Translate the following chemical equation into a sentence. CH4+2O2 CO2+2H2O

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

3 years ago

How do solids and liquids compare with each other?

Marizza181 [45]

I do believe it’s C if I’m wrong Myb fam

5 0

3 years ago

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41 to

tia_tia [17]

Answer:

d = 0.9 g/L

Explanation:

Given data:

Number of moles = 1 mol

Volume = 24.2 L

Temperature = 298 K

Pressure = 101.3 Kpa (101.3/101 = 1 atm)

Density of sample = ?

Solution:

PV = nRT (1)

n = number of moles

number of moles = mass/molar mass

n = m/M

Now we will put the n= m/M in equation 1.

PV = m/M RT (2)

d = m/v

PM = m/v RT ( by rearranging the equation 2)

PM = dRT

d = PM/RT

The molar mass of neon is = 20.1798 g/mol

d = 1 atm × 20.1798 g/mol / 0.0821 atm. L/mol.K × 273K

d = 20.1798 g/22.413 L

d = 0.9 g/L

4 0

3 years ago

Use the table to answer the question.

Alex17521 [72]

Answer:

High levels of moisture are contained within an air mass.

Tell me if I'm correct, plz!

Explanation:

6 0

3 years ago

Read 2 more answers

Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h

bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

We know that,

The relation between the $C_p\text{ and }C_v$ for an ideal gas are :

$C_p-C_v=R$

As we are given :

$C_p=28.253J/K.mole$

$28.253J/K.mole-C_v=8.314J/K.mole$

$C_v=19.939J/K.mole$

Now we have to calculate the entropy change of the gas.

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

$\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J$

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. $(w=-pdV)$

(C) Heat during the process will be,

$q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J$

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

7 0

3 years ago