we need to use the stoichiometry of the balanced equation. From the equation, we can see that 4 moles of KO2 reacts with 2 moles of CO2. We can convert the volume of CO2 to moles using the ideal gas law and then use the mole ratio to find the mass of KO2:
1. Convert 8.90 L of CO2 to moles (using the ideal gas law)
2. Use the mole ratio to find moles of KO2
3. Convert moles of KO2 to grams using the molar mass of KO2
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Which of the following species has
(a) equal numbers of neutrons and electrons;
(b) protons, neutrons, and electrons in the ratio 9:11:8;
(c) a number of neutrons equal to the number of protons plus one-half the number of electrons?
24Mg2+, 47Cr, 60Co3+, 35Cl-, 124Sn2+, 2266Th, 90Sr Petrucci, Ralph H.. General Chemistry (p. 63). Pearson Education. Kindle Edition.
Answer:
see explanation...
Explanation:
Mg⁺²-24 Co⁺³-60 Clˉ-35
Protons (p⁺) 12 27 17
Neutrons (n⁰) 12 33 18
Electrons (eˉ) 10 24 18
(c) (b) (a)
12/2 : 12/2 : 10/2 27/3 : 33/3 : 24/3 #n⁰ = 18
6 : 6 : 5 9 : 11 : 8 #eˉ = 18
The specs that match the conditions are 24Mg2+ for equal neutrons and electrons, 47Cr for the 9:11:8 proton:neutron:electron ratio, and 35Cl- for having neutrons equal to protons plus one-half the electrons.
Explanation:The question is asking for the species that match specific sets of conditions related to the number of its protons, electrons, and neutrons. In atomic structure, the number of protons defines the atomic number of an element, while the number of electrons defines its charge. The number of neutrons can be found by subtracting the atomic number from the atomic mass.
(a) For a species with equal numbers of neutrons and electrons, we can choose 24Mg2+. Its atomic number is 12, so it has 12 protons. Since it’s a +2 ion, it has lost 2 electrons thus leaving it with 10 electrons. Also, it has 24-12=12 neutrons.
(b) For a species that has protons, neutrons, and electrons in the ratio 9:11:8, we can choose 47Cr. Chromium (Cr) has an atomic number of 24 so it has 24 protons, and 47-24=23 neutrons, and as neutral atom it has 24 electrons. Although not exact, it’s the closest among the choices given.
(c) Finally, a species with a number of neutrons equal to the number of protons plus one-half the number of electrons would be 35Cl-. Chlorine ordinarily has 17 protons and 17 electrons, but as a -1 ion, it has gained 1 extra electron, giving a total of 18 electrons. Half of this is 9. The 17 protons plus the 9 gives 26, which is the number of neutrons in chlorine-35 (35-17=18).
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Who determines the crude oil specifications at production facilities?
Answer:
I believe it is the overseer of the operation
In sodium chloride, the distance between the center of the sodium ion and the centerof an adjacent chloride ion is 2.819 angstroms. Calculate the density in g/cm3of an ideal NaCl crystal from this information and what you learned from this lab.Hints: To calculate mass, determine how many equivalent ions are in a unit cell. To determinevolume of the unit cell, start by determining the length of on side of the unit cell.
Answer:
[tex]\boxed{\text{2.17 g/cm}^{3}}[/tex]
Explanation:
1. Ions per unit cell
(a) Chloride
8 corners + 6 faces
[tex]\text{No. of Cl$^{-}$ ions}\\\\= \text{8 corners} \times \dfrac{\frac{1 }{ 8} \text{ ion}} {\text{1 corner}} + \text{6 faces}\times \dfrac{\frac{1}{2} \text{ ion}}{\text{1 face}} = \text{1 ion + 3 ions = 4 ions}}[/tex]
(b) Chloride
12 edges + 1 centre
[tex]\text{No. of Na$^{+}$ ions}\\\\= \text{12 edges} \times \dfrac{\frac{1 }{ 4} \text{ ion}} {\text{1 edge}} + \text{1 centre}\times\dfrac{\text{1 ion}}{\text{1 centre}} = \text{3 ions + 1 ion = 4 ions}[/tex]
There are four formula units of NaCl in a unit cell.
2. Mass of unit cell
m = 4 × NaCl = 4 × 58.44 u = 233.76 u
[tex]m = \text{233.76 g} \times \dfrac{\text{1 g} }{6.022 \times 10^{23} \text{ u} } = 3.882 \times 10^{-22}\text{ g}[/tex]
3. Volume of unit cell
(a) Edge length
[tex]a = 2d_{\text{Na-Cl}} = 2 \times \text{2.819 \AA} = 5.638 \times10^{-10} \text{ m} = 5.638 \times10^{-8} \text{ cm}[/tex]
(b) Volume
[tex]V = a^{3} = \left( 5.638 \times (10^{-8} \text{ cm}\right)^{3} = 1.792 \times10^{-22} \text{ cm}^{3}[/tex]
4. Density
[tex]\rho = \dfrac{\text{mass}}{\text{volume}} = \dfrac{3.882 \times 10^{-22}\text{ g}}{1.792 \times10^{-22} \text{ cm}^{3}}} = \text{2.17 g/cm}^{3}\\\\\text{The density of NaCl is }\boxed{\textbf{2.17 g/cm}^{3}}[/tex]
calculate how many moles of NO2 form when each of the following completely reacts. 2N2O5(g)→4NO2(g)+O2(g) Part A 1.0 mol N2O5 Express your answer using two significant figures. nothing mol m o l Request Answer Part B 5.4 mol N2O5 Express your answer using two significant figures.
In the given equation, 2 moles of N2O5 will react to form 4 moles of NO2. Therefore, 1 mole of N2O5 will form 2 moles of NO2, and 5.4 moles of N2O5 will form 11 moles of NO2.
Explanation:When determining the amount of a product made from a reactant in a chemical equation, we use stoichiometry, guided by the coefficients in the balanced chemical equation. In this equation, 2N2O5(g)→4NO2(g)+O2(g), the coefficient in front of N2O5 is 2 and in front of NO2 is 4. This tells us that for every 2 moles of N2O5, 4 moles of NO2 will be produced.
Part A: If 1 mole of N2O5 fully reacts, it will form 2 moles of NO2. (Ratio is 2:4, therefore halved to 1:2)
Part B: If 5.4 moles of N2O5 fully reacts, it will form 10.8 (which rounds to 11 with two significant figures) moles of NO2. (Ratio 5.4 * 2 moles of NO2/mole of N2O5)
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Consider the following unbalanced equation for the combustion of hexane: αC6H14(g)+βO2(g)→γCO2(g)+δH2O(g) Part A Balance the equation. Give your answer as an ordered set of numbers α, β, γ, ... Use the least possible integers for the coefficients. α α , β, γ, δ = nothing Request Answer Part B Determine how many moles of O2 are required to react completely with 5.6 moles C6H14. Express your answer using two significant figures. n n = nothing mol Request Answer Provide Feedback
Answer:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
α =2
β = 19
γ = 12
δ = 14
53.2moles of O₂
Explanation:
Proper equation of the reaction:
αC₆H₁₄ + βO₂ → γCO₂ + δH₂O
This is a combustion reaction for a hydrocarbon. For the combustion of a hydrocarbon, the combustion equation is given below:
CₓHₙ + (x + [tex]\frac{n}{4}[/tex])O₂ → xCO₂ + [tex]\frac{n}{2}[/tex]H₂O
From the given combustion equation, x = 6 and n = 14
Therefore:
β = x + [tex]\frac{n}{4}[/tex] = 6 + [tex]\frac{14}{4}[/tex] = 6 + 3.5 = 9[tex]\frac{1}{2}[/tex]
γ = 6
δ = [tex]\frac{n}{2}[/tex] = [tex]\frac{14}{2}[/tex] = 7
The complete reaction equation is therefore given as:
C₆H₁₄ + 9[tex]\frac{1}{2}[/tex]O₂ → 6CO₂ + 7H₂O
To express as whole number integers, we multiply the coefficients through by 2:
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O
Problem 2
From the reaction:
2 moles of hexane are required to completely react with 19 moles of O₂
∴ 5.6 moles of hexane would react with k moles of O₂
This gives: 5.6 x 19 = 2k
k = [tex]\frac{5.6 x 19}{2}[/tex]
k = 53.2moles of O₂
A student performs a reaction that makes aluminum oxide. According to her calculations, she should expect to make 115.2 grams. She actually produces 66.9 grams. What is her percent yield?
Answer:
The percentage of the student is 58.17%.
Explanation:
Expected yield of aluminum oxide = 115.2 g
Actual yield of aluminum oxide produced =66.9 g
The percentage yield is calculated by dividing actual yield by expected yield and then multiplying it with hundred.
Percentage yield:
[tex]\frac{\text{Actual yield}}{\text{Expected yield}}\times 100[/tex]
[tex]\% Yield=\frac{66.9 g}{115.2 g}\times 100=58.17\%[/tex]
The percentage of the student is 58.17%.
CH4 (g) yields C(g) + 4H (g) (reaction for expansion)
Compare the reaction for the "expansion" of methane with the reverse of the reaction that represents the standard enthalpy of formation. Which properties are the same for both reactions and which are different?
1. the number of atoms of each element, 2. the state of the methane molecules, 3. the enthalpy change of the reaction, 4. the state of the carbon atoms.
Answer:
Here's what I find.
Explanation:
[tex]\rm CH$_{4}$(g) $\, \rightleftharpoons \,$ C(g) + 4H(g)[/tex]
[tex]\text{Same} =\begin{cases}1. & \text{The number of atoms of each element}\\2. & \text{The state of the methane molecules}\\4. &\text{The state of the carbon atoms}\\\end{cases}[/tex]
[tex]\text{Different} =\begin{cases}3. & \text{The enthalpy change of the reaction}\\\end{cases}[/tex]
The sign of ΔH changes when you reverse the reaction.
Answer:
Same:
-the number of atoms
of each element
-the state of the
methane molecules
Different:
-the state of the
carbon atoms
-the enthalpy change
of the reaction
Explanation:
The enthalpy changes are different for the two reactions even though all the same elements are involved in equal numbers. This illustrates that the enthalpy change of a reaction is dependent on the states of the reactants and products, as well as how the atoms are bonded.
At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s2HIg=H2g+I2g Suppose a vessel contains HI at a concentration of 1.28M . Calculate how long it takes for the concentration of HI to decrease to 17.0% of its initial value. You may assume no other reaction is important. Round your answer to 2 significant digits.
Answer:
[tex]\boxed{\text{28.0 s}}[/tex]
Explanation:
Whenever a question asks you, "How long does it take to reach a certain concentration?" or something like that, you must use the appropriate integrated rate law expression.
The integrated rate law for a first-order reaction is
[tex]\ln \left (\dfrac{[A]_{0}}{[A]} \right ) = kt[/tex]
Data:
[A]₀ = 1.28 mol·L⁻¹
[A] = 0.17 [A]₀
k = 0.0632 s⁻¹
Calculation:
[tex]\begin{array}{rcl}\ln \left (\dfrac{[A]_{0}}{0.170[A]_{0}} \right ) & = & 0.0632t\\\\\ln \left (5.882) & = & 0.0632t\\1.772 & = & 0.0632t\\\\t & = & \dfrac{1.772}{0.0632}\\\\t & = & \textbf{{28.0 s}}\\\end{array}\\\text{It will take } \boxed{\textbf{28.0 s}} \text{ for [HI] to decrease to 17.0 \% of its original value.}[/tex]
In acidic solution, the nitrate ion can be used to react with a number of metal ions. One such reaction is NO3−(aq)+Sn2+(aq)→NO2(aq)+Sn4+(aq) Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: NO3−(aq)+Sn2+(aq)+ −−−→NO2(aq)+Sn4+(aq)+ −−− Part A What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.
Answer:
[tex]\boxed{\text{2, 1, 4, 2, 1, 2}}[/tex]
Explanation:
NO₃⁻ + Sn²⁺ + __ → NO₂ + Sn⁴⁺ + __
Step 1: Separate into two half-reactions.
NO₃⁻ ⟶ NO₂
Sn²⁺ ⟶ Sn⁴⁺
Step 2: Balance all atoms other than H and O.
Done
Step 3: Balance O.
NO₃⁻ ⟶ NO₂ + H₂O
Sn²⁺ ⟶ Sn⁴⁺
Step 4: Balance H
NO₃⁻ + 2H⁺ ⟶ NO₂ + H₂O
Sn²⁺ ⟶ Sn⁴⁺
Step 5: Balance charge.
NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O
Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻
Step 6: Equalize electrons transferred.
2 × [NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O]
1 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]
Step 7: Add the two half-reactions.
2 × [NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O]
1 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]
2NO₃⁻ + Sn²⁺ + 4H⁺ ⟶ 2NO₂ + Sn⁴⁺ + 2H₂O
Step 8: Check mass balance.
On the left: 2 N, 6 O, 1 Sn, 4H
On the right: 2 N, 6 O, 1 Sn, 4H
Step 9: Check charge balance.
On the left: -2 + 6 = +4
On the right: +4
The equation is balanced.
[tex]\text{The coefficients are }\boxed{\textbf{2, 1, 4, 2, 1, 2}}[/tex]
The balanced equation for the redox reaction in an acidic solution between nitrate ions and tin ions is: 4H+(aq) + NO3−(aq) + 3Sn2+(aq) → NO2(aq) + 2H2O(l) +3Sn4+(aq). Here, nitrate ions get reduced to nitrite, while tin ions get oxidized.
Explanation:The reaction you have provided is a redox reaction taking place in an acidic medium. The goal is to balance this equation, including the terms representing water (H2O) and hydrogen ion (H+). Using the half-reaction method, the balanced equation becomes: 4H+(aq) + NO3−(aq) + 3Sn2+(aq) → NO2(aq)+ 2H2O(l) +3Sn4+(aq).
In this reaction, the nitrate ion is reduced to nitrite (NO2−), while the tin ions (Sn2+) are oxidized to Sn4+. The acidic medium provides the necessary hydrogen ions (H+) and water is also a product of the reaction.
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Consider the oxidation of sodium metal to sodium oxide described by the balanced equation: 4 Na(s) + O2(g) → 2 Na2O(s) As you can see there are letters in parentheses after each substance. Show you know the meaning of these letters by filling in what they stand for:
Final answer:
The letters in parentheses after each substance represent their states of matter in the reaction equation. In this reaction, sodium in solid state reacts with oxygen in gas state to form sodium oxide in solid state.
Explanation:
In the balanced equation 4 Na(s) + O2(g) → 2 Na2O(s), the letters in parentheses after each substance represent their states of matter. (s) stands for solid, (g) stands for gas. In this reaction, sodium (Na) in solid state reacts with oxygen (O2) in gas state to form sodium oxide (Na2O) in solid state.
The letters in parentheses denote states of matter: (s) for solid, (g) for gas, at room temperature and standard pressure.
The letters in parentheses after each substance in the chemical equation represent the state of matter for each reactant and product at room temperature and standard pressure. Here is what they stand for:
Na(s) stands for sodium in the solid state.
O₂(g) stands for oxygen in the gaseous state.
Na₂O(s) stands for sodium oxide in the solid state.
The balanced chemical equation provided is:
[tex]\[ 4 \text{Na}(s) + \text{O}_2(g) \rightarrow 2 \text{Na}_2\text{O}(s) \][/tex]
This equation indicates that four moles of solid sodium react with one mole of gaseous oxygen to produce two moles of solid sodium oxide.
If you add 5.00 mL of 0.100 M sodium hydroxide to 50.0 mL of acetate buffer that is 0.100 M in both acetic acid and sodium acetate, what is the pH of the resulting solution? Acetic Acid: Ka = 1.8. x 10-5
Answer:
see explanation ...
Explanation:
5.00ml(0.100M NaOH) + 0.100M HOAc/NaOAc Bfr
5.00ml(0.100M NaOH) = 0.005(0.100) mole NaOH = 0.0005 mole NaOH = 0.0005 mole OH⁻ in 55 ml Bfr solution (50ml + 5ml)=> [OH⁻] = (0.0005/0.055)M OH⁻ = 0.0091M OH⁻ ≈ 0.010M OH⁻ added into Bfr solution. The amount of OH⁻ added must be removed by H⁺ in the HOAc equilibrium; that is, H⁺ + OH⁻ → H₂O leaving a void at the H⁺ position in the HOAc equilibrium. HOAc then decomposes to replace the H⁺ removed by the excess OH⁻ giving new H⁺ and OAc⁻ concentrations. Reaction shifts right => subtract 0.01M from HOAc side of equilibrium and add 0.01M to OAc⁻side of equilibrium and recompute the H⁺ concentration and new pH.
HOAc ⇄ H⁺ + OAc⁻
C(i) 0.10M ~0M* 0.10M =>pH=-log(Ka)=-log(1.85x10⁻⁵)=4.73
ΔC -0.01M +x +0.01M
C(eq) 0.09M x 0.11M => New HOAc equil. conc.
Ka = [H⁺][OAc⁻]/[HOAc]
=> x(0.11)/(0.09) = 1.85x10⁻⁵
=> x = [H⁺] = 0.09(1.85x10⁻⁵)/0.11 = 1.52x10⁻⁵M (after adding NaOH)
=> pH = -log[H⁺] -log(1.52x10⁻⁵) = 4.82 ( pH shifts to more basic value b/c of OH⁻ addition )
If you add 5.00 mL of 0.100 M sodium hydroxide to 50.0 mL of acetate buffer. The pH of the resulting solution is 4.82.
What is pH?pH is a measurement scale, used to measure the acids and the bases
The pH of the resulting solution
[tex]\rm Ka = \dfrac{ [H^+][OAc^-]}{[HOAc]}[/tex]
[tex]x \dfrac{ (0.11)}{(0.09)} = 1.85 \times 10^-^5[/tex]
[tex]\rm x = [H^+] = 0.09 \dfrac{(1.85x10^-^5)}{0.11} = 1.52 \times 10^-^5 M[/tex]
pH = -log[H⁺] -log(1.52x10⁻⁵) = 4.82
( pH shifts to more basic value b/c of OH⁻ addition )
Thus, the pH of the resulting solution is 4.82.
Learn more about pH
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