A balloon is filled with 3.50 L of water at 24.0°C. What is the volume of the water at 307 K?

Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a

Luden [163]

Answer : The mass of $NaN_3$ at temperature $20^oC$ 28.47 g.

The mass of $NaN_3$ at temperature $10^oC$ 29.51 g.

Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = $20^oC=273+20=293K$ $(0^oC=273K)$

Volume of gas = $25\times 25\times 20cm=12500cm^3=12.5L$ $(1L=1000cm^3)$

Molar mass of $NaN_3$ = 65 g/mole

Part 1 : First we have to calculate the moles of gas at temperature $20^oC$ . The gas produced in the given reaction is $N_2$ .

Using ideal gas equation,

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)$

By rearranging the terms, we get the value of 'n'

$n=0.7015moles$

The moles of $N_2$ = 0.7015 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.7015 moles of $N_2$ produced from $\frac{20}{32}\times 0.7015=0.438$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g$

Therefore, the mass of $NaN_3$ needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (283K)$

By rearranging the terms, we get the value of 'n'

$n=0.726moles$

The moles of $N_2$ = 0.726 moles

The given balanced reaction is,

$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.726 moles of $N_2$ produced from $\frac{20}{32}\times 0.726=0.454$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.454moles)\times (65g/mole)=29.51g$

Therefore, the mass of $NaN_3$ needed are 29.51 g.

7 0

3 years ago

Please help.

Fed [463]

<h2>Answer: O: 1s22s22p4</h2>

Explanation: In writing the electron configuration for oxygen the first two electrons will go in the 1s orbital. Since 1s can only hold two electrons the next 2 electrons for O go in the 2s orbital. The remaining four electrons will go in the 2p orbital. Therefore the O electron configuration will be 1s22s22p4.

8 0

3 years ago

Phosphorus pentachloride decomposes according to the chemical equation PCl 5 ( g ) − ⇀ ↽ − PCl 3 ( g ) + Cl 2 ( g ) K c = 1.80 a

mezya [45]

Answer:

See explanation for answer

Explanation:

First, let's write the equation again:

PCl5 <---------> PCl3 + Cl2 Kc = 1.8

The expression for equilibrium is the following:

Kc = [PCl3] [Cl2] / [PCl5]

We only know the data for PCl5 at the beggining, and to solve this, we need to write what happens before and after the reaction.

At first, we only have the 0.342 moles of PCl5 and nothing of the products. This is logic, because there is no reaction yet.

Concentration of PCl5 at this point (With a volume of 3.65 L) is:

[PCl5] = 0.342 / 3.65 = 0.094 M

Keep in mind, that this is not the concentration in equlibrium. This is only the concentration at the beggining. To know it's concentration in equilibrium, we'll do the following:

PCl5 <---------> PCl3 + Cl2 Kc = 1.8

i) 0.094 0 0

eq) 0.094 - x x x

Now, we will replace this values in the equilibrium equation:

1.8 = x * x / 0.094 - x

Solving for x we have:

1.8(0.094 - x) = x²

0.1692 - 1.8x = x²

x² + 1.8x - 0.1692 = 0

Using the general equation for x we have:

x = -1.8 ± √(1.8)² - 4 * 1 * (-0.1692) / 2

x = -1.8 ± √3.9168 / 2

x = -1.8 ± 1.98 / 2

x1 = -1.8 + 1.98 / 2 = 0.09

x2 = -1.8 - 1.98 / 2 = -1.89

This means that the value of x is 0.09 therefore, the concentrations in equilibrium for each species is:

[PCl5] = 0.094 - 0.09 = 0.04 M

[PCl3] = [Cl2] = 0.09 M

7 0

4 years ago

2H2 + O2 --> 2H2O<br> How many grams of oxygen gas are required to produce 90. g of water?

Travka [436]

Answer:

80g

Explanation:

2H2 + O2 —> 2H2O

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

Mass conc. of H2O from the balanced equation = 2 x 18 = 36g

MM of O2 = 16 x 2 = 32g/mol

From the equation,

32g of O2 reacted to produce 36g of H2O.

Therefore Xg of O2 will react to produce 90g of H2O i.e

Xg of O2 = (32x90)/36 = 80g

8 0

3 years ago

Wind patterns are impacted by which of the following?

mrs_skeptik [129]

Answer:

last one is your answer.....

8 0

3 years ago