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2 years ago

9.0 mol Na2S can from 9.0 mol CuS and 8.0 mol CuSO4 can form 8.0 mol Cus.

Chemistry

1 answer:

ICE Princess25 [194]2 years ago

4 0

Answer:

765.0 grams CuS

Explanation:

The limiting reagent is the reactant which completely reacts before the other reactant(s) is used up. When 9.0 moles Na₂S and 8.0 moles CuSO₄ react, it appears that CuSO₄ is the limiting reagent. You can tell because it results in the production of less product.

You can determine the mass of CuS by multiplying the moles by the molar mass. It is important to arrange the ratio in a way that allows for the cancellation of units.

Molar Mass (CuS): 95.62 g/mol

8.0 moles CuS 95.62 g
------------------------- x ----------------------- = 765.0 grams CuS
1 mole

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4 years ago

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what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C

svetoff [14.1K]

Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

$\frac{V}{T}=k$

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

$\frac{V1}{T1}=\frac{V2}{T2}$

In this case:

V1= 0.730 L
T1= 28 °C= 301 °K (0°C= 273°K)
V2= ?
T2= 0°C= 273 °K

Replacing:

$\frac{0.730 L}{301K}=\frac{V2}{273K}$

Solving:

$V2=273K*\frac{0.730L}{301K}$

V2=0.662 L

The volume of the air is 0.662 L

6 0

3 years ago

When we have stomach disorders, we take antacids to solve the problem? how does it help? support your answer with an equation.

Sedbober [7]

Answer:

Antacid Neutralises the acid in the stomach.

Explanation:

Antacid contains ingredients such as aluminum, calcium, magnesium and sodium bicarbonate which act as bases. These help neutralise the pH levels in the stomach and makes the contents of the stomach less corrosive.

4 0

3 years ago

A concentration cell consisting of two hydrogen electrodes (PH2 = 1 atm), where the cathode is a standard hydrogen electrode and

Lunna [17]

The pH of the unknown solution is 3.07.

Explanation:

1.Find the cell potential as a function of pH

From the Nernst Equation:

Ecell=E∘cell−RT /zF × lnQ

where

R denotes the Universal Gas Constant

T denotes the temperature

z denotes the moles of electrons transferred per mole of hydrogen

F denotes the Faraday constant

Q denotes the reaction quotient

Substitute the values,

E∘cell=0 lnQ=2.303logQ

E0cell=−2.30/RT /zF × log Q

Solving the equation,

2. Find the Q value

Q=[H+]2prod pH₂, product/ [H+]2reactpH₂, reactant

Q=[H+]^2×1/1×1=[H+]2

Taking the log

logQ= log[H+]^2=2log[H+]=-2pH

From the formula,

Ecell=−2.303RT /zF× logQ

E cell= 2.303 × 8.314 CK mol (inverse) × 298.15

K × 2pH /2×96 485 C⋅mol

( inverse)

E cell= 0.0592 V × pH

3. Finding the pH value

E cell= 0.0592 V × pH

pH = E cell/ 0.0592 V= 0.182V/ 0.0592V

pH=3.07

The pH of the unknown solution is 3.07.

7 0

3 years ago

An alkanol contains 37.5 carbon

kow [346]

Answer:

Empirical formula is CH₄

Molecular formula = C₂H₈

Explanation:

Mass of carbon = 37.5 g

Mass of hydrogen = 12.5 g

Molecular weight = 32 g/mol

Molecular formula = ?

Empirical formula = ?

Solution:

Number of gram atoms of C = 37.5 g /12g/mol = 3.125

Number of gram atoms of H = 12.5 g / 1.008 g/mol= 12.4

Atomic ratio:

C : H

3.125/3.125 : 12.4 /3.125

1 : 4

C : H : = 1 : 4

Empirical formula is CH₄

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 32 / 16

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 ( CH₄)

Molecular formula = C₂H₈

3 0

3 years ago