Copper(II) sulfate forms a bright blue hydrate with the formula CuSO 4 ⋅ n H 2 O ( s ) . If this hydrate is heated to a high enough temperature, H 2 O ( g ) can be driven off, leaving the grey‑white anhydrous salt CuSO 4 ( s ) . A 14.220 g sample of the hydrate was heated to 300 ∘ C . The resulting CuSO 4 ( s ) had a mass of 8.9935 g . Calculate the val

Answer:

In 7.80 g of styrene, we have 3.60×10²³ atoms of H

Explanation:

Empirical formula of styrene is CH

Molecular formula of styrene is C₈H₈

So, 1 mol of styrene has 8 moles of C and 8 moles of H and 1 mol weighs 104.14 grams. Let's make a rule of three:

104.14 g (1 mol of C₈H₈) have 8 moles of H

Then 7.80 g would have ( 7.80  .8) / 104.14 = 0.599 moles

As we know, 1 mol of anything has NA particles (Avogadro's Number,  6.02×10²³), so 0.599 moles will have (mol . NA) particles

0.599 mol . 6.02×10²³ atoms / 1 mol = 3.60×10²³ atoms

Answer:

27oz

Explanation:

Let the volume quantity of the 10% solution = x

The volume quantity of the 15% alcohol solution is known as = 18oz

Therefore the volume of the 12% product = 18oz + x

Therefore

0.10 × x + 0.15 × 18oz = 0.12 × (18oz+x)

0.15×18oz - 0.12×18oz = 0.12×x - 0.10×x

.

0.03*18oz = 0.02x

Dividing both sides by 0.02

x = 27oz

Verifying the answer, we have

0.1 × 27 + 18 × 0.15 = 0.12 × 45

2.7 + 2.7 = 5.4

Answer:

The answer to your question is  21 g of CO₂  

Explanation:

Balanced Reaction

                               C₃H₈  +  6O₂    ⇒    3CO₂   +   4H₂O

Data

mass of C₃H₈ = 7 g

mass of O₂  = 98 g

Determine the limiting reactant

Molecular mass of C₃H₈ = (12 x3) + (1 x 8) = 36 + 8 = 44 g

Molecular mass of Oxygen = 16 x 12 = 192 g

Theoretical proportion  C₃H₈ / O₂ = 44 / 192 = 0.23

Experimental proportion  C₃H₈ / O₂ = 7 / 98 = 0.07

As the proportion diminishes in the experiment, the excess reagent is the oxygen and the limiting reactant is propane.

- Calculate the mass of CO₂

                     44 g of C₃H₈   ---------------- 3(44) of CO₂

                       7 g of C₃H₈    ---------------   x

                       x = (7 x 3(44)) / 44

                       x = 924 / 44

                       x = 21 g of CO₂                        

Compounds have the lowest percentage of gold content by weight is Aui3

M(Au) =197g/mol

M(O) =16g/mol

M(H)=1g/mol

M(Cl)=35g/mol

M(I)=53 g/mol

1) M=214g/mol =197+1+16

197/214= 0.921

2) M=248g/mol =[tex]197 +3 \times ( 1+16)[/tex]

197/248=0.794

3) M=302g/mol = [tex]197+3 \times 35[/tex]

197/302=0.652

4) M=356g/mol =[tex]197+3 \times 53[/tex]

197/356=0.5533

Explanation:

Iodine is the heaviest out of I, Cl and OH. The items on your list hold less and less gold by mass as you go down. The gold dissolution percentage in a hypochlorite-iodide mix-up is completely conditioned upon the solution pH. So the iodine has the lowest gold content percentage.

Answer: AuI3

Explanation:

Final answer:

The concentration after 17.0 minutes will be approximately 0.067 M.

Explanation:

The rate constant for a reaction is a constant that relates the rate of the reaction to the concentration of the reactants. The rate law for a reaction is given by the equation: rate = k[A]^[x][B]^[y], where k is the rate constant, [A] and [B] are the concentrations of the reactants, and x and y are the reaction orders concerning A and B, respectively.

In this case, since the rate constant (k) is given as 5.40 × 10^(-3) s^(-1) and the initial reactant concentration is 0.100 M, we can use the integrated rate law for a first-order reaction to find the concentration after a certain time:

[A] = [A]_0 * e^(-kt)

Substituting the given values, we have:

[A] = (0.100 M) * e^(-5.40 × 10^(-3) s^(-1) * (17.0 * 60 s))

Simplifying this equation gives:

[A] ≈ 0.067 M

The concentration of the reactant after 17.0 minutes is approximately 0.000405 M.

This was determined using the formula for a first-order reaction. Initial concentration, rate constant, and time were given and plugged into the formula for accuracy.

To solve this problem, we need to use the formula for a first-order reaction:

[A] = [A]0 e-kt

Where:

Given:

Plug these values into the formula:

Simplify the exponent:

Using a calculator to find e-5.508:

The concentration of the reactant after 17.0 minutes is approximately 0.000405 M.

The student can use Graham's Law of diffusion to find the molar mass of the unknown hydrocarbon gas. They can then identify plausible molecular formulas that correspond to this molar mass to determine the unknown hydrocarbon gas. The rate of diffusion for the known and unknown gases in the apparatus serves as the initial data for this determination.

The question is about finding the molecular formula of an unknown hydrocarbon gas based on its rate of diffusion compared to HBr gas in a given diffusion apparatus. In general, the lighter a gas molecule, the faster it diffuses, and this rate of diffusion can be related to its molecular mass by Graham's Law of diffusion. This law illustrates that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass.

According to the information given, we can use the formula derived from Graham's law: r1/r2 = sqrt(M2/M1), where r1 and r2 are the rates of diffusion of the two gases, and M1 and M2 are their respective molar masses.

Here, HBr (Hydrobromic acid) gas diffused at a rate of 15.0 mL/min and its molar mass is about 81 g/mol. The unknown gas diffused at a rate of 20.3 mL/min. Plugging these values into the formula, we can solve for the molar mass of the unknown gas. After finding the molar mass, we can then determine plausible hydrocarbon molecular formulas that correspond to this molar mass, thus identifying the unknown hydrocarbon gas.

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Answer: Thus volume in liters is [tex]6.14\times 10^3L[/tex]

Explanation:

Molarity is defined as the number of moles of solute dissolved per liter of the solution.

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of silver oxide}}{\text{Volume of solution in L}}[/tex]     .....(1)

Molarity of silver oxide solution = [tex]7.73\times 10^{-5}mmol/L=7.73\times 10^{-2}\mu mol/L[/tex]      [tex]1mmol=1000\mu mol[/tex]

Moles of silver oxide = [tex]475\mu mol[/tex]      

Volume of solution in L = ?

Putting values in equation 1, we get:

[tex]7.73\times 10^{-2}\mu mol\L=\frac{475\mu mol}{\text{Volume of solution in L}}\\\\\{\text{Volume of solution in L}}=\frac{475\mu mol}{7.73\times 10^{-2}\mu mol\L}=6.14\times 10^3L[/tex]

Thus volume in liters is [tex]6.14\times 10^{3}L[/tex]

Explanation:

When we are working in a laboratory then it is necessary that certain appropriate measures have to be taken in order to avoid any king of accident or mistake.

So, when we are mixing liquids in a separatory funnel then following safety measures should be followed.

1. Everything should be done in your hood

2. You should wear gloves

3. Your hood sash should be between your face and the separatory funnel

4. Don't aim the separatory funnel at anyone

5. You should have a firm grip on the stopper and vent the funnel frequently by using the stopcock

Therefore, we can conclude that all the given statements are true about mixing liquids in a separatory funnel.

Answer : The partial pressure of [tex]NO_2[/tex] is, 12.34  atm

Explanation :

For the given chemical reaction:

[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]

The expression of [tex]K_p[/tex] for above reaction follows:

[tex]K_p=\frac{(P_{NO_2})^2}{P_{N_2O_4}}[/tex]         ........(1)

The equilibrium reaction is:

                     [tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]

Initial             x                    0

At eqm       (x-y)                 2y

Putting values in expression 1, we get:

[tex]70.9=\frac{(2y)^2}{(x-y)}[/tex]           ..............(2)

As we are given that, 25.8 % of [tex]N_2O_4[/tex] remains at equilibrium. That means,

[tex]25.8\% \times x=(x-y)[/tex]

[tex]\frac{25.8}{100}\times x=(x-y)[/tex]

[tex]0.258\times x=(x-y)[/tex]

[tex]0.742x=y[/tex]       ..............(3)

Now put equation 3 in 2, we get the value of 'x'.

[tex]70.9=\frac{(2\times 0.742x)^2}{(x-0.742x)}[/tex]

[tex]x=8.31[/tex]

Now put the value of 'x' in equation 3, we get:

[tex]0.742x=y[/tex]

[tex]0.742\times 8.31=y[/tex]

[tex]y=6.17[/tex]

Now we have to calculate the new partial pressure of [tex]NO_2[/tex] at equilibrium.

Partial pressure of [tex]NO_2[/tex] = (2y) = (2\times 6.17) = 12.34  atm

Hence, the partial pressure of [tex]NO_2[/tex] is, 12.34  atm

To determine the partial pressure of NO2 at equilibrium, use the equilibrium expression and substitute the given values into the equation.

To determine the partial pressure of NO2 at equilibrium, we will use the equilibrium expression for the reaction:

Kp = (P(NO2))^2 / P(N2O4)

Given that the equilibrium constant Kp is 70.9 and the equilibrium concentration of N2O4 is 25.8% of its initial concentration, we can calculate the partial pressure of NO2 at equilibrium:

P(NO2) = sqrt(Kp * P(N2O4))

Substituting the values, we get:

P(NO2) = sqrt(70.9 * (1 - 0.258))

P(NO2) = sqrt(70.9 * 0.742)

Answer:

6.1 h = 6 h and 8 min

Explanation:

First, let's found the rate of disappearing of N2O5. Knowing that it's a first-order reaction, it means that the rate law is:

rate = k*pN2O5

Where k is the rate constant, and pN2O5 is the initial pressure of N2O5 (2.0 atm), so:

rate = 3.4x10⁻⁵*2.0

rate = 6.8x10⁻⁵ atm/s

Thus, at each second, the partial pressure of the reagent decays 6.8x10⁻⁵ atm. The rate is also the variation of the pressure divided by the time. Because it is decreasing, we put a minus signal in the expression.

1 atm = 760 torr, so 380torr/760 = 0.5 atm

rate = -Δp/t

6.8x10⁻⁵ = -(0.5 - 2.0)/t

t = 1.5/6.8x10⁻⁵

t = 22,058 s (÷60)

t = 368 min (÷60)

t = 6.1 h = 6 h and 8 min

To determine the time it takes for a sample of N2O5 to reach a certain partial pressure, we can use the first-order rate equation and solve for time.

The decomposition of N2O5(g) into NO2(g) and O2(g) at 25°C follows first-order kinetics. The rate constant, k, is given as 3.4×10−5 s−1.

To determine how long it will take for a sample initially containing 2.0 atm of N2O5 to reach a partial pressure of 380 torr, we can use the first-order rate equation:

ln([N2O5]t/[N2O5]0) = -kt

Rearranging the equation and substituting the given values, we get:

ln(380/760) = -(3.4×10−5 t)

Solving for t, we find that it will take approximately 2,370 seconds or about 39.5 minutes.

The given question is incomplete. The complete question is as follows.

The value of Henry's law constant [tex]k_{H}[/tex] for oxygen in water at [tex]25^{o}C[/tex] is [tex]1.66 \times 10^{-6}[/tex] M/torr.

Calculate the solubility of oxygen in water at [tex]25^{o}C[/tex] when the total external pressure is 1 atm and the mole fraction of oxygen in the air is 0.20 atm.

Explanation:

Formula to calculate partial pressure of a gas is as follows.

   Partial pressure of oxygen = mole fraction of oxygen x total pressure

Putting the given values into the above equation as follows.

       = [tex]0.20 \times 760[/tex] = 152 torr

Therefore, solubilty (concentration) of oxygen in water  will be calculated as follows.

        Solubility = Henry's law constant x partial pressure of oxygen

                       = [tex]1.66 \times 10^{-6} M/torr \times 152 torr[/tex]

                       = [tex]2.52 \times 10^{-4}[/tex] M

Thus, we can conclude that solubility of given oxygen is [tex]2.52 \times 10^{-4}[/tex] M.

The solubility of oxygen in water at 25°C, when the total external pressure is 1 atm and the mole fraction of oxygen in the air is 0.2, is approximately [tex]\( 0.000260 \, \text{mol/L} \)[/tex].

The solubility of oxygen in water at 25°C, when the total external pressure is 1 atm and the mole fraction of oxygen in the air is 0.2, can be calculated using Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The relationship can be expressed as:

[tex]\[ C = k \cdot p \][/tex]

where:

- C  is the concentration of the gas in the liquid (solubility of oxygen in water, in this case),

- k  is the Henry's Law constant for oxygen in water at the given temperature,

-  p is the partial pressure of oxygen above the liquid.

Given that the total external pressure is 1 atm and the mole fraction of oxygen is 0.2, the partial pressure of oxygen [tex](\( p_{O_2} \))[/tex] can be calculated as:

[tex]\[ p_{O_2} = \text{total pressure} \times \text{mole fraction of oxygen} \] \[ p_{O_2} = 1 \, \text{atm} \times 0.2 \] \[ p_{O_2} = 0.2 \, \text{atm} \][/tex]

The Henry's Law constant for oxygen in water at 25°C [tex](\( k_{O_2} \)) is approximately \( 769.23 \, \text{L} \cdot \text{atm/mol} \).[/tex]

Now, we can calculate the solubility of oxygen in water:

[tex]\[ C_{O_2} = k_{O_2} \cdot p_{O_2} \]\[ C_{O_2} = 769.23 \, \text{L} \cdot \text{atm/mol} \times 0.2 \, \text{atm} \] \[ C_{O_2} = 153.846 \, \text{L} \cdot \text{atm/mol} \][/tex]

To express the solubility in terms of molarity (mol/L), we divide the partial pressure by the Henry's Law constant:

[tex]\[ C_{O_2} = \frac{p_{O_2}}{k_{O_2}} \]\[ C_{O_2} = \frac{0.2 \, \text{atm}}{769.23 \, \text{L} \cdot \text{atm/mol}} \]\[ C_{O_2} \approx 0.000260 \, \text{mol/L} \][/tex]

The answer is: [tex]0.000260 \, \text{mol/L}.[/tex]

Answer:

The non-cyclic photophosphorylation which is the light-requiring part of photosynthesis in some higher plants, in which an electron donor is required, and oxygen is also produced as a waste product. this consists of two photoreactions, resulting in the synthesis of ATP and NADPH 2.

Explanation:

   A. When photosystem II absorbs light, an electron excited to a higher energy level in the reaction center chlorophyll (P680) is captured by the primary electron acceptor.  The oxidized chlorophyll is now a very strong oxidizing agent; its electron “hole” must be filled.

  B.  An enzyme extracts electrons from water and supplies them to P680, replacing the electrons that the chlorophyll molecule lost when it absorbed light energy.  This reaction splits a water molecule into two hydrogen ions and an oxygen atom, which immediately combines with another oxygen atom to form O2.  This splitting of water is responsible for the release of O2 into the air.

  C.  Each photoexcited electron (energized by light) passes from the primary electron acceptor in photosystem II to photosystem I via an electron transport chain.  This electron transport chain is very similar to the one in cellular respiration; however, the carrier proteins in the chloroplast ETC are different from those in the mitochondrial ETC.

  D.  As electron move down the chain, their exergonic “fall”to a lower energy level is harnessed by the thylakoid membrane to produce ATP (by chemiosmosis).  The production of ATP in the chloroplast is called photophosphorylation because the energy harnessed in the process originally came from light.  This process of ATP production is called non-cyclic photophosphorylation.  The ATP generated in this process will provide the energy for the synthesis of glucose during the Calvin cycle (light independent reactions).

  E.  When an electron reaches the “bottom” of the electron transport chain, it fills an electron “hole” in the chlorophyll a molecule in the reaction center of photosystem I (P700).  The hole was created when light energy drives an electron from P700 to the primary electron acceptor of photosystem I.

   F. The primary electron acceptor of photosystem I passes the excited electrons to a second electron transport chain which transmits them to an iron-containing protein.  An enzyme reaction transfers the electrons from the protein to NADP+ that forms NADPH (which has high chemical energy due to the energy of the electrons).  NADPH is the reducing agent needed for the synthesis of glucose in the Calvin cycle.

Answer:

The answer to your question is 432 g of CO₂

Explanation:

Data

CaCO₃  = 983 g

CaO = 551 g

CO₂ = ?

Balanced reaction

                               CaCO₃ (s)   ⇒   CaO (s)   +  CO₂ (g)

This reaction is balanced, to solve this problem just remember the Lavoisier Law of conservation of mass that states that the mass of the reactants is equal to the mass of the products.

                    Mass of reactants = Mass of products

                    Mass of CaCO₃   = Mass of CaO + Mass of CO₂

Solve for CO₂

                    Mass of CO₂  = Mass of CaCO₃ - Mass of CaO                    

                     Mass of CO₂ = 983 g - 551 g

Simplification

                     Mass of CO₂ = 432 g                        

Answer:

The answer to your question is Empirical formula  InCl₃

Explanation:

Data

InCl = 0.5 g

Cl = 0.2404 g

Empirical formula = ?

Process

1.- Calculate the mass of Indium

    Total mass = mass of Indium + mass of Chlorine

     0.50 = mass of Indium + 0.2404

     mass of Indium = 0.50 - 0.2404

     mass of Indium = 0.2596 g

2.- Calculate the moles of Indium and Chloride

Atomic mass Indium = 115 g

Atomic mass Chlorine = 35.5 g

                       115 g of In ------------------ 1 mol

                     0.2596 g of In ------------- x

                        x = (0.2596 x 1) / 115

                        x = 0.0023 moles of Indium

                        35.5 g of Cl ------------- 1 mol

                         0.2404 g    --------------- x

                          x = (0.2404 x 1) / 35.5

                          x = 0.0068 moles of Cl

3.- Divide by the lowest number of moles

Indium     0.0023 / 0.0023 = 1

Chlorine  0.0068 / 0.0023 = 3

4.- Write the empirical formula

                                       InCl₃

The empirical formula of the indium compound is [tex]InCl_3[/tex]

Mass of indium = 0.500-0.2404

= 0.2596g

And,  

Mass of chlorine = 0.2404g

Now  

Divide each mass by atomic mass

[tex]In = 0.2596\div 114.818 = 0.002261\\\\Cl = 0.2404\div 35.453 = 0.00678[/tex]

Divide by smaller:

In = 1

Cl = 3

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Answer:Use the K L M N shell orbit

Explanation: Valence electrons are those electrons on the outermost shell of an atomic nucleus.

The K L M N electron shell is named by a spectroscopist called Charles G. Barkla, who discovered the A X-rays ( high energy rays) and the B X-rays ( low energy rays), he later renamed the A X-rays K X-rays which is the Energy level orbit, and the B X-rays was also renamed to be the L M N and so on.

PLEASE SEE PICTURE ATTACHED

THE PICTURE SHOWS HOW TO DETERMINE VALENCE ELECTRON USING THE As AND B's WHICH ARE THE K L M N SHELL ORBIT, USING SILICON AS AN EXAMPLE.

The maximum number of electron in the K orbit is 2

The maximum number of electron in the L M N shell is 8

Therefore the number of electron found in the outermost shell becomes the number of Valence electron in that element.

Final answer:

The molarity of the KNO3 solution is calculated by first finding the molar mass of KNO3, then determining the number of moles of KNO3 dissolved, and finally dividing the moles by the volume of the solution in liters, resulting in a molarity of 0.4118 M.

Explanation:

To calculate the molarity of a KNO3 solution made by dissolving 76.6 g of KNO3 in a total solution volume of 1.84 L, follow these steps:

Therefore, the molarity of the KNO3 solution is 0.4118 M.

Answer:

The answer to your question is 2 ml

Explanation:

Data

Initial volume = ?

Initial concentration = 220 mg/ml

Final volume = 10 ml

Final concentration = 43 mg/ml

Equation

    Initial volume x Initial concentration = Final volume x Final concentration

Solve for Initial volume

 Initial volume = (Final volume x Final concentration) / Initial concentration

Substitution

 Initial volume = (10 x 43) / 220

Simplification

 Initial volume = 430 / 220

Result

 Initial volume = 1.95 ml ≈ 2.0 ml

Answer:

In a neutral solution the concentration of hydrogen ions is equal to the concentration of hydroxide ions

Explanation:

When:

[H⁺] > [OH⁻]  the pH is acid, so the solution is practically acid.

When

[H⁺] < [OH⁻], the pH is basic, so the solution is practically basic

In neutral solutions:

[H⁺] = [OH⁻], so the pH is neutral (7)

pH > 7 → BASIC SOLUTIONS

pH < 7 → ACID SOLUTION

In a neutral solution,d) the concentration of hydrogen ions is equal to the concentration of hydroxide ions. This balance comes from water's autoionization process, with discrepancy coming from the addition of acids or bases.

In a neutral solution, d) the concentration of hydrogen ions (H+) is equal to the concentration of hydroxide ions (OH-).

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Answer:

The answer to your question is a) pure substance- compound

Explanation:

a) pure substance-compound.  A pure substance can be an element or compound that is not mixed with another substance, glucose is a compound and also a pure substance.

b) mixture-heterogeneous.  This option is wrong because glucose is not mixed with another substance.

c) pure substance-element.  Glucose is a pure substance but it is a compound because it is composed of carbon, hydrogen, and oxygen. This option is incorrect.

d) mixture-homogeneous.  Glucose is not a mixture, it is a pure substance. This option is wrong.

e) none of the above. This option is wrong because the first option is correct.

Answer: A. Pure substance - compound

Explanation: Sugars or sacharrides are complex compounds having Carbon,Hydrogen and Oxygen atoms on its composition. Since a combination of two or more different elements forms a compound and compounds are pure substances.

Answer:

-375.9_KJ/(mol)

Explanation:

H(T2 ) ≈ H(T1)+CPΔT

Specific heat of Carbon is 0.71 J/g K.

At 283.15 the heat capacity is 37.12 J/(mol*K)

Kirchhoff's law

H(T2 ) ≈ H(T1)+CPΔT

Where

H(T1) and H(T2 ) are the heat of formation of CO2 at temperatures T1 and T2

CP is the heat capacity

Thus we have and ΔT is the temperature change

H(T2 ) ≈ -393.51×10^3+CP×(500-25)

= -393.51×10^3+37.12×(500-25)

= -375878 J/(mol)

= -375.9KJ/(mol)

To find the enthalpy of formation of CO2 at 500°C, we can use the equation ΔH = ΔH° + CpΔT, where ΔH° is the enthalpy change at 25°C, Cp is the heat capacity, and ΔT is the change in temperature. Assuming a constant heat capacity over the temperature range, we can calculate the change in enthalpy using the average heat capacity and the given values. The enthalpy of formation at 500°C is approximately -380.76 kJ/mol.

The enthalpy of formation of CO2 at 25°C is -393.51 kJ/mol. To find the enthalpy of formation at 500°C, we need to consider the change in enthalpy with temperature. One approach is to use the equation ΔH = ΔH° + CpΔT, where ΔH is the enthalpy change at the higher temperature, ΔH° is the enthalpy change at 25°C, Cp is the heat capacity, and ΔT is the change in temperature. However, since the question does not provide the heat capacity, an alternative method is to assume that the heat capacity is constant over the temperature range and use the average heat capacity to calculate the change in enthalpy.

For CO2, the average heat capacity over this temperature range is approximately 29.0 J/(mol·K). So, to find the enthalpy of formation at 500°C, we can use the equation ΔH = ΔH° + CpΔT, where ΔH° = -393.51 kJ/mol, Cp = 29.0 J/(mol·K), and ΔT = 500 - 25 = 475 K. Plugging in these values, we get: ΔH = -393.51 kJ/mol + (29.0 J/(mol·K) × 475 K) = -380.76 kJ/mol.

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Answer: ATP Adenosine triphosphate

Explanation: During light reaction plants produce glucose and as well as ATP and NADPH as another products. Now during the Dark or light independent reaction there is no presence of light energy to fuel the process so ATP is used as an alternative to fix CO2 molecules.

Answer:

4 for C in CH4

+4 for C in CO2

-2 for O in all substances

+1 for H in both CH4 and H2O

option 1,2,3 and 4 are correct. Option 5 is not correct

Explanation:

Step 1: Data given

Oxidation number of H = +1

Oxidation number of O = -2

Step 2: The balanced equation

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: The oxidation numbers

-4 for C in CH4

 ⇒ Oxidation number of H = +1

    ⇒ 4x +1 = +4

C has an oxidation number of -4

This is correct.

+4 for C in CO2

 ⇒ Oxidation number of O = -2

    ⇒ 2x -2 = -4

C has an oxidation number of +4

This is correct.

-2 for O in all substances

⇒ this is correct, the oxidation number of O is always -2 (except in H2O2 and Na2O2)

This is correct.

+1 for H in both CH4 and H2O

⇒ this is correct, the oxidation number of H is always +1 (except in metal hydrides).

This is correct.

+4 for O in H2O

 ⇒ Oxidation number of H = +1

    ⇒ 2x +1 = +2

The oxidation number of O is -2

This is not correct

Final answer:

The correct statements regarding oxidation states in the methane combustion reaction are 4 for C in CH4

+4 for C in CO2

-2 for O in all substances

+1 for H in both CH4 and H2O

Explanation:

The student's question relates to the oxidation states of elements in a chemical reaction, specifically in the combustion of methane represented by the equation CH4 + 2O2 → CO2 + 2H2O. Let's evaluate the statements:

Step 1: Data given

Oxidation number of H = +1

Oxidation number of O = -2

Step 2: The balanced equation

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: The oxidation numbers

-4 for C in CH4

⇒ Oxidation number of H = +1

   ⇒ 4x +1 = +4

C has an oxidation number of -4

This is correct.

+4 for C in CO2

⇒ Oxidation number of O = -2

   ⇒ 2x -2 = -4

C has an oxidation number of +4

This is correct.

-2 for O in all substances

⇒ this is correct, the oxidation number of O is always -2 (except in H2O2 and Na2O2)

This is correct.

+1 for H in both CH4 and H2O

⇒ this is correct, the oxidation number of H is always +1 (except in metal hydrides).

This is correct.

+4 for O in H2O

⇒ Oxidation number of H = +1

   ⇒ 2x +1 = +2

The oxidation number of O is -2

This is not correct

Answer:

See explanation for answer

Explanation:

You are missing some important data, in this case, the mass and temperature of the water.

In order to solve for this and explain to you how to do it, I'm going to assume some values for the mass and temperature for water, and then, you only need to replace these values with the values you have to get an accurate answer.

Let's assume the mass of water we have is 0.150 kg of water at 40 °C.

Now, let's remember that the water and ice are exerting heat and the sum of these heats must be zero always:

Q = Qw + Qi  = 0 (1)

In the case of the heat exerted by the ice, let's remember that the ice is passing through several stages. First, it's temperature changes but not it's phase, it remains solid. Second, it's when the ice begins to melt and ends when it's totally melted. And third, the change of temperature of water when the ice melted. So for these three stages, the heat of ice can be calculated with the following expression:

Qi = (mi * Ci * ΔTi) + (mi * Lf) + (mi * Cw * ΔTm)   (2)

The values of Ci, Lf and Cw are tabulated and reported data, and these are the following:

Ci: Specific heat of ice =  2100 J/kg °C

Cw: Specific heat of water = 4190 J/kg °C

Lf: heat of fusion of water =  3.34x10⁵ J/kg

For the case of heat of water it's just the specific heat of water, it's mass and the difference of temperature and the expression is:

Qw = mw * Cw * ΔT  (3)

Now, let's calculate firt the heat of water:

Qw = 0.150 * 4190 * (27.2 - 40)

Qw = -8044.8 J (a)

We have the heat of water, now, let's calculate heat of ice in function of the mass of ice:

Qi = [mi * 2100 * (0 + 10.9)] + (mi * 3.34x10⁵) + [mi * 4190 * (27.2 - 0)]

Qi = 22,890mi + 3.34x10⁵mi + 113,968mi

Qi = 470,858mi (b)

Finally, replace (a) and (b) in equation (1) to solve for mass of ice:

0 = 470,858mi - 8044.8

8044.8 = 470,858mi

mi = 8044.8 / 470,858

mi = 0.0171 kg or 17.1 g of ice

This is the mass required to drop the temperature of the system to 27.2 °C. Now, remember to replace the value of mass of water and temperature in this procedure to get the real and accuraten answer.

Answer:

The equation in function notation is:

= [tex]f(q)=\frac{1}{2}s-\frac{3}{2}[/tex]

Explanation:

Given equation:

[tex]6q = 3s-9[/tex]

Where q is the independent variable.

To write the equation in function notation we will do the following:

Step 1 : divide 6 on both sides:

[tex]\frac{6q}{6}=\frac{3s-9}{6}[/tex]

[tex]q=\frac{1}{2}s-\frac{3}{2}[/tex]

The function notation will become:

[tex]f(q)=\frac{1}{2}s-\frac{3}{2}[/tex]

Answer:

D. f(q)=2q+3

Explanation:

The answer should be D. f(q)=2q+3

If you need a explannation, comments below and I will edit the explannation  

(a) The percentage of gold in the jewelry is  100%.

(b) The purity of the gold jewelry in karats is 24 karats.

(a) The percentage of gold in the jewelry is calculated as;

The given parameters include;

Density of gold (ρg) = 19.3 g/cm³

Density of silver (ρs) = 10.5 g/cm³

Volume of the jewelry (Vj) = 0.675 cm³

The total volume of the jewelry is the sum of the volumes of gold and silver:

Vj = Vg + Vs

Vg = Vj (since we are assuming the total volume is the sum of gold and silver volumes)

Vs = Vj - Vg

= 0.675 cm³ - 0.675 cm³

= 0 cm³

Mg = Vg x ρg

= 0.675 cm³ x 19.3 g/cm³

= 13.03 g

Ms = Vs x ρs

= 0 cm³ x 10.5 g/cm³

= 0 g

The percentage of gold is;

Percentage of gold = (Mg / (Mg + Ms)) x 100

= (13.03 g / (13.03 g + 0 g)) x 100

=  100%

(b) Since the jewelry is 100% gold by mass, its purity in karats is 24 karats (the highest purity since it's pure gold).

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The steps associated with energy entering the system during the dissolution process involve the endothermic separation of solvent molecules and the separation of solute molecules, which require energy to overcome intermolecular forces.

The steps associated with energy entering the system during the dissolution process are:

Both the separation of solvent molecules (step 1) and the separation of solute molecules (step 2) are endothermic processes that require energy input. These processes involve overcoming the intermolecular attractive forces holding the respective particles together. Conversely, the mixing of solute and solvent molecules (step 3) is an exothermic process where energy is released, thus not associated with energy entering the system.

Whether the dissolution process is overall exothermic or endothermic depends on the relative magnitudes of the energy changes during these steps. For example, if the energy released during solvation is greater than the energy required for separating solute and solvent molecules, the overall process is exothermic. Conversely, if more energy is consumed in overcoming the solute-solute and solvent-solvent attractive forces than is released during solvation, the overall process is endothermic.

Answer:

Al2(SO4)3(aq) + 3BaCl2(aq) → 3BaSO4(s) + 2AlCl3(aq)

Explanation:

The question wants you to use the lowest possible coefficients to balance the equation.

According to the question the reaction is as follows ;

Generally, writing the chemical formula requires one to exchange the charge between the cations and anions involved. Example aluminum sulfate has Al3+ and (SO4)2- . cross multiply the charges to get Al2(SO4)3

aluminum sulfate → Al2(SO4)3

barium chloride → BaCl2

barium sulfate → BaSO4

aluminum chloride → AlCl3

Al2(SO4)3(aq) + BaCl2(aq) → BaSO4(s) + AlCl3(aq)

To balance a chemical equation one have to make sure the number of atom of element on the reactant side(left) is equal to the number of atom of elements on the product side(right).

Now, the equation can be be balance with the lowest coefficient as follows;

Al2(SO4)3(aq) + 3BaCl2(aq) → 3BaSO4(s) + 2AlCl3(aq)

The bold numbers is the coefficient use to balance the equation.

The number of atom on the reactant side is equal to the product side. Using aluminium atom as a case study the number of aluminium atom on the reactant side is 2 and the on the product  side it is also  2.

When aqueous aluminum sulfate and barium chloride are mixed, aluminum chloride forms as a product, while barium sulfate precipitates.

When aqueous solutions of aluminum sulfate and barium chloride are mixed, a double replacement reaction occurs. The aluminum sulfate dissociates into aluminum ions (Al³⁺) and sulfate ions (SO₄²⁻), while the barium chloride separates into barium ions (Ba²⁺) and chloride ions (Cl⁻).

The aluminum ions react with the chloride ions, forming aluminum chloride (AlCl₃) as a product. The barium ions react with the sulfate ions, resulting in the formation of solid barium sulfate (BaSO₄) as a precipitate. This reaction can be represented by the chemical equation: 3BaCl₂ + Al₂(SO₄) ₃ → 3BaSO₄ + 2AlCl₃.

In this reaction, the aluminum chloride remains in the aqueous solution, while the barium sulfate precipitates and can be separated from the solution.

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Molar mass of C_3H_6 = 36 g mol^-1 +6g mol^-1 =42g mol^-1

Empirical formula of Sample = (CH_3)

Explanation:

Mass sample having only C and H = 42g

Mass of C = 12g mol^-1

Molar mass of C = 12g mol^-1

Moles of C = 36 g / 12g mol^-1

= 3 mol

Mass of H in sample = 6g

Molar mass of H = 1 g mol^-1

Moles of H = 6g / 1g mol^-1

= 6 mol

Therefore in 42g sample 3 mol of C + 6 mol of H present, that is C_3H_6

Molar mass of C_3H_6 = 36 g mol^-1 +6g mol^-1 =42g mol^-1

Empirical formula of Sample = (CH_3)

Answer:

We can solve the question 'What is the empirical formula of the compound?' The answer is CH2

Explanation:

A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0g of H. Which of the following questions about the compound can be answered using the results of the analysis?

A) What was the volume of the sample?

B) What is the molar mass of the compound?

C) What is the chemical stability of the compound?

D) What is the empirical formula of the compound?

Step 1: Data given

Mass of the compound = 42.0 grams

⇒ 36.0 grams = carbon

⇒ 6.0 grams = hydrogen

Molar mass of carbon = 12.01 g/mol

Molar mass of hydrogen = 1.01 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles Carbon = 36.0 grams / 12.01 g/mol

Moles carbon = 3.00 moles

Moles hydrogen = 6.0 grams / 1.01 g/mol

Moles hydrogen = 5.95 moles

Step 3: Calculate mol ratio

We divide by the smallest number of moles

Carbon: 3.00 / 3.00 = 1

Hydrogen: 5.95 / 3.00 = 2

The empirical formula is CH2

Answer:

5

Explanation:

because im a bot :)