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A 500.0 g block of dry ice (solid CO2, molar mass = 44.0 g) vaporizes at room temperature. Calculate the volume of gas produced

Damm [24]

Considering the ideal gas law, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.

<h3>Definition of ideal gas</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

<h3>Ideal gas law</h3>

An ideal gas is characterized by absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:

P×V = n×R×T

<h3>Volume of gas</h3>

In this case, you know:

P= 1.50 atm
V= ?
n= 500 g× $\frac{1 mole}{44 g}$ = 11.36 moles, being 44 $\frac{g}{mole}$ the molar mass of CO₂
R= 0.082 $\frac{atmL}{molK}$
T= 25 C= 298 K (being 0 C=273 K)

Replacing in the ideal gas law:

1.50 atm×V = 11.36 moles×0.082 $\frac{atmL}{molK}$ × 298 K

Solving:

V= (11.36 moles×0.082 $\frac{atmL}{molK}$ × 298 K) ÷ 1.50 atm

Finally, the volume of gas produced at 25.0 °C and 1.50 atm is 184.899 L.

Learn more about the ideal gas law:

<u>brainly.com/question/4147359?referrer=searchResults</u>

4 0

2 years ago

Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 h

Genrish500 [490]

Explanation:

The given data is:

The half-life of gentamicin is 1.5 hrs.

The reaction follows first-order kinetics.

The initial concentration of the reactants is 8.4 x 10-5 M.

The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

$k=\frac{0.693}{t_1_/_2}$

Where k = rate constant

t1/2=half-life

So, the rate constant k value is:

$k=\frac{0.693}{1.5 hrs}$

The expression for the rate constant is :

$k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}$

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

$\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6$

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

5 0

3 years ago

24) What is momentum? *

sergiy2304 [10]

Answer:

Momentum is the measure of the motion of an object found by multiplying the objects mass and velocity.

Symbol: p

Units: kg x m/s

Explanation:

......

6 0

2 years ago

A 2.5 L flask is filled with 0.25 atm SO3, 0.20 atm SO2, and 0.40 atm O2, and allowed to reach equilibrium. Assume at the temper

Anit [1.1K]

Explanation:

Reaction equation for the given chemical reaction is as follows.

$2SO_{3} \rightleftharpoons 2SO_{2} + O_{2}$

Equation for reaction quotient is as follows.

Q = $\frac{P^{2}_{SO_{2}} \times P_{O_{2}}}{P^{2}_{SO_{3}}}$

= $\frac{(0.20)^{2} \times 0.40}{(0.25)^{2}}$

= 0.256

As, Q > K (= 0.12)

The effect on the partial pressure of $SO_{3}$ as equilibrium is achieved by using Q, is as follows.

This means that there are too much products.

Equilibrium will shift to the left towards reactants.

More $SO_{3}$ is formed.

Partial pressure of $SO_{3}$ increases.

4 0

3 years ago

what is the mass of water vapor produced when 3.2 liters reacts with 8.7 liters of oxygen gas at STP?

uranmaximum [27]

Answer:

2.57g

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

2H2 + O2 —> 2H2O

Next let us determine the limiting reactant. This is achieved as follows:

From the equation,

2L H2 required 1L of O2.

Therefore, 3.2L of H will require = 3.2/2 = 1.6L of O2

From the calculation above, O2 is excess because the volume of O2 given from the question is far greater than the volume of O2 obtained from our calculation. Therefore, H2 is the limiting reactant.

Now let us covert 3.2L of H2 to mole. This is illustrated below:

1mole of a gas occupy 22.4L at stp

Therefore, Xmol of H2 will occupy 3.2L i.e

Xmol of H2 = 3.2/22.4 = 0.143mol

From the equation,

2moles of H2 produced 2moles of H2O.

Therefore, 0.143mol of H2 will also produce 0.143moles of H2O.

Now, we can obtain the mass of the water vapour produced by convert 0.143mol of H2O to gram. This is illustrated below:

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Number of mole of H2O = 0.143mol

Mass of H2O =?

Mass = mole x Molar Mass

Mass of H2O = 0.143 x 18 = 2.57g

The mass of water vapour produce is 2.57g

8 0

2 years ago