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2 years ago

As a gas changes to a solid, energy A. is not used. B. is released. C. is absorbed.

Chemistry

1 answer:

pogonyaev2 years ago

7 0

Answer:

Explanation:

Energy is released, when a gas changes to solid

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How much heat is needed to change 5g of ice at 0°C to<br> water at 50°C?

natta225 [31]

Answer:

400cal

Explanation:

Since ice melted at 0 Celesius then the heat gained by ice (latent heat) will melt it so you should substitute in that law

Q=mlf ..where Q is the heat required to convert ice to water , m is the mass of ice and lf is the latent heat of fusion

Q=5∗80=400cal

6 0

2 years ago

Using the equation MgCl2 -> Mg + Cl2, if 5.00 grams of MgCl2 is given, how many grams of Mg is produced?

motikmotik

Answer:

MgCl2-> Mg+Cl

mass 5.00g Mass=1.263g

RFM Mg=24+35.5 ×2= 95 RFM= 24

moles= 5.00÷ 95= 0.0526 Moles=0.0526

Explanation:

mas of Mg= 1.263 grams

4 0

3 years ago

the pressure of 5.0 L of gas increases from 1.50atm to 1240 mmgh. what is the final volume of the gas, assuming no change in mol

NikAS [45]

Answer:

V₂ = 4.7 L

Explanation:

Given data:

Initial volume = 5.0 L

Initial pressure = 1.50 atm

Final pressure = 1240 mmHg (1240/760 = 1.6 atm)

Final volume = ?

Solution:

P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂

V₂ = 1.50 atm ×5.0 L/1.6 atm

V₂ = 7.5 atm. L /1.6 atm

V₂ = 4.7 L

6 0

2 years ago

What is the density of air at 75 F and latm pressure? Express in lbm/ft3 and kg/m3

ser-zykov [4K]

Answer : The density of air in $lbm/ft^3$ and $kg/m^3$ is, $0.0743lbm/ft^3$ and $1.19kg/m^3$ respectively.

Explanation :

$PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

where,

P = pressure of air = 1 atm

V = volume of air

T = temperature of air = 297 K

The conversion used for the temperature from Fahrenheit to degree Celsius is:

$^oC=(^oF-32)\times \frac{5}{9}$

$^oC=(75-32)\times \frac{5}{9}=24^oC$

The conversion used for the temperature from degree Celsius to Kelvin is:

$K=273+^oC$

$K=273+24=297K$

n = number of moles

m = mass of air

M = average molar mass of air = 28.97 g/mole

$\rho$ = density of air = ?

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

Now put all the given values in the above formula, we get:

$\rho=\frac{PM}{RT}$

$\rho=\frac{(1atm)\times (28.97g/mole)}{(0.0821L.atm/mol.K)\times (297K)}$

$\rho=1.19g/L$

Now we have to calculate density in $lbm/ft^3$ .

Conversion used :

$1g/L=0.0624lbm/ft^3$

So,

$1.19g/L=\frac{1.19g/L}{1g/L}\times 0.0624lbm/ft^3=0.0743lbm/ft^3$

The density of air in $lbm/ft^3$ is, $0.0743lbm/ft^3$

Now we have to calculate density in $kg/m^3$ .

Conversion used :

$1g/L=1kg/m^3$

So,

$1.19g/L=1.19kg/m^3$

The density of air in $kg/m^3$ is, $1.19kg/m^3$

8 0

2 years ago

Consider the burning of gasoline and the evaporation of gasoline. which process represents a chemical change and which represent

mart [117]

Burning gas is a chemical change and evaporation of gas is physical change.

5 0

2 years ago