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Karo-lina-s [1.5K]
3 years ago
7

Can someone help me with this?

Chemistry
1 answer:
Rudik [331]3 years ago
4 0
Yeah so you have to start of with converting your first two values into moles (forget the third one)

97.5 g NO * 1 mol/30.01 g NO = 3.25 moles NO

88.0 g O2 * 1 mol/16.00 g O2 = 5.5 moles O2

now we can find the limiting reactant by checking for the amount of product each reactant should give us by using molar ratios

3.25 mol NO * 2 mol NO2/2 mol NO = 3.25 mol NO2

5.5 mol O2 * 2 mol NO2/ 1 mol O2 = 11

so NO is the limiting reactant since it produces less product/gets used up quicker

3.25 mol NO * 2 mol NO2/2molNO = 3.25 mol NO2

so this is our theoretical yield and the question provides us with the actual yield (2.68 moles). since the actual yield is given in moles, we don't have to convert to grams. our percent yield formula goes like: actual yield/theoretical yield * 100

2.68 mol/3.25 mol * 100 = 82.46%
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Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}      ......(1)

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Putting values in equation 1, we get:

\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M

To calculate the molarity of the diluted solution, we use the equation:

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M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

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0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M

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Molarity of glucose solution = 0.0602 M

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Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g

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Answer:

kp= 3.1 x 10^(-2)

Explanation:

To solve this problem we have to write down the reaction and use the ICE table for pressures:

                                2SO2      +        O2         ⇄              2SO3

Initial                      3.4 atm           1.3 atm                         0 atm

Change                    -2x                    - x                                + 2x

Equilibrium            3.4 - 2x            1.3 -x                          0.52 atm

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2x = 0.52

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                               2SO2             +          O2              ⇄              2SO3

Equilibrium        3.4 - 0.52                1.3 - 0.26                     0.52 atm

Equilibrium        2.88 atm                 1.04 atm                      0.52 atm

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