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3 years ago

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Chemistry

1 answer:

Licemer1 [7]3 years ago

3 0

Answer:

The answer to your question is:

Vol of NO2 = 11.19 L

Vol of O2 = 2.8 L

Explanation:

Data

N2O5 = 56 g

STP T = 0°C = 273°K

P = 1 atm

MW N2O5 = 216 g

Gases law = PV = nRT

Process

216 g of N2O5 ---------------- 1 mol

54 g ----------------- x

x = (54 x 1) / 216

x = 0.25 mol of N2O5

2 mol of N2O5 ----------------- 4 mol of NO2

0.25 mol ------------------ x

x = (0.25 x 4) / 2 = 0.5 mol of NO2

V = nRT/P

V = (0.5)(0.082)(273) / 1 = 11.19 L

2 mol of N2O5 ----------------- 1 O2

0.25 N2O5 ---------------------- x

x = (0.25 x 1) / 2 = 0.125 mol

Vol = (0.125)((0.082)(273) / 1 = 2.8 L

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A gas under a pressure of 74 mmHg and at a temperature of 75°C occupies a 500.0-L container. How many moles of gas are in the co

Leona [35]

Answer : The number of moles of gas present in container are 1.697 mole.

Explanation :

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of gas = 74 mmHg = 0.097 atm

conversion used : (1 atm = 760 mmHg)

V = volume of gas = 500.0 L

T = temperature of gas = $75^oC=75+273=348K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get the number of moles of gas in the container.

$(0.097atm)\times (500.0L)=n\times (0.0821L.atm/mole.K)\times (348K)$

$n=1.697mole$

Therefore, the number of moles of gas present in container are 1.697 mole.

4 0

3 years ago

How would you prepare 500 ml of the following solutions : Sodium succinate buffer (0.1 mol/dm3 pH 5.64)

Nana76 [90]

Answer:

8.10g of sodium succinate must be added and 247mL of 0.1M HCl adding enough water until make 500mL

Explanation:

Succinic acid has a pKa₂ of 5.63

To solve this question we must find the amount of sodium succinate and 0.1M HCl that we have to add using H-H equation:

pH = pKa + log [A-] / [HA]

5.64 = 5.63 + log [Na₂Succ.] / [HSucc⁺]

0.01 = log [Na₂Succ.] / [HSucc⁺]

1.0233 = [Na₂Succ.] / [HSucc⁺] (1)

As:

0.1M = [Na₂Succ.] + [HSucc⁺] (2)

Replacing (2) in (1):

1.0233 = 0.1M - [HSucc⁺] / [HSucc⁺]

1.0233[HSucc⁺] = 0.1M - [HSucc⁺]

2.033[HSucc⁺] = 0.1M

[HSucc⁺] = 0.0494M

[Na₂Succ] = 0.0506M

Both [Na₂Succ⁺] and [HSucc⁺] ions comes from the same sodium succinate we have to find the moles of sodium succinate in 500mL of 0.1M. Then, based on the reaction:

Na₂Succ + HCl → HSucc⁺ + Cl⁻

The moles of HCl added = Moles HSucc⁺ we need:

Moles Na₂Succ:

0.500L * (0.1mol/L) = 0.0500 moles

Mass -Molar mass sodium succinate: 162.05g/mol-:

0.0500mol * (162.05g/mol) = 8.10g of sodium succinate must be added

Moles HCl:

0.0494M * 0.500L = 0.0247 moles HCl * (1L / 0.1mol) = 0.247L =

And 247mL of 0.1M HCl adding enough water until make 500mL

7 0

3 years ago

Find the volume of 56.0 grams of O2 at stp

BaLLatris [955]

The molar mass of O2 is 32g/mol. So the mol amount of these O2 is 56/32=2 mol. STP stands for the standard temperature and pressure which means the temperature is 0 ℃ and pressure is 100 kPa. And the molar volume of gas is 22.7 L/mol under STP. So the answer is 22.7*2=45.4 L

6 0

4 years ago

A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

Answer: The mass of glucose in final solution is 1.085 grams

Explanation:

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)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

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:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ 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:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

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

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

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

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

Sedimentary rocks that are made up of large pebbles and stones have a coarse-granied?

umka2103 [35]

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3 0

3 years ago