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3 years ago

What is the volume of 33.25g of butane gas at 293 C and 10.934 kPa?

Chemistry

1 answer:

PSYCHO15rus [73]3 years ago

6 0

Answer:

V = 240.79 L

Explanation:

Given data:

Volume of butane = ?

Temperature = 293°C

Pressure = 10.934 Kpa

Mass of butane = 33.25 g

Solution:

Number of moles of butane:

Number of moles = mass/ molar mass

Number of moles = 33.25 g/ 58.12 g/mol

Number of mole s= 0.57 mol

Now we will convert the temperature and pressure units.

293 +273 = 566 K

Pressure = 10.934/101 = 0.11 atm

Volume of butane:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

V = nRT/P

V = 0.57 mol × 0.0821 atm.L/ mol.K ×566 K / 0.11 atm

V = 26.49 L/0.11

V = 240.79 L

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What is the mass of 3.6 x 1024 atoms of Zinc (Zn)?

AlladinOne [14]

Answer is: mass od zinc is 392,28 g.
N(Zn) = 3,6·10²⁴.
n(Zn) = N(Zn) ÷ Na.
n(Zn) = 3,6·10²⁴ ÷ 6·10²³ 1/mol.
n(Zn) = 6 mol.
m(Zn) = n(Zn) · M(Zn).
m(Zn) = 6 mol · 65,38 g/mol.
m(Zn) = 392,28 g.
Na - Avogadro number.
n - amount of substance.
M - molar mass.

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3 years ago

What is the molarity of a solution that contains 87.75g of NaCI in 500.ml of solution

Brut [27]

Answer:

M = 3.0 mol/L.

Explanation:

We can calculate the molarity of a solution using the relation:

M = (mass x 1000) / (molar mass x V)

M is the molarity "number of moles of solute per 1.0 L of the solution.

mass is the mass of the solute (g) (m = 87.75 g of NaCl).

molar mass of NaCl = 58.44 g/mol.

V is the volume of the solution (ml) (V = 500.0 ml).

∴ M = (mass x 1000) / (molar mass x V) = (87.75 g x 1000) / (58.44 g/mol x 500.0 ml) = 3.0 mol/L.

6 0

3 years ago

The substances below are listed by increasing specific heat capacity value. Starting at 30 Celsius, they absorb 100 kJ of therma

Georgia [21]

Answer:

Silver.

Explanation:

To obtain the right answer to question, let us calculate the change in temperature for each substance assuming they all have the same mass as 100g.

This is illustrated below:

1. For Siver:

Mass (M) = 100g

Specific heat capacity (C) = 0.239J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 0.239)

ΔT = 4184°C

2. For Aluminium:

Mass (M) = 100g

Specific heat capacity (C) = 0.921J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 0.921)

ΔT = 1086°C

3. For Lithium:

Mass (M) = 100g

Specific heat capacity (C) = 3.56J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 3.56 )

ΔT = 281°C

4. For water:

Mass (M) = 100g

Specific heat capacity (C) = 4.184J/g°C

Heat (Q) = 100 kJ = 100000J

Change in temperature (ΔT)

Q = MCΔT

ΔT = Q/MC

ΔT = 100000/(100 x 4.184)

ΔT = 239°C

Summary

Temperature change of each substance is given below

1. Silver => 4184°C

2. Aluminum => 1086°C

3. Lithium => 281°C

4. Water => 239°C

From the calculations made above, Silver has the highest rise in temperature.

4 0

3 years ago

Consider the reaction N2(g) + 2O2(g)2NO2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surr

zysi [14]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$N_2+2O_2\rightarrow 2NO_2$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(NO_2(g))})]-[(1\times \Delta S^o_{(N_2(g))})+(2\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(NO_2(g))}=240.06J/K.mol\\\Delta S^o_{(O_2)}=205.14J/K.mol\\\Delta S^o_{(N_2)}=191.61J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (240.06))]-[(1\times (191.61))+(2\times (205.14))]\\\\\Delta S^o_{rxn}=-121.77J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(-121.77) J/K = 121.77 J/K

We are given:

Moles of nitrogen gas reacted = 1.90 moles

By Stoichiometry of the reaction:

When 1 mole of nitrogen gas is reacted, the entropy change of the surrounding will be 121.77 J/K

So, when 1.90 moles of nitrogen gas is reacted, the entropy change of the surrounding will be = $\frac{121.77}{1}\times 1.90=231.36 J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of nitrogen gas is reacted is 231.36 J/K

7 0

3 years ago

Which description refers to cirrus clouds?

Vesna [10]

Answer:

Explanation:

8 0

3 years ago

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