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3 years ago

What is the temperature of a gas that is expanded from 3.75 L at 37 degrees Celsius to 5.6 L?

Chemistry

1 answer:

deff fn [24]3 years ago

3 0

Answer:

190 °C

Step-by-step explanation:

The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:

V₁/T₁ = V₂/T₂ Invert both sides of the equation.

T₁/V₁ = T₂/V₂ Multiply each side by V₂

T₂ = T₁ × V₂/V₁

=====

V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K

V₂ = 5.6 L; T₂ = ?

=====

T₂ = 310.15 × 5.6/3.75

T₂ = 310.15 × 1.49

T₂ = 463 K

t₂ = 463 – 273.15

t₂ = 190 °C

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A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the

Viefleur [7K]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V$

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V$

The mole fraction of nitrogen in the mixture is:

$X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33$

Learn more: brainly.com/question/2060778

5 0

3 years ago

Organize the following list of atoms from lowest to highest Electronegativity:

Anon25 [30]

Answer:

Barium<Strontium<Calcium <Magnesium< Beryllium

Explanation:

Electronegativity is defined as the ability of an atom in a bond to attract the shared electrons of the bond towards itself.

Electronegativity is a periodic trend that decreases down the group and increases across the period.

Hence, if i want to arrange Beryllium, Barium, Strontium, Magnesium, Calcium in order of increasing electronegativity, i will have;

Barium<Strontium<Calcium <Magnesium< Beryllium

5 0

3 years ago

An actacide tablet containing Mg(OH)2 (MM = 58.3g / (mol)) is titrated with a 0.100 M solution of HNO3. The end point is determi

PtichkaEL [24]

Answer:

0.0583g

Explanation:

The equation of the reaction is;

2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)

From the question, number of moles of HNO3 reacted= concentration × volume

Concentration of HNO3= 0.100 M

Volume of HNO3 = 20.00mL

Number of moles of HNO3= 0.100 × 20/1000

Number of moles of HNO3 = 2×10^-3 moles

From the reaction equation;

2 moles of HNO3 reacts with 1 mole of Mg(OH)2

2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2

But

n= m/M

Where;

n= number of moles of Mg(OH)2

m= mass of Mg(OH)2

M= molar mass of Mg(OH)2

m= n×M

m= 1×10^-3 moles × 58.3 gmol-1

m = 0.0583g

6 0

3 years ago

The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions

mamaluj [8]

Answer:

a) 62.1 kJ/mol

b) 2.82 kJ/mol

c) 270.91 kJ/mol

d) -851.5 kJ/mol

Explanation:

The enthalpy change for a reaction in standard conditions (ΔH°rxn) can be calculated by:

ΔH°rxn = ∑n*ΔH°f, products - ∑n*ΔH°f, reagents

Where n is the number of moles in the stoichiometry reaction, and ΔH°f is the enthalpy of formation at standard conditions. ΔH°f = 0 for substances formed by only a single element. The values can be found in thermodynamics tables.

a) 2Ag₂O(s) → 4Ag(s) + O₂(g)

ΔH°f, Ag₂O(s) = -31.05 kJ/mol

ΔH°rxn = 0 - (2*(-31.05)) = 62.1 kJ/mol

b) SnO(s) + CO(g) → Sn(s) + CO₂(g)

ΔH°f,SnO(s) = -285.8 kJ/mol

ΔH°f,CO(g) = -110.53 kJ/mol

ΔH°f,CO₂(g) = -393.51 kJ/mol

ΔH°rxn = [-393.51] - [-110.53 - 285.8] = 2.82 kJ/mol

c) Cr₂O₃(s) + 3H₂(g) → 2Cr(s) + 3H₂O(l)

ΔH°f,Cr₂O₃(s) = -1128.4 kJ/mol

ΔH°f,H₂O(l) = -285.83 kJ/mol

ΔH°rxn = [3*(-285.83)] - [( -1128.4)] = 270.91 kJ/mol

d) 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s)

ΔH°f,Fe₂O₃(s) = -824.2 kJ/mol

ΔH°f,Al₂O₃(s) = -1675.7 kJ/mol

ΔH°rxn = [-1675.7] - [-824.2] = -851.5 kJ/mol

3 0

3 years ago

Atoms with different numbers of neutrons in their nucleus are:.

bonufazy [111]

Answer: They are referred to as isotopes, which are a form of an element that has different numbers of neutrons in their nucleus.

5 0

2 years ago