A compound composed of only C and H contains 79.88 % C by mass. What is its empirical formula?

enot [183]

Answer:

C

Explanation:

A gas is a state of matter that has no fixed shape and no fixed volume gases have lower density than other states of matter such as solids and liquids.

7 0

3 years ago

NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te

makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y (1)

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol× $\frac{1mol}{80,043g}$ X

Y = 321J/g X (2)

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

X = 35,2g

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

6 0

3 years ago

A 2.00-g sample of a large biomolecule was dissolved in 15.0 g carbon tetrachloride. the boiling point of this solution was dete

Katarina [22]

We will use boiling point formula:

ΔT = i Kb m

when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35

and Kb is the boiling point constant =5.03

and m = molality

i = vant's Hoff factor

so by substitution, we can get the molality:

1.35 = 1 * 5.03 * m

∴ m = 0.27

when molality = moles / mass Kg

0.27 = moles / 0.015Kg

∴ moles = 0.00405 moles

∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol

5 0

4 years ago

A 15.0-L vessel contains 0.50 mol CH4 with a pressure of 1.0 atm. After 0.50 mol C2H6 is added to the vessel, what is the partia

Kazeer [188]

The partial pressure of methane in the mixture of methane and ethane has been 1 atm.

Partial pressure has been the pressure exerted by a gas in the solution or mixture. The partial pressure of each gas has been the total pressure of the gaseous mixture.

The partial pressure of the gas has been dependent on the volume, temperature, and concentration of the gas.

The given methane has a partial pressure of 1 atm in the 15 L vessel. The addition of ethane results in the change in the total pressure of the mixture, as there have been additional moles of solute that contributes to the solution pressure.

However, since there has been no change in the concentration and volume of methane, the pressure exerted by methane has been the same. Thus, the partial pressure of methane has been 1 atm.

For more information about the partial pressure, refer to the link:

brainly.com/question/14623719

7 0

3 years ago

Read 2 more answers

Consider the equilibrium reaction. 2 A + B − ⇀ ↽ − 4 C After multiplying the reaction by a factor of 2, what is the new equilibr

vredina [299]

Answer : The correct expression for equilibrium constant will be:

$K_c=\frac{[C]^8}{[A]^4[B]^2}$

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$4A+2B\rightleftharpoons 8C$

The expression of $K_c$ will be,

$K_c=\frac{[C]^8}{[A]^4[B]^2}$

Therefore, the correct expression for equilibrium constant will be, $K_c=\frac{[C]^8}{[A]^4[B]^2}$

4 0

3 years ago