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Ostrovityanka [42]
3 years ago
9

What are the effects of convection currents in our surroundings?

Chemistry
1 answer:
miskamm [114]3 years ago
8 0
Land breeze sea breeze,ventilation,etc
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There are two common oxides of copper; one is 80% copper and the other is 89% copper by weight. Calculate the formula and name e
irakobra [83]

80% copper (Cu)

Cu: 80 :  63.546 = 1.259

O: 20 : 16 = 1.25

Cu:O = 1 : 1

the formula: CuO

89% copper (Cu)

Cu: 89 :  63.546 = 1.4

O: 11 : 16 = 0.6875

Cu:O = 2:1

the formula: Cu₂O

8 0
2 years ago
12.70 L of a gas has a pressure of 0.63 atm. What is the volume of the gas at 105kPa?
Zielflug [23.3K]

Answer: 7.693 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=0.63atm\\V_1=12.70L\\P_2=105kPa=1.04atm(1kPa=0.009atm)\\V_2=?

Putting values in above equation, we get:

0.63\times 12.70mL=1.04\times V_2\\\\V_2=7.693L

Thus new volume of the gas is 7.693 L

6 0
3 years ago
Read 2 more answers
Which of the following examples includes only chemical changes that occur in
Paul [167]
I don’t think you can get it for
6 0
3 years ago
Calculate the maximum amount of useful work that can be obtained and comment on the spontaneity for the reaction at 25C :
harkovskaia [24]

Answer:

1.41 *10^{3}  kJ/mol

Explanation:

First, we find in the tables the ΔH of formation of each compound. As you can see in the (image 1)

Then we solve the ecuation for ΔH°reaction

ΔH°reaction=∑ΔH°f(products)−∑ΔH°f(Reactants)

ΔH°reaction= (-2* 393.5 - 2*285.8) - (52.4 + 0) kJ/mol

ΔH°reaction = -1.41 *10^3  kJ/mol

3 0
3 years ago
State the definition of the partial molar Gibbs energy.
balu736 [363]

Explanation :

As we know that the Gibbs free energy is not only function of temperature and pressure but also amount of each substance in the system.

G=G(T,P,n_1,n_2)

where,

n_1\text{ and }n_2 is the amount of component 1 and 2 in the system.

Partial molar Gibbs free energy : The partial derivative of Gibbs free energy with respect to amount of component (i) of a mixture when other variable (T,P,n_j) are kept constant are known as partial molar Gibbs free energy of i^{th} component.

For a substance in a mixture, the chemical potential (\mu) is defined as the partial molar Gibbs free energy.

The expression will be:

\bar{G_i}=\mu_i=\frac{\partial G}{\partial n_i}_{(T,P,n_j)}

where,

T = temperature

P = pressure

n_i\text{ and }n_j is the amount of component 'i' and 'j' in the system.

4 0
3 years ago
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