Aluminum fluoride contains only aluminum and fluorine. what is the formula for this compound?

The stopcock connecting a 3.06 L bulb containing methane gas at a pressure of 9.61 atm, and a 6.65 L bulb containing oxygen gas

Contact [7]

Answer : The final pressure in the system is 4.22 atm.

Explanation :

First we have to calculate the moles of methane.

$PV=n_1RT$

where,

P = pressure of gas = 9.61 atm

V = volume of gas = 3.06 L

T = temperature of gas = T

$n_1$ = number of moles of methane gas = ?

R = gas constant

Now put all the given values in the ideal gas equation, we get:

$(9.61atm)\times (3.06L)=n_1\times RT$

$n_1=\frac{29.4}{RT}$

Now we have to calculate the moles of oxygen gas.

$PV=n_2RT$

where,

P = pressure of gas = 1.75 atm

V = volume of gas = 6.65 L

T = temperature of gas = T

$n_2$ = number of moles of oxygen gas = ?

R = gas constant

Now put all the given values in the ideal gas equation, we get:

$(1.75atm)\times (6.65L)=n_2\times RT$

$n_2=\frac{11.6}{RT}$

Now we have to determine the final pressure in the system after mixing the gases.

$P_{total}=(n_1+n_2)\times \frac{RT}{V_{total}}$

where,

$P_{total}$ = final pressure of gas = ?

$V_{total}$ = final volume of gas = (3.06 + 6.65)L = 9.71 L

T = temperature of gas = T

R = gas constant

Now put all the given values in the ideal gas equation, we get:

$P_{total}=(\frac{29.4}{RT}+\frac{11.6}{RT})\times \frac{RT}{9.71L}$

$P_{total}=4.22atm$

Therefore, the final pressure in the system is 4.22 atm.

4 0

3 years ago

1. According to the equation, what mass of hydrogen fluoride is necessary to produce 2.3 g of sodium fluoride?

Luba_88 [7]

Answer:

1.09 grams

Explanation:

According to the following chemical equation:

HF + NaNO₃ -> HNO₃ + NaF

1 mol of hydrogen fluoride (HF) produces 1 mol of sodium fluoride (NaF). Thus, we first convert from mol to grams by using the molar mass (MM) of each compound:

MM(HF)= (1 g/mol x 1 H) + (19 g/mol x 1 F) = 20 g/mol HF

1 mol HF x 19.9 g/mol HF = 20 g

MM(NaF) = (23 g/mol x 1 Na) + (19 g/mol x 1 F) = 42 g/mol NaF

1 mol NaF x 42 g/mol NaF = 42 g

Thus, from 20 g of HF are produced 42 g of NaF ⇒ 20 g HF/42 g NaF. We multiply this stoichiometric ratio by the mass of NaF produced to calculate the required mass of HF:

20 g HF/42 g NaF x 2.3 g NaF = 1.09 g HF

Therefore, 1.09 grams of HF are necessary to produce 2.3 g of NaF.

5 0

3 years ago

A firefighter applies a force of 2 N to lift the fire hose 3 meters up a ladder to put out a fire. How much work has the firefig

Paraphin [41]

Answer:

6 J

Explanation: Work done(J) = Force(N) × Distance(m) along the direction of force.

Given, Force=2 N, Distance=3 meters

∴ Wok done (J) = 2 × 3 = 6 J.

5 0