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Irina18 [472]
2 years ago
10

"The elementary reaction 2 NO2(g) → 2 NO(g) + O2(g) is second order in NO2 and the rate constant at 600 K is 6.77 × 10-1 M-1s-1.

The reaction half-life at this temperature when [NO2]0 = 0.45 M is ________ s."
Chemistry
1 answer:
blsea [12.9K]2 years ago
7 0

Answer : The half-life at this temperature is, 3.28 s

Explanation :

To calculate the half-life for second order the expression will be:

t_{1/2}=\frac{1}{k\times [A_o]}

When,

t_{1/2} = half-life = ?

[A_o] = initial concentration = 0.45 M

k = rate constant = 6.77\times 10^{-1}M^{-1}s^{-1}

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

t_{1/2}=\frac{1}{6.77\times 10^{-1}M^{-1}s^{-1}\times 0.45M}

t_{1/2}=3.28s

Therefore, the half-life at this temperature is, 3.28 s

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120 grams of calcium nitrite Ca(NO2)2 is dissolved in a 240 mL solution. What is the molarity of the solution???
expeople1 [14]

Hey there!:

Molar mass Ca(NO2)2 = 132.089 g/mol

Mass of solute = 120 g

Number of moles:

n = mass of solute / molar mass

n = 120 / 132.089

n = 0.0009084 moles of Ca(NO2)2

Volume in liters of solution :

240 mL / 1000 => 0.24 L

Therefore:

Molarity = number of moles / volume of solution

Molarity = 0.0009084 / 0.24

Molarity = 0.003785 M

Hope that helps!

5 0
2 years ago
Calculate the empirical formula for each of the following substances. (Express answer as a chemical formula) 1) 2.90 g of Ag and
exis [7]

Answer:

1) Ag3N

2)Na2S

3)NaHSO4

4) KNO3

Explanation:

We divide each mass by the element's relative atomic mass

1) 2.90/108-Ag, 0.125/14-N

0.027-Ag, 0.0089-N

Divide by the lowest ratio

0.027/0.0089-Ag, 0.0089/0.0089 N

3-Ag, 1-N

Empirical formula- Ag3N

2)2.22/23-Na, 1.55/32-S

0.097-Na, 0.048-S

Divide by the lowest ratio

0.097/0.048-Na, 0.048/0.048-S

2-Na, 1-S

Empirical formula- Na2S

3) 2.11/23-Na, 0.0900/1-H, 2.94/32-S,5.86/16-O

0.09-Na, 0.09-H, 0.09-S,0.366-O

Divide by the lowest ratio

0.09/0.09-Na, 0.09/0.09-H, 0.09/0.09-S, 0.366/0.09-O

1-Na, 1-H, 1-S, 4-O

Empirical formula- NaHSO4

4)1.84/39, 0.657/14-N, 2.25/16-O

0.047-K, 0.047-N, 0.14-O

Divide through by the lowest ratio

0.047/0.047-K, 0.047/0.047-N, 0.14/0.047-O

1-K, 1-N, O-3

Empirical formula- KNO3

4 0
2 years ago
Candace is finding the boiling point of water. From her research, she knows the boiling point of pure water is 212°F. When she r
Anestetic [448]

I would say that Candace's answer is d. wide-ranging. she didn't get the exact / precise (they mean the same thing) answer.

6 0
2 years ago
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water’s molar mass is 18.01 g/mol. The molar mass of glycerol is 92.09 g/mol. At 25 celsius, glycerol is more viscous than water
Gemiola [76]

Answer is: glycerol because it is more viscous and has a larger molar mass.

Viscosity depends on intermolecular interactions.

The predominant intermolecular force in water and glycerol is hydrogen bonding.

Hydrogen bond is an electrostatic attraction between two polar groups in which one group has hydrogen atom (H) and another group has highly electronegative atom such as nitrogen (like in this molecule), oxygen (O) or fluorine (F).

4 0
2 years ago
How many mL of a 5.0% glucose solution provide 80.0 g of glucose?
Llana [10]
Maybe
A. 400 ml of 5.0% glucose solution
4 0
2 years ago
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