What does D represent?

A sample of argon gas has a volume of 795 mL at a pres-sure of 1.20 atm and a temperature of 116 ∘C. What is the final volume of

jek_recluse [69]

<u>Answer:</u> The volume when the pressure and temperature has changed is $1.6\times 10^2mL$

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

Let us assume:

$P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K$

Putting values in above equation, we get:

$\frac{1.20atm\times 795mL}{389K}=\frac{0.55atm\times V_2}{348K}\\\\V_2=\frac{1.20\times 795\times 348}{0.55\times 389}=1.6\times 10^3mL$

Hence, the volume when the pressure and temperature has changed is $1.6\times 10^2mL$

5 0

3 years ago

When the temperature goes up, the volume will also go up

Tju [1.3M]

Answer:

Yes it will

An example is ice, when it melts the volume goes up which means it occupies much more space

4 0

3 years ago

if 100. mL of 0.800 M Na2SO4 is added to 200. mL of 1.20 M NaCl, what is the concentration of Na+ ions in the final solution? As

Black_prince [1.1K]

Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻

1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions

the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
= 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol

NaCl ----> Na⁺ + Cl⁻
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
= 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol

total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive,
therefore total volume = 100 ml + 200 ml = 300 ml
the concentration of Na⁺ ions = number of moles / volume
= 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³

5 0

3 years ago

What is the only nonmetal with four valence electrons.

Alja [10]

Carbon is a non metal and also has four valence electrons.

3 0

2 years ago

You want to determine ΔH o for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) To do so, you first determine the heat capacity

Assoli18 [71]

Answer:

(A) The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

(B) ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

Explanation:

Solution

Calculate the heat actually evolved.

q = mcΔt

Finding the mass of the reactants in grams we have.

Use density. (50 mL + 50 mL ) = 100 mL of solution.

100 mL X 1.04g/mL = 104 grams of solution. (mass = Volume X Density)

Find the temperature change.

Δt =tfinal - tinitial = 30.4°C – 16.9°C = 13.5°C

q = mcΔt

= 104grams × 3.93J/g°C × 13.5°C = 5.51772×103J

= 5.51772 × 103 J

This is the heat lost in the reaction between HCl and NaOH, therefore q = -5.52 × 103 J.

this is an exothermic heat producing reaction.

To calculate the total heat of the reaction or heat per mole we have

50.0 mL of HCl X 2.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

The same quantity of base, 0.100 mole NaOH, was used.

The energy per unit mole is given by

i.e. molar enthalpy = J/mol = -5.52 × 103J / 0.100 mol

= -5.52 × 104 J/mol

= -55177.2 J/mol

= -55.177 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -55.177 kJ/mol

Heat absorbed by the calorimeter = −57.32kJ − 55.177 kJ = −2.1428KJ

The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

(B) For the ZnCl we have

Calculate the heat actually evolved.

q = mcΔt

Finding the mass of the reactants in grams we have.

Use density. 100 mL of solution of HCl

100 mL X 1.015g/mL = 101.5 grams of solution. (mass = Volume X Density)

Find the temperature change.

Δt =tfinal - tinitial = 20.5°C – 16.8°C = 3.7 °C

q = mcΔt

= 101.5grams × 3.95J/g°C × 3.7°C = 1483.422×103J

= -1483.422×103J

This is the heat lost in the reaction between HCl and NaOH, therefore q = -1.483 × 103 J.

this is an exothermic heat producing reaction.

To calculate the total heat of the reaction or heat per mole we have

100.0 mL of HCl X 1.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

The energy per unit mole is given by

i.e. molar enthalpy = J/mol = -1.483 × 103J / 0.100 mol

= -1.483 × 104 J/mol

= -14834.22 J/mol

= -14.834 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -14.834 kJ/mol

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

= -14.834 kJ –(0.1587KJ/°C×3.7°C) = -15.42KJ

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

5 0

3 years ago