Answer:
it is acidified with concentrated hydrochloric acid.
Hydrogenotrophy is Choose one: A. an award given to fuel companies that have the most fuel-efficient vehicles. B. the oxidation of water during photosynthesis to liberate electrons, protons, and oxygen gas. C. the use of hydrogen gas as an electron donor. D. the generation of hydrogen gas by methanogens in syntrophy.
Answer:
C. the use of hydrogen gas as an electron donor.
Explanation:
Hydrogenotrophy is the convertion of hydrogen gas to other compounds as part of its metabolism.
Polar molecules have a partial positive charge on one side and a partial negative charge on the other side, a separation of charge called a dipole. Which of the following molecules has this kind of a dipole?a. IonicB. Covalent
Answer:Covalent
Explanation:
Covalent bonds are formed by sharing of electrons between two atoms. The shared electrons are normally situated between the nuclei of the both atoms. However, atoms of some elements posses an usual ability to attract the shared electrons of a bond towards itself. We say that such atoms have a high electronegativity. High electronegativity leads to the existence of polar covalent bonds since the shared electron pair is now closer to one of the bonding atoms than the other. The atom to which the electron pair is closer becomes partially negative while the other becomes partially positive. This is only possible in a covalent bond where electrons are shared between bonding atoms. The charge separation is known as a dipole.
Note that ionic bonds involve a complete transfer of electrons leading to the formation of ions and is not applicable here.
The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochloric acid (HCl) is 3.32 × 10−10 s−1 at 25°C. Calculate the rate constant at 48.5°C if the activation energy is 116 kJ/mol.
Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
CH3Cl + H2O → CH3OH + HClat T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
K(T) = A e∧(-Ea/RT)Ln K = Ln(A) - [(Ea/R)(1/T)]∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
The rate constant at 48.5°C if the activation energy is 116 kJ/mol is [tex]1.017 E-8 s^{-1}[/tex]
Chemical reaction:CH₃Cl + H₂O → CH₃OH + HCl
At T₁ = 25°C (298 K) ⇒ K₁ = 3.32 E-10 s-1
At T₂ = 48.5°C (321.5 K) ⇒ K₂ = ?
According to Arrhenius equation:
[tex]K(T) = A*e^{(-Ea/RT)}\\\\ln K = ln(A) - [(Ea/R)(1/T)][/tex]
where,
A = frequency factor
R = 8.314 E-3 KJ/K.mol
[tex]ln K_1 = ln(A) - [Ea/R)*(1/T_1)][/tex]..........(1)
[tex]ln K_2 = ln(A) - [(Ea/R)*(1/T_2)][/tex]...........(2)
On dividing equation 1 by 2:
ln (K₁/K₂) = (Ea/R)* (1/T₂ - 1/T₁)
ln (K₁/K₂) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
ln (K₁/K₂) = (13952.37 K)*(- 2.453 E-4 K-1)
ln (K₁/K₂) = - 3.422
(K₁/K₂)= [tex]e^{(-3.422)}[/tex]
[tex](3.32 E-10 s^{-1})[/tex] / K₂ = 0.0326
K₂ = [tex](3.32 E-10 s^{-1})[/tex] /0.0326
K₂ = [tex]1.017 E-8 s^{-1}[/tex]
Thus, the rate constant at 48.5°C if the activation energy is 116 kJ/mol is 1[tex]0.017 E-8 s^{-1}[/tex]
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When a nonvolatile solute is added to a solvent, it tends to ___________________ the freezing point and ___________________ the boiling point, compared to the pure solvent.
Answer:
decreases and increases
Explanation:
the addition of solutes can alter the freezing point and the boiling point .the changes in solute content where there is an addition of solutes results in the dropping of freezing point and increase in boiling point.this is known as freezing point depression and boiling point depression.
Sodium chloride as a compound does not truly exist in the ocean. True or False
. The molar heat of vaporization of acetone, C3H6O, is 30.3 kJ/mol at its boiling point. How many kilojoules of heat would be liberated by the condensation of 5.00 g of acetone?
2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone
Explanation:
To convert grams into moles
[tex]moles = grams \times \frac{1 mole}{grams}[/tex]
We have 5.00 g acetone
[tex]moles = 5 \times \frac{1}{58.1}[/tex]
[tex]moles = 0.0861[/tex]
Heat liberated = moles [tex]\times[/tex] heat of vapourization
=0.0861 mol x 30.3 kJ/mol
= 2.61 kJ
Therefore, 2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone
A calorimeter contains 32.0 mLmL of water at 15.0 ∘C∘C . When 1.20 gg of XX (a substance with a molar mass of 54.0 g/molg/mol ) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq)X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 29.0 ∘C∘C . Calculate the enthalpy change, ΔHΔHDelta H, for this reaction per mole of XX. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)J/(g⋅∘C)], that density of water is 1.00 g/mLg/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.
The enthalpy change [tex](\( \Delta H \))[/tex] for this reaction per mole of [tex]\( X \)[/tex] is approximately 88.27 kj/mol.
To calculate the enthalpy change[tex](\( \Delta H \))[/tex] for the reaction per mole of X , we can use the formula:
[tex]\[ \Delta H = \frac{q}{n} \][/tex]
where:
- [tex]\( q \)[/tex] is the heat absorbed or released by the reaction,
- [tex]\( n \)[/tex] is the moles of [tex]\( X \).[/tex]
First, let's find the moles of \( X \):
[tex]\[ \text{moles of } X = \frac{\text{mass}}{\text{molar mass}} \][/tex]
[tex]\[ \text{moles of } X = \frac{1.20 \, \text{g}}{54.0 \, \text{g/mol}} \][/tex]
[tex]\[ \text{moles of } X \approx 0.0222 \, \text{mol} \][/tex]
Now, we need to find the heat [tex](\( q \))[/tex] absorbed or released by the reaction. The heat can be calculated using the formula:
[tex]\[ q = mc\Delta T \][/tex]
where:
- m is the mass of the solution [tex](water + \( X \)),[/tex]
- c is the specific heat of the solution,
- [tex]\( \Delta T \)[/tex]is the change in temperature.
First, calculate the mass of the solution:
[tex]\[ \text{mass of solution} = \text{mass of water} + \text{mass of } X \][/tex]
[tex]\[ \text{mass of solution} = 32.0 \, \text{g} + 1.20 \, \text{g} \][/tex]
[tex]\[ \text{mass of solution} = 33.20 \, \text{g} \][/tex]
Now, calculate [tex]\( \Delta T \)[/tex]:
[tex]\[ \Delta T = T_f - T_i \][/tex]
[tex]\[ \Delta T = 29.0^\circ \text{C} - 15.0^\circ \text{C} \][/tex]
[tex]\[ \Delta T = 14.0^\circ \text{C} \][/tex]
Now, plug in the values into the heat formula:
[tex]\[ q = (33.20 \, \text{g}) \times (4.18 \, \text{J/(g} \cdot ^\circ \text{C})) \times (14.0 \, ^\circ \text{C}) \][/tex]
[tex]\[ q \approx 1961.12 \, \text{J} \][/tex]
Now, use the first formula to find [tex]\( \Delta H \):[/tex]
[tex]\[ \Delta H = \frac{1961.12 \, \text{J}}{0.0222 \, \text{mol}} \][/tex]
[tex]\[ \Delta H \approx 88272 \, \text{J/mol} \][/tex]
Therefore, the enthalpy change[tex](\( \Delta H \))[/tex] for this reaction per mole of X is approximately[tex]\( 88.27 \, \text{kJ/mol} \).[/tex]
Calculate the heat absorbed using q = mcΔT and convert to kJ. Determine the moles of substance X, then compute the enthalpy change per mole. The enthalpy change (ΔH) for the reaction per mole of X is 84.34 kJ/mol.
To calculate the enthalpy change (ΔH) for the reaction per mole of substance X, follow these steps:
Determine the heat absorbed by the solution using the formula q = mcΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature.The mass of the solution is the volume of water (32.0 mL) converted to grams (32.0 g) and the specific heat capacity is 4.18 J/g°C. The temperature change (ΔT) is 29.0°C - 15.0°C = 14.0°C.Calculate q: q = 32.0 g * 4.18 J/g°C * 14.0°C = 1874.56 J.Convert q to kJ: 1874.56 J * (1 kJ / 1000 J) = 1.87456 kJ.Calculate the moles of substance X: Moles = 1.20 g / 54.0 g/mol = 0.02222 mol.Finally, calculate the ΔH per mole of X: ΔH = 1.87456 kJ / 0.02222 mol = 84.34 kJ/mol.The size of a population increases if the number of individuals added to the population is equal to the number of individuals leaving the population, is this true
Answer:
The answer to the question
The size of a population increases if the number of individuals added to the population is equal to the number of individuals leaving the population, is this true?
is No
Explanation:
The size of a population increases if the number of individuals added to the population is more than the number of individuals leaving the population
In biology, a population is the total number of organism of a particular species or group living together at a given geographical location. It is the number of inhabitants of a place.
Population growth is the increase in the number count of the specie or group in a population.
In order to get high-quality energy, you must use high-quality energy. This rule is the result of which of the following? a.the law of gravityb.the universal law of energyc.the second law of thermodynamicsd.the law of diminishing returnse.the first law of thermodynamics
Answer: C. The Second Law of Thermodynamics.
Explanation:
The law of gravity is related to the force interaction due to gravitation between two bodies (i.e. a person and planet Earth, a star and a planet).
The law of diminishing return is not used in Physics, but in Economics and describes the diminishing of marginal returns in time.
The first law of Thermodynamics analyses the energy interactions of a system in a quantitative manner and it is a generalized form of the Principle of Energy Conservation. It is oriented to the analyses of energy inflows and outflows.
The second law of Thermodynamics analyses the energy interactions of a system in a qualitative manner and it is based on thermodynamic property named Entropy, which may helpful to measure irreversibilities associated with system and irreversibility generation as well, provided that a comparison with another equivalent system exists.
Irreversibilities depends on the characteristics of system and nature of energy sources. Empirically, it is known that heat offers a lower quality than electricity in order to get a available work. That is to say, there is a lower of obtaining available work from heat than from electricity.
Hence, the statement is a consequence of using the second law of Thermodynamics and, therefore, the correct answer is C.
Describe a simple chemical test that could be used to distinguish between members of each of the following pairs of compounds (a) 4-Chlorophenol and 4-chloro-1-methylbenzene (c) 4-Methylphenol and 2,4,6-trinitrophenol (b) 4-Methylphenol and 4-methylbenzoic acid (d) Ethyl phenyl ether and 4-methylphenol
Answer:
Explanation:
(A) Phenol gives violet color by complexation with Fe3+ solution . It is best identification...
(B) For b , same phenolic test can be done ...But other esterification is also possibility...
(C) Picric acid has a particular identification test ,
(D) Here , like the first one , phenolic test with FeCl3 gives violet color for 4-ethylphenol and no color for ethyl phenyl ether...
What mass of silver chloride can be produced from 1.30 L of a 0.245 M solution of silver nitrate? The reaction described in Part A required 3.36 L of calcium chloride. What is the concentration of this calcium chloride solution?
Please explain, I don't know where to begin.
Answer:
Explanation:
Given parameters:
Volume of AgNO₃ = 1.3L
Molarity or concentration AgNO₃ = 0.245M
Unknown:
Mass of AgCl produced = ?
Solution:
To solve this problem, we have to work from the known compound to the unknown one using the mole concept.
The known compound is the one that we can obtain the value of the number of moles from. Here it is the given AgNO₃ that can furnish us with this piece of information;
Now let us establish the balanced reaction equation;
2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂
Now;
Find the number of moles of AgNO₃;
Number of moles of AgNO₃ = concentration x volume
= 1.3 x 0.245
= 0.32moles
From the balanced reaction equation;
2 mole of AgNO₃ produced 2 moles of AgCl
0.32 moles of AgNO₃ will produce 0.32moles of AgCl
Now that we know the number of moles of the AgCl, we can find the mass;
Mass of AgCl = number of moles x molar mass
Molar mass of AgCl = 107.9 + 35.5 = 143.4g/mol
Mass of AgCl = 0.32 x 143.4 = 45.89g
A small toy car draws a 0.50-mA current from a 3.0-V NiCd (nickel-cadmium) battery. In 10 min of operation, (a) how much charge flows through the toy car, and (b) how much energy is lost by the battery? 4. (Resistance and Ohm’s law, Prob. 17.16, 1.0 point) How
Answer:
0.3 Coulomb charge flows through the toy car.
0.9 Joules of energy is lost by the battery
Explanation:
[tex]Current(I)=\frac{charge(Q)}{Time (T)}[/tex]
a)
Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)
milli Ampere = 0.001 Ampere
Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s
1 min = 60 seconds
Charge flows through the toy car be Q
[tex]0.50\times 0.001 A=\frac{Q}{600 s}[/tex]
[tex]Q=0.50 \times 0.001 A\times 600 s=0.3 C[/tex]
0.3 Coulomb charge flows through the toy car.
b)
[tex]Heat(H)=Voltage(V)\times Current(I)\times Time(T)[/tex]
Voltage of the battery = V = 3.0 V
Current drawn from the battery = 0.50 mA = (0.50 × 0.001 A)
Duration of time current drawn = T = 10 min = 10 × 60 s = 600 s
[tex]H=V\times I\time T=3.0 V\times 0.50 \times 0.001 A\times 600 s[/tex]
H = 0.9 Joules
0.9 Joules of energy is lost by the battery
A sample of ethane (C2H6)was combusted completely and the water that formed has a mass of 1.61 grams.How much ethane, in grams, was in the sample?Put your answer in thespace provided belo socratic.org
Answer:
0.89 g of ethane
Explanation:
The balanced reaction equation is
C2H6(g) + 7/2 O2(g) -----------> 2CO2(g) + 3H2O(g)
From this balanced reaction
30g of ethane yields 54g of water
Therefore mass of ethane necessary to obtain 1.61g of water we have:
30 × 1.61/54 = 0.89 g of ethane
Phosphine, PH₃, is a colorless, toxic gas that is used in the production of semiconductors as well as in the farming industry. When heated, phosphine decomposes into phosphorus and hydrogen gases.
4 PH₃(g) ⟶ P₄(g) + 6 H₂(g)
This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial pressure of hydrogen gas that is present after 70.5 s if a 2.20 L vessel containing 2.29 atm of phosphine gas is heated to 953 K.
1.72 atm of hydrogen gas is created when 2.29 atm of phosphine breaks down at 953 K for 70.5 seconds.
Explanation:
The first step in this calculation is to use the first-order decay equation to find the remaining moles of phosphine. For a first-order reaction with a half-life of 35.0 s, after 70.5 s (or two half-lives), 25% of the phosphine would remain. To find the remaining moles of phosphine, we multiply 2.29 atm by 25%, yielding 0.57 atm.
From the balanced equation for the decomposition of phosphine (4PH3 -> P4 + 6H2), we see that 1 mole of phosphine yields 3/2 moles of H2. Thus, the number of moles of H2 produced is 75% of the original moles of PH3 (1.72 atm).
The sum of the partial pressures of phosphine and hydrogen is 0.57 + 1.72 = 2.29 atm, which is the final pressure in the 2.20 L vessel after 70.5 s.
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A sample of carbon dioxide occupies 2.77 L at a pressure of 18.1 psi and a temperature of 56ºC. What pressure will the carbon dioxide exert if the volume is increased to 4.81 L while the temperature is held constant?
Answer:
10.4psi
Explanation:
Since the problem involves pressure and volume at constant temperature, Boyle's law is the appropriate formula to use.
Boyle's law States that at a constant temperature, the volume of a given mass of a bass is inversely proportional to its pressure.
Using Boyle's law
P1v1 = p2v2
P1 is 18.1psi
V1 is 2.77L
P2 =?
V2 is 4.81
Temperature is constant
P2 =P1v1/ v2
= 18.1 x 2.77 / 4.82
= 10.4psi
Choose the correct statement about melting points. * Melting point can tell us if we have a mixture of compounds (for example, compound of interest impurity). Melting point can tell us the identity of the components of a mixture. If we have a sample with a lowered melting range as compared to our standards, we can do mixed melting point determinations, combining our sample with each standard, to determine the identity of our sample. All of these are true. None of these are true.
Answer: All of these statements are true
Explanation:
Melting point help us to determine if a mixture is pure or has impurities by the virtues of it melting range..
20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator would be most suitable for this titration
20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator would be most suitable for this titration? The pKa of CH3CH2NH3+ is 10.75.
Answer:
Bromocresol green, color change from pH = 4.0 to 5.6
Explanation:
The equation for the reaction is as follows:
[tex]C_2H_5NH_{2(aq)[/tex] + [tex]H^+_{(aq)[/tex] ⇄ [tex]C_2H_5NH_{3(aq)}^+[/tex]
Given that concentration of [tex]C_2H_5NH_{2(aq)[/tex] = 10%
i.e 10 g of [tex]C_2H_5NH_{2(aq)[/tex] in 100 ml solution
molar mass = 45.08 g/mol
number of moles = [tex]\frac{10}{45.08}[/tex]
= 0.222 mol
Molarity of [tex]C_2H_5NH_{2(aq)[/tex] = 0.222 × [tex]\frac{1000}{100}mL[/tex]
= 2.22 M
However, number of moles of [tex]C_2H_5NH_{2(aq)[/tex] in 20 mL can be determined as:
number of moles of [tex]C_2H_5NH_{2(aq)[/tex] = 20 mL × 2.22 M
= [tex]44*10^{-3} mole[/tex]
Concentration of [tex]C_2H_5NH_{2(aq)[/tex] = [tex]\frac{44*10^{-3}*1000}{20}[/tex]
= 2.22 M
Similarly, The pKa Value of [tex]C_2H_5NH_{2(aq)[/tex] is given as 10.75
pKb value will be: 14 - pKa
= 14 - 10.75
= 3.25
Finally, the pH value at equivalence point is:
pH= [tex]\frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C][/tex]
pH = [tex]\frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22][/tex]
pH = 5.21
∴
The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6.
For titration of the weak base ethylamine with a strong acid like HCl, the pH at the equivalence point will be less than 7, so an indicator that changes color in an acidic pH range such as methyl orange would be most suitable.
Explanation:The appropriate indicator in this case would depend on the equivalence point of the titration. Ethylamine, CH3CH2NH2, is a weak base, and it's being titrated with HCl, a strong acid. The reaction between them generates water, and ethylammonium chloride which functions as a weak acid. This setup indicates that the pH at the equivalence point will be less than 7.
Because the pH at the equivalence point will be in the acidic range (below 7), we should select an indicator which changes color around this pH. A good choice would be methyl orange, which changes color from red to yellow across a pH range of approximately 3.1 to 4.4.
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When powdered zinc is heated with sulfur, a violent reaction occurs, and zinc sulfide forms: Some of the reactants also combine with oxygen in air to form zinc oxide and sulfur dioxide. When 83.2 g of Zn reacts with 52.4 g of S8, 104.4 g of ZnS forms. (a) What is the percent yield of ZnS? (b) If all the remaining reactants combine with oxygen, how many grams of each of the two oxides form?
Answer:
Explanation:
The detailed steps and step by step calculations is as shown in the attached file.
To calculate the percent yield of ZnS, use the given masses of Zn and S8 and compare to the theoretical yield. For the oxides formed, determine the moles of remaining Zn and S8 using the balanced chemical equation.
Explanation:To calculate the percent yield of ZnS, we need to compare the actual yield (104.4 g) to the theoretical yield. The theoretical yield can be calculated using the stoichiometry of the reaction. The balanced chemical equation is: Zn + S8 → ZnS. The molar ratio between Zn and ZnS is 1:1, which means that the moles of ZnS formed will be equal to the moles of Zn used. Convert the given masses of Zn and S8 to moles using their molar masses, then compare the moles to calculate the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction. Once you have determined the limiting reactant and calculated the theoretical yield, you can calculate the percent yield using the formula: (actual yield / theoretical yield) x 100.
For part (b), if all the remaining reactants combine with oxygen, we need to determine the remaining moles of Zn and S8 after the reaction with ZnS is complete. Convert these moles to masses using their molar masses, and then use the balanced chemical equation to determine the moles of each oxide formed. The balanced chemical equation for the reaction of zinc with oxygen is: 2Zn + O2 → 2ZnO. The molar ratio between Zn and ZnO is 2:2, so the moles of ZnO formed will be equal to the moles of Zn remaining. The balanced chemical equation for the reaction of sulfur with oxygen is: S8 + 8O2 → 8SO2. The molar ratio between S8 and SO2 is 1:8, so the moles of SO2 formed will be 8 times the moles of S8 remaining. Convert the moles of each oxide formed to masses using their molar masses.
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How many grams of potassium fluoride can form if 4.00 grams of potassium are reacted with 3.00 grams of fluorine gas according to the reaction: 2K (s) + F2 (g) → 2KF (s)
Answer:
We can for 5.93 grams potassium fluoride
Explanation:
Step 1: Data given
Mass of potassium = 4.00 grams
Mass of fluorine = 3.00 grams
Molar mass potassium = 39.10 g/mol
Molar mass fluorine gas =38.00 g/mol
Step 2: The balanced equation
2K (s) + F2 (g) → 2KF (s)
Step 3: Calculate moles potassium
Moles potassium = 4.00 grams / 39.10 g/mol
Moles potassium = 0.102 moles
Step 4: Calculate moles F2
Moles F2 = 3.00 grams / 38.00 g/mol
Moles F2 = 0.0789 moles
Step 5: Calculate limiting reactant
Potassium is the limiting reactant. There will react 0.102 moles
Fluorine gas is in excess. There will react 0.102/ 2 = 0.051 moles
There will remain 0.0789 - 0.051 = 0.0279 moles
Step 6: Calculate moles potassium fluoride
For 2 moles potassium we need 1 mol fluorine to produce 2 moles potassium fluoride
For 0.102 moles K we need 0.102 moles KF
Step 7: Calculate mass KF
Mass KF = moles KF * molar mass KF
Mass KF = 0.102 moles * 58.10 g/mol
Mass KF = 5.93 grams
Final answer:
By calculating the moles of potassium and fluorine used in the reaction from their given masses, and considering the stoichiometry of the balanced equation, it's determined that 4.59 grams of potassium fluoride can be produced.
Explanation:
The question asks how many grams of potassium fluoride can form when 4.00 grams of potassium react with 3.00 grams of fluorine gas according to the balanced chemical equation 2K (s) + F2 (g) → 2KF (s). To solve this, we first find the molar mass of potassium (K) and fluorine (F2), which are approximately 39.10 g/mol and 38.00 g/mol, respectively. Then, we calculate the moles of K and F2 available by dividing the given masses by their respective molar masses. With potassium: 4.00 g / 39.10 g/mol = 0.102 moles of K; with fluorine: 3.00 g / 38.00 g/mol = 0.079 moles of F2. As the equation shows a 2:1 ratio, fluorine limits the reaction. Therefore, we can form 0.079 moles of KF. The molar mass of KF is approximately 58.10 g/mol (39.10 g/mol for K + 19.00 g/mol for F). Thus, the mass of KF that can form is 0.079 moles × 58.10 g/mol = 4.59 grams.
On the periodic table, calcium and bromine are located in period 4. What is one difference between the two elements?
A. Calcium atoms are more reactive than bromine atoms.
B. Bromine and calcium atoms will each combine with the same elements to form compounds.
C. Calcium atoms combine more easily with other atoms in Group 2.
D. Bromine atoms have more electrons in their outer shell than calcium atoms.
Explanation:
it's is d cause bromine can even react with water so
Bromine atoms have more electrons in their outer shell than calcium atoms. So the correct option is D.
What is valency?In chemistry, valence, usually spelled valency, is the characteristic of an element that establishes the maximum number of other atoms that one atom of the element may combine with. The phrase, which was first used in 1868, is used to indicate both the broad power of a combination of elements as well as its numerical value.
For 19th-century chemists, the explanation and systematization of valence posed significant difficulties. The majority of the work was focused on developing empirical guidelines for finding the valences of the elements because there was no good hypothesis for its origin.
The amount of hydrogen atoms that an element's atom may mix with or replace in a compound is used to quantify the characteristic valences of the various elements. But it soon became clear that various compounds had varied valences for numerous elements.
The identification of the chemical bond of organic compounds with a pair of electrons held jointly by two atoms and acting to hold them together in 1916 by the American chemist G.N. Lewis represented the first significant step in the development of a satisfactory explanation of valence and chemical combination. The German scientist W. Kossel studied the nature of the chemical connection between electrically charged atoms (ions) in the same year.
The theory of valence was reconstructed in terms of electronic structures and interatomic forces following the advent of the precise electronic theory of the periodic system of the elements. In response to the many ways that atoms interact, new notions such as ionic valence, covalence, oxidation number, coordination number, and metallic valence were developed.
Therefore the correct option is D.
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Carbon, nitrogen, oxygen and hydrogen are considered _______ because they are required in relatively large quantities, whereas manganese, zinc, and copper are considered _____ because they are required in very small quantities.
Answer:
macro-nutrients , micro-nutrients
Explanation:
For plant growth soil needs some basic elements which is basic need for plant growth and metabolic synthesis.
plant need approx 17 essential nutrients for growth and for completing its life cycle among which some are found in water and air. Nitrogen, carbon, hydrogen and oxygen needed the most and in large quantity as nitrogen helps in rapid growth and green color and is mobile and its deficiency lead to reduced growth and yellow foliage.
What is true about the chemiosmotic synthesis of atpt?
Explanation:
The chemiosmotic theory
The theory suggests that the actual production of adenosine triphosphate ATP in cellular respiration of respiring cells takes place through the process of chemiosmosis that makes the electrochemical gradient across the inner and outer membranes of mitochondria.Breaking down of energy-rich molecules such as glucose the energy is released in the form of NADH and FADH2.It is found that for each glucose molecule 36 molecules of ATP can be produced during cellular respiration. These molecules are generated by the Krebs cycle, the electron transport system, and chemiosmosis.Two ATP molecules are produced through glycolysis, so the total number of molecules formed is 38 molecules of ATP.For the reaction 2 H2O(g) equilibrium reaction arrow 2 H2(g) + O2(g), K = 2.4 ✕ 10−3 at a given temperature. At equilibrium in a 2.7 L container, it is found that [H2O(g)] = 2.0 ✕ 10−1 M and [H2(g)] = 2.5 ✕ 10−2 M. Calculate the moles of O2(g) present under these conditions.
To calculate the moles of O2(g) present at equilibrium, use the equilibrium constant expression and the given concentrations of H2O(g) and H2(g). The moles of O2(g) present at equilibrium is approximately 7.23 × 10-3 mol.
Explanation:To calculate the moles of O2(g) present at equilibrium, we can use the equilibrium constant expression and the given concentrations of H2O(g) and H2(g). The equilibrium constant expression for the given reaction is K = [H2]^2/[H2O]. Since we have the values for [H2] and [H2O], we can rearrange the expression to solve for [O2], which is the unknown.
By substituting the known values into the expression and solving for [O2], we get [O2] ≈ 2.67 × 10-3 M. This represents the concentration of O2(g), which we can then convert to moles of O2 using the volume of the container.
Given that the container has a volume of 2.7 L, we can multiply the concentration of O2 by the volume to get the moles of O2(g). In this case, the moles of O2(g) present at equilibrium would be approximately 7.23 × 10-3 mol.
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To calculate the moles of O2(g) present under the given conditions, divide the concentration of H2(g) by 2 and use the ideal gas law to find the number of moles in the 2.7 L container. The number of moles of O2(g) in the 2.7 L container is equal to the number of moles in any other volume multiplied by the ratio of the volumes.
Explanation:To calculate the moles of O2(g) present under the given conditions, we can first start by calculating the concentration of O2(g) using the given equilibrium concentrations of H2O(g) and H2(g). Since the reaction is 2 H2O(g) <=> 2 H2(g) + O2(g), the concentration of O2(g) is equal to half the concentration of H2(g). Therefore, the concentration of O2(g) is 2.5 x 10^-2 M / 2 = 1.25 x 10^-2 M.
Next, we can use the ideal gas law to calculate the number of moles of O2(g) in the 2.7 L container. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin. In this case, we don't have the pressure or the temperature, so we can't directly calculate the number of moles. However, since the pressure and temperature are not changing, we can assume that the pressure and temperature are constant, and therefore the number of moles is directly proportional to the volume. So, if we know the number of moles of O2(g) in the 2.7 L container, we can calculate the number of moles in any other volume using the equation: n1V1 = n2V2.
Therefore, if we let n1 be the number of moles of O2(g) in the 2.7 L container and n2 be the number of moles of O2(g) in another volume, we can set up the equation: n1 x 2.7 = n2 x V2. Substituting the values, we get: n1 x 2.7 = 1.25 x 10^-2 x V2. Solving for n2, we get: n2 = n1 x 2.7 / (1.25 x 10^-2). Plugging in the value of n1 as 1.25 x 10^-2 M x 2.7 L, we can calculate n2.
Therefore, the moles of O2(g) present under the given conditions can be calculated as follows: n2 = (1.25 x 10^-2 M) x (2.7 L) / (1.25 x 10^-2) = 2.7 moles.
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N2O5 decomposes to form NO2 and O2 with first-order kinetics. How long does it take for the N2O5 concentration to decrease from its initial value of 2.75 M to its final value of 1.85 M, if the rate constant, k, equals 5.89 × 10−3?
Answer:
67.3 s
Explanation:
The equation for the reaction can be represented as:
N₂O₅ ⇄ NO₂ + O₂
Rate (k) = [tex]5.89 * 10^{-3[/tex]
Rate law for first order is expressed as:
In [A] = -kt + In [A]₀
Given that:
[A] = Final Concentration = 1.85 M
[A]₀ = Initial Concentration = 2.75 M
time-taken = ???
substituting our given data; we have:
In[1.85] = -[5.89 × 10⁻³](t) + In [2.75]
t = [tex]\frac{In(2.75)-In(1.85)}{(5.89*10^{-3}}[/tex]
t = [tex]\frac{0.3964}{5.89*10^{-3}}[/tex]
t = 67.3 s
Calculate the molecular mass or formula mass (in amu) of each of the following substances: (a) BrN3 amu (b) C2H6 amu (c) NF2 amu (d) Al2S3 amu (e) Fe(NO3)3 amu (f) Mg3N2 amu (g) (NH4)2CO3 amu
Answer:
Explanation:
The atomic unit of mass (a.m.u) is the unit used to compare the relative masses of atoms. An atomic mass unit is a twelfth of the mass of a carbon-12 atom.
The calculation of the molecular mass of a compound is done by adding the relative atomic masses of the atoms of the formula of said substance, taking into account its abundance in the compound. The atomic mass of the atoms that make up the molecule are obtained in the Periodic Table.
So, in these cases you have:
BrN₃:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
Br: 79.9 amu
N: 14 amu
I take into account the abundance of each element in the compound you get:
BrN₃=79.9 amu + 3* 14 amu= 121.9 amu
C₂H₆:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
C: 12 amu
H: 1 amu
I take into account the abundance of each element in the compound you get:
C₂H₆= 2*12 amu + 6*1 amu= 30 amu
NF₂:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
N: 14 amu
F: 19 amu
I take into account the abundance of each element in the compound you get:
NF₂= 14 amu + 2*19 amu= 52 amu
Al₂S₃:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
Al: 27 amu
S: 32 amu
I take into account the abundance of each element in the compound you get:
Al₂S₃= 2*27 amu + 3*32 amu= 150 amu
Fe(NO₃)₃:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
Fe: 55.85 amu
N: 14 amu
O: 16 amu
I take into account the abundance of each element in the compound you get:
Fe(NO₃)₃= 55.85 amu + 3*(14 amu + 3*16 amu)= 241.85 amu
Mg₃N₂:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
N: 14 amu
Mg: 24 amu
I take into account the abundance of each element in the compound you get:
Mg₃N₂= 3*24 amu + 2*14 amu= 100 amu
(NH₄)₂CO₃:You know the atomic masses of the elements that make up the molecule, obtained from the periodic table:
N: 14 amu
H: 1 amu
C: 12 amu
O: 16 amu
I take into account the abundance of each element in the compound you get:
(NH₄)₂CO₃: 2*(14 amu + 4*1 amu) + 12 amu+3*16 amu= 96 amu
The molecular mass or formula mass of the listed substances have been calculated using atomic masses from the periodic table.
Explanation:The molecular mass or formula mass of a substance can be calculated by summing up the atomic masses of all atoms in the substance. Using the atomic masses from the periodic table, we have:
BrN3 = 1(Br) + 3(N) = 79.9 + 3(14.0) = 122.9 amu C2H6 = 2(C) + 6(H) = 2(12.0) + 6(1.0) = 30.0 amu NF2 = 1(N) + 2(F) = 14.0 + 2(19.0) = 52.0 amu Al2S3 = 2(Al) + 3(S) = 2(26.98) + 3(32.06) = 150.16 amu Fe(NO3)3 = 1(Fe) + 3(1(N) + 3(O)) = 55.85 + 3(1(14.0) + 3(16.0)) = 241.85 amu Mg3N2 = 3(Mg) + 2(N) = 3(24.3) + 2(14.0) = 100.9 amu (NH4)2CO3 = 2(1(N) + 4(H)) + 1(C) + 3(O) = 2(1(14.0) + 4(1.0)) + 12.0 + 3(16.0) = 96.0 amu Learn more about Molecular Mass Calculation here:https://brainly.com/question/33791791
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C8H18 + O2à CO2 + H2O is the unbalanced combustion reaction between octane and oxygen. What is the coefficient for carbon dioxide in the balanced equation?
A.2
B.8
C.16
D.18
Answer:
16. Option c.
Explanation:
The reaction is:
C₈H₁₈ + O₂ → CO₂ + H₂O
Let's count, we have 8 C, 18 H and 2 O in reactant side
We have 1 C, 2 H and 3 O in product side
In order to balance the C, we add 8 to CO₂ and to balance the H, we add a 9 to H₂O
Now we have 8 C on both sides and 18 H On both sides. Finally we have 25 O on product side.
C₈H₁₈ + O₂ → 8CO₂ + 9H₂O
To balance the O in product side we must add 25/2, but as it is a rational number, we must multiply x2 to get an integer number (x2 in all the stoichiometry). The balanced reaction is :
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
with 16 C, 36 H and 50 O, on both sides
The initial concentration of C2H6N2 in the first-order reaction C2H6N2→N2+C2H6 is 0.427 M. After 30. seconds, the concentration of C2H6N2 is 0.395 M. What is the rate constant k for the reaction?
Answer:
2.6×10^-3 mols-1
Explanation:
From
k= 2.303/t × log[A]o/[A]t
Time taken t= 30 seconds
Initial concentration [A]o= 0.427M
Concentration at time (t) [A]t= 0.395M
k= 2.303/30 ×log[0.427]/[0.395]=2.6×10^-3 mols-1
Of the reactions below, only ________ is not spontaneous. Mg (s) 2HCl (aq) MgCl2 (aq) H2(g) 2Ag (s) 2HNO3 (aq) 2AgNO3 (aq) H2 (g) 2Ni (s) H2SO4 (aq) Ni2SO4 (aq) H2 (g) 2Al (s) 6HBr (aq) 2AlBr3 (aq) 3H2 (g) Zn (s) 2HI (aq) ZnI2(aq) H2 (g)
Answer:
2Ag (s) + 2HNO₃ (aq) 2AgNO₃ (aq) + H₂ (g)
Explanation:
2Ag (s) + 2HNO₃ (aq) 2AgNO₃ (aq) + H₂ (g)
Ag is below H₂ in reactivity series. Therefore Ag does not spontaneously replace H₂ from any compound.
Answer:
The answer to the question is
2Ag (s) +2HNO₃ (aq) →2AgNO₃ (aq)+ H₂ (g)
Explanation:
The position of a cell on the reduction potential table determines if a reduction reaction will be spontaneous
The higher the positive reduction potential the higher the spontaneity of the reduction reaction. that is if the E⁰ cell is positive the cell is a spontaneous voltaic cell, however if the E⁰ cell is negative the cell is a electrolytic non-spontaneous cell
On the standard potential table silver has a low positive potential of
Ag⁺ + e⁻ → Ag = 0.799
It is an oxidizing agent tending to scarcely release electrons, therefore for the reaction to take place, there has to be some external supply of electromotive force.
As their name suggests, calcium ion channel blockers block calcium ion channels in the sarcolemma of pacemaker and contractile cardiac muscle cells and slow calcium ion entry into the cell during an action potential. What effects would these drugs have on the myocardial cells? View Available Hint(s)
Answer:
Effects of Calcium channel blockers on the myocardial cells include lower excitability of the cells due lower influx of calcium ions during an action potential. This effect helps to regulate abnormal rapid heart rhythms (arrhythmias).
Calcium channel blockers also reduce blood pressure by dilating the arteries thereby reducing pressure in the arteries. This in turn also reduces the workload on the heart and therefore allows for adequate oxygen supply to heart in the treatment of Angina.
Examples of calcium ion channel blockers are amlodipine, nimodipine, verapamil, diltiazem etc.
Copper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f = 6.3 × 10 10 Calculate the solubility (in g·L−1) of CuBr ( s ) ( K sp = 6.3 × 10 − 9 ) in 0.76 M NH 3 ( aq ) .
Answer:
53.18 gL⁻¹
Explanation:
Given that:
[tex]Cu^{2+}_{(aq)}[/tex] [tex]+[/tex] [tex]2NH_{3(aq)}[/tex] [tex]------>[/tex] [tex][Cu(NH_3)_2]^+_{(aq)}[/tex] ------equation (1)
where;
Formation Constant [tex](k_f) =[/tex] [tex]6.3*10^{10}[/tex]
However, the Dissociation of [tex]CuBr_{(s)[/tex] yields:
[tex]CuBr_{(s)}[/tex] ⇄ [tex]Cu^{+}_{(aq)}[/tex] [tex]+[/tex] [tex]Br^-_{(aq)}[/tex] -------------- equation (2)
where;
the Solubility Constant [tex](k_{sp})[/tex] [tex]= 6.3 *10^{-9[/tex]
From equation (1);
[tex](k_f) =[/tex] [tex]\frac{[[Cu(NH_3)_2]^+]}{[Cu^{2+}][NH_3]^{2}}[/tex] --------- equation (3)
From equation (2)
[tex](k_{sp})[/tex] [tex]= [Cu^+][Br^-][/tex] --------- equation (4)
In [tex]NH_3[/tex], the net reaction for [tex]CuBr_{(s)[/tex] can be illustrated as:
[tex]CuBr_{(s)[/tex] [tex]+[/tex] [tex]2NH_{3(aq)}[/tex] ⇄ [tex][Cu(NH_3)_2]^+_{(aq)}[/tex] [tex]+ Br^-_{(aq)}[/tex]
The equilibrium constant (K) can be written as :
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]
If we multiply both the numerator and the denominator with [tex][Cu^+][/tex] ; we have:
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}*\frac{[Cu^+]}{[Cu^+]}[/tex]
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+}{[NH_3]^2[Cu^+]}*{[Cu^+][Br^-]}[/tex]
[tex]K = k_f *k_{sp}[/tex]
[tex]K= (6.3*10^{10})*(6.3*10^{-9})[/tex]
[tex]K= 3.97*10^2[/tex]
[tex]K[/tex] ≅ [tex]4.0*10^2[/tex]
Now; we can re-write our equilibrium constant again as:
[tex]K=[/tex][tex]\frac{[[Cu(NH_3)_2]^+][Br^-]}{[NH_3]^2}[/tex]
[tex]4.0*10^2 = \frac{(x)(x)}{(0.76-2x)^2}[/tex]
[tex]4.0*10^2 = \frac{(x)^2}{(0.76-2x)^2}[/tex]
[tex]4.0*10^2 = (\frac{(x)}{(0.76-2x)})^2[/tex]
By finding the square of both sides, we have
[tex]\sqrt {4.0*10^2} = \sqrt {(\frac{(x)}{(0.76-2x)})^2[/tex]
[tex]2.0*10 = \frac{x}{(0.76-2x)}[/tex]
[tex]20(0.76-2x) =x[/tex]
[tex]15.2 -40x=x[/tex]
[tex]15.2 = 40x +x[/tex]
[tex]15.2 = 41x[/tex]
[tex]x = \frac{15.2}{41}[/tex]
[tex]x = 0.3707 M[/tex]
In gL⁻¹; the solubility of [tex]CuBr_{(s)[/tex] in 0.76 M [tex]NH_3[/tex] solution will be:
[tex]= \frac{0.3707 mole of CuBr}{1L}*\frac{143.45 g}{mole of CuBr}[/tex]
= 53.18 gL⁻¹
The problem involves finding the solubility of copper(I) bromide in a solution of ammonia. The solution entails expressing the formation of complex ions and dissolution of CuBr in terms of equilibrium expressions. Solving the system of expressions for solubility 's' can eventually give the solubility in g·L-1.
Explanation:In this problem, we are looking for the solubility of copper(I) bromide (CuBr) in a solution with 0.76 M NH3. Since CuBr does not have high solubility in water, we are using ammonia which forms complex ions with copper, enhancing its solubility. We will utilize the given equilibrium constant (Kf) which helps us to understand how far the forward reaction (complex ion formation) proceeds.
To start, we'll write the reaction of Cu+ ion with NH3 and the corresponding Kf expression:
Cu+ (aq) + 2NH3 (aq) <=> Cu(NH3)2+ (aq); Kf = [Cu(NH3)2+] / [Cu+] [NH3]^2
Next, we need to find an expression for [Cu+] in terms of solubility and solve for solubility. Let's denote the solubility of CuBr as 's'. The dissolution of CuBr can be represented as:
CuBr (s) <=> Cu+ (aq) + Br- (aq)
We can see that for every mole of CuBr that dissolves, one mole of Cu+ ion is produced. Thus, [Cu+] = s.
Now using the Kf expression, we can solve the system of equations to find s, which effectively gives us the solubility of CuBr in 0.76 M NH3. After that, by taking the molecular weight of CuBr into account, we can finally express the solubility in g·L−1.
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