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grandymaker [24]
3 years ago
10

Based on the chemical equation, use the drop-down menu to choose the coefficients that will balance the chemical equation:

Chemistry
1 answer:
Leni [432]3 years ago
5 0

Answer: (1)CaSO4 -> (2)O2 + (1)CaS

Explanation: edge 2020 chem

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If a gas displays a solubility of 0.00290M at a partial pressure of 125 kPa, what is the proportionality constant for this gas i
alexgriva [62]

Answer:

The proportionality constant ( Henry’s constant) = 2.32 * 10^-5 M/kPa

Explanation:

Here in this question, we are concerned with calculating the proportionality constant for this gas.

Mathematically, we can get this from Henry law

From Henry law;

Concentration = Henry constant * partial pressure

Thus Henry constant = concentration/partial pressure

Henry constant = 0.00290 M/125 kPa = 2.32 * 10^-5 M/kPa

5 0
3 years ago
. If sulphur (IV) oxide and methane are
zubka84 [21]

Answer:

The answer is C. 1:2

Explanation:

5 0
3 years ago
What type of reaction is the following chemical reaction? SeO2 + 3Cl2 -> SeCl6 + O2​
dem82 [27]

Answer:

This is a synthesis reaction.

Explanation:

This is because nothing is being swapped out, reactants are being combined and then form a product.

4 0
3 years ago
The half-life of the first-order decay of radioac- . 14C . b l:Ive 1s a out 5720 years. Calculate the rate constant for the reac
Alinara [238K]

Answer:

a. Rate constant: 1.2118x10⁻⁴ yrs⁻¹

b. The age of the object is 20750 years

Explanation:

a. We can solve the rate constant in an isotope decay by using Half-Life, as follows:

K = Ln 2 / Half-life

K = ln 2 / 5720 years =

<h3>1.2118x10⁻⁴ yrs⁻¹</h3><h3 />

b. The general equation of isotope decay is:

Ln [A] = -kt + Ln [A]₀

<em>Where [A] is concentration of the isotope after time t, </em>

<em>k is rate constant</em>

<em>and [A]₀ initial concentration of the isotope.</em>

<em />

Computing the values of the problem:

Ln [0.89x10⁻¹⁴] = -1.2118x10⁻⁴ yrs⁻¹t + Ln [1.1x10⁻¹³]

-2.5144 = -1.2118x10⁻⁴ yrs⁻¹t

20750 years = t

The age of the object is 20750 years

<em />

4 0
3 years ago
The rate of effusion of nitrogen gas (N2) is 1.253 times faster than that of an unknown gas. What is the molecular weight of the
ollegr [7]

Answer:

43.96

Explanation:

Graham's law was applied and the rates of effusion of nitrogen and the unknown gas were compared as shown in the image. The unknown gas is heavier than hydrigen hence it effuses slower than hydrogen as anticipated by Graham's law.

5 0
3 years ago
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