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2 years ago

Assuming it behaves as an ideal gas, calculate the density of sulfur dioxide, so2, at stp.

2 answers:

6 0

Hello!

At Standard Pressure and Temperature, an ideal gas has a molar density of 0,04464 mol/L.

So, we need to apply a simple conversion factor to calculate the density of Sulfur Dioxide using the molar mass of Sulfur Dioxide.

$\frac{0,04464 mol SO_2}{1 L SO_2}* \frac{64,066 g SO_2}{1 mol SO_2}=2,8599 g/L$

So, the Density of Sulfur Dioxide (SO₂) at STP is 2,8599 g/L

Have a nice day!

kolbaska11 [484]2 years ago

4 0

Answer:

2.86 g/L

Explanation:

Density can be defined as the ratio of mass to volume, that is, Mass/volume.

The molar mass of sulfur dioxide, SO2 can be calculated and it is given as: 32.07 + (16×2) = 64.07 grams per mole.

At standard temperature and pressure(step), one mole of an ideal gas is equal to the volume 22.4 litres. Hence, the gas density can be calculated using the formula below;

Density of sulfur dioxide,SO2 at stp= [molar mass(in grams per mole)÷ 1 mole] × [ 1 mol÷ 22.4 litres] ----------------------------------------------(***).

substitute in the values into the equation above, we have;

==> (64.07 grams per mol/ 1 mol) × 1 mol/22.4 litres.

===> 2.8603 grams per litre.

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BlackZzzverrR [31]

Answer:

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Explanation:

5 0

2 years ago

A solution of sugar contains 35 gramsof sucrose, C12H22O11in 100 mL of water. What is the percent composition of the solution?

KATRIN_1 [288]

Answer:

Percent composition of the solution is 26 % of sucrose and 74 % of water

Explanation:

Percent composition is the mass of solute, either of solvent in 100 g of solution.

Mass of solution = Mass of solvent + Mass of solute

Mass of solute = 35 g

Mass of solvent = 100 g

As we know, water density = 1g/mL

So 1g/mL . 100 mL = 100 g

35 g + 100 g = 135 g → Mass of solution

(Mass of solute / Mass of solution) . 100 =

(35 g / 135 g) . 100 = 26 %

(Mass of solvent / Mass of solution) . 100 =

(100 g / 135 g) . 100 = 74 %

8 0

2 years ago

The following reactions have the indicated equilibrium constants at a particular temperature: N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 ×

Anuta_ua [19.1K]

Answer:

$Kc=~1.49x10^3^4}$

Explanation:

We have the reactions:

A: $N_2_(_g_) + O_2_(_g_) 2NO_(_g_)~~~~~~Kc = 4.3x10^-^2^5$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~~~Kc = 6.4x10^9$

Our target reaction is:

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)$

We have $NO_(_g_)$ as a reactive in the target reaction and $NO_(_g_)$ is present in A reaction but in the products side. So we have to flip reaction A.

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

Then if we add reactions A and B we can obtain the target reaction, so:

A: $2NO_(_g_) N_2_(_g_) + O_2_(_g_) ~Kc =\frac{1}{4.3x10^-^2^5}$

B: $2NO_(_g_)+~O_2_(_g_)~2NO_2_(_g_)~Kc=6.4x10^9$

For the final Kc value, we have to keep in mind that when we have to add chemical reactions the total Kc value would be the multiplication of the Kc values in the previous reactions.

$4NO_(_g_) N_2_(_g_) + 2NO_2_(_g_)~~~Kc=\frac{6.4x10^9}{4.3x10^-^2^5}$

$Kc=~1.49x10^+^3^4}$

3 0

2 years ago

A scientist measures the speed of sound in a monatomic gas to be 449 m/s at 20∘C. What is the molar mass of this gas?

Ivan

Answer:

The molar mass of the gas is 36.25 g/mol.

Explanation:

To solve this problem, we can use the mathematical relation:

ν = $\sqrt{3RT/M}$

Where, ν is the speed of light in a gas (ν = 449 m/s),

R is the universal gas constant (R = 8.314 J/mol.K),

T is the temperature of the gas in Kelvin (T = 20 °C + 273 = 293 K),

M is the molar mass of the gas in (Kg/mol).

ν = $\sqrt{3RT/M}$

(449 m/s) = √ (3(8.314 J/mol.K) (293 K) / M,

by squaring the two sides:

(449 m/s)² = (3 (8.314 J/mol.K) (293 K)) / M,

∴ M = (3 (8.314 J/mol.K) (293 K) / (449 m/s)² = 7308.006 / 201601 = 0.03625 Kg/mol.

∴ The molar mass of the gas is 36.25 g/mol.

7 0

3 years ago

Enter an equation for the formation of NO2(g) from its elements in their standard states. Express your answer as a chemical equa

kkurt [141]

The standard state of the elements Nitrogen and Oxygen are N2 and O2, knowing that they are diatomic elements. With that piece of information, the unbalanced equation for the formulation of NO2(g) should be as follows -

N2 + O2 ---> NO2

And if you include their states -

N2 ( g ) + O2 ( g ) ---> NO2 ( g )

To balance this chemical equation consider the number of reactants and products on other side of the equation. If you were to include a coefficient of one - half with respect to N2 on the reactant side, it would balance the reactants and products -

7 0

3 years ago