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xz_007 [3.2K]
4 years ago
14

In an experiment,which variable changes In response to the manipulation of another variable

Chemistry
1 answer:
BlackZzzverrR [31]4 years ago
8 0
The dependent variable<span />
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An important reaction sequence in the industrial production of nitric acid is the following:
Nataly_w [17]

Answer:

25.0 mol O₂ are required in the second reaction

Explanation:

N₂ (g) + 3H₂ (g)  →  2NH₃ (g)

4NH₃ (g)  +  5O₂ (g)  →  4NO (g) + 6 H₂O(l)

Molar ratio in first reaction is 1:2

For every mol of N₂. I make 2 moles of ammonia. If I have 20 moles of N₂, i'm going to get, 40 moles of ammonia.

In the second reaction, molar ratio between products is 4:5.

If I obtained 40 moles of ammonia in first step, let's prepare the rule of three.

4 moles of ammonia react with 5 moles of O₂

40 moles of ammonia react with ( 40.5) /4 = 25moles

5 0
4 years ago
Instant cold packs, often used to ice athletic injuries on the field, contain ammonium nitrate and water separated by a thin pla
Pavel [41]

Answer:

ΔH = +26.08 kJ/mol

Explanation:

The change in enthalpy (ΔH) is given in J/mol, and can be calculated for  dissolution by the equation:

ΔH = m(water)*Cp*ΔT/n(solute)

The mass of water is the density multiplied by the volume

m = 1g/mL * 25.0mL = 25.0 g

The number of the moles is the mass divided by the molar mass. Knowing the molar masses of the elements:

N = 14 g/mol x 2 = 28

H = 1 g/mol x 4 = 4

O = 16 g/mol x 3 = 48

NH₄NO₃ = 80 g/mol

n = 1.25/80 = 0.015625 mol

So,  

ΔH = 25*4.18*(25.8 - 21.9)/0.015625

ΔH = 26,083.2 J/mol

ΔH = +26.08 kJ/mol

5 0
3 years ago
Atmospheric chemistry involves highly reactive, odd-electron molecules such as the hydroperoxyl radical HO2, which decomposes in
dem82 [27]

Answer:

Rate = k [HO2]

Rate constant = 0.8456us-1

Explanation:

Time(us) 0 0.6 1.0 1.4 1.8 2.4

[HO2](uM) 8.5 5.1 3.6 2.6 1.1 1.1

The rate law is given as;

Rate = k [HO2]^x

Where x signify the order of reaction.

For an order of reaction, the rate constant is constant for all concentrations. We are going to use this to obtain the order of reaction.

Zero Order:

[A] = [A]o -kt

5.1 = 8.5 - k(0.6)

-k (0.6) = 5.1 - 8.5

k = 5.67

3.6 = 5.1 - k(0.4)

-k (0.4) = 3.6 - 5.1

k = 3.75

The fact that the rate constant was not constant means the reaction is not a zero order reaction.

First Order:

ln[A] = ln[A]o -kt

(5.1) = ln(8.5) - k(0.6)

-k (0.6) = ln(5.1) - ln(8.5)

k = 0.8524

ln(3.6) = ln(5.1) - k(0.4)

-k (0.4) = ln(3.6) - ln(5.1)

k = 0.8708

ln(2.6) = ln(3.6) - k (0.4)

-k (0.4) = ln(2.6) - ln(3.6)

k = 0.8136

From the three calculations we see that the value of the rate constant is fairly constant in the range of 0.8 This means our reaction is a first order reaction.

The rate law is given as;

Rate = k [HO2]

We can represent the rate constant as the average of the three rate constants calculated above;

Rate constant = (0.8136 + 0.8708 + 0.8524 / 3)

Rate constant = 0.8456us-1

6 0
3 years ago
15.35 L of gas is at a pressure of 198.0 kpa. At what pressure would the gas have a volume of 13.78
Ivahew [28]

Answer:

P₂ = 220.56KPa

Explanation:

Boyles Law P ∝ 1/V => Inverse relationship => Decreasing Volume => Increasing Pressure.

P₁ = 198 KPa           P₂ = ?

V₁ = 15.35 Liters     V₂ = 13.78 Liters

P₁V₁ = P₂V₂ => P₂ = P₁(V₁/V₂) = 198KPa(15.35 L/ 13.78L) = 220.6KPa

5 0
3 years ago
If you weighed out 0.38 g of calcium and reacted it with an excess of hcl, how many moles of h2(g) would you expect to produce?
Hitman42 [59]
Answer is: 0,0095 mol of hydrogen gas will be produced in reaction.
Chemical reaction: Ca + 2HCl → CaCl₂ + H₂.
m(Ca) = 0,38 g.
n(H₂) = ?
n(Ca) = m(Ca) ÷ M(Ca).
n(Ca) = 0,38 g ÷ 40 g/mol
n(Ca) = 0,0095 mol.
from reaction: n(Ca) : n(H₂) = 1 : 1.
n(H₂) = n(Ca) = 0,0095 mol.
n - amount of substance.
4 0
3 years ago
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