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morpeh [17]
2 years ago
6

Plssssssssssssssssssmejepleplele

Chemistry
2 answers:
Firlakuza [10]2 years ago
4 0
Yes that answer is correct
Leona [35]2 years ago
3 0

Answer:

Yes the two of the answer is True

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Calculate the poh of this solution. round to the nearest hundredth. ph = 1.90 poh =
Viktor [21]

Answer:

The answer is 5.10

Explanation:

<h3><u>Given</u>;</h3>
  • pH = 1.9
<h3><u>To </u><u>Find</u>;</h3>
  • pOH = ?

We know that

pH + pOH = 7

pOH = 7 – pH

pOH = 7 – 1.90

pOH = 5.10

Thus, The pOH of the solution is 5.10

6 0
2 years ago
8. A sample of chloroform is found to contain 24.0 g of carbon,212.8 g of chlorine, and 2.02 g of hydrogen. If a second sample o
Misha Larkins [42]

Answer:

595.5

Explanation:

chloroform with 24.0 g C was 238.2 g

24g/238.2g= 60g/x

595.5

4 0
3 years ago
All organic molecules contain carbon and hydrogen. what parts of an organic molecule may contain oxygen, nitrogen, or phosphorus
Minchanka [31]
Answer is: <span>functional groups.
</span>Functional groups<span> are specific </span>groups<span> that are responsible for the characteristic chemical properties of molecule.</span>
<span>Proteins have nitrogen and oxygen in functional group.
Nucleic acids and some lipids have phosphorus in fuctional group.
Carbohydrates have oxygen in functional group for example.</span>
7 0
3 years ago
Read 2 more answers
A sample of gas occurs 9.0mL at a pressure of 500mmHg. A new volume of the same sample is at 750mmHg. Use two significant figure
Anit [1.1K]

Answer:

V₂ = 6.0 mL

Explanation:

Given data:

Initial volume = 9.0 mL

Initial pressure = 500 mmHg

Final volume = ?

Final pressure = 750 mmHg

Solution:

According to Boyle's Law

P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂

V₂ = 500 mmHg × 9.0 mL / 750 mmHg

V₂ = 4500 mmHg .mL / 750 mmHg

V₂ = 6.0 mL

5 0
3 years ago
Can someone help?? This is really hard.
Sergeu [11.5K]

Answer:

See Explanation

Explanation:

Let us consider the first two reactions, the initial concentration of CO was held constant and the concentration of Hbn was doubled.

2.68 * 10^-3/1.34 * 10^-3 = 6.24 * 10^-4/3.12 * 10^-4

2^1 = 2^1

The rate of reaction is first order with respect to Hbn

Let us consider the third and fourth reactions. The concentration of Hbn is held constant and that of CO was tripled.

1.5 * 10^-3/5 * 10^-4 = 1.872 * 10^-3/6.24 * 10^-4

3^1 = 3^1

The reaction is also first order with respect to CO

b) The overall order of reaction is 1 + 1=2

c) The rate equation is;

Rate = k [CO] [Hbn]

d) 3.12 * 10^-4 = k [5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4  /[5 * 10^-4] [1.34 * 10^-3]

k = 3.12 * 10^-4/6.7 * 10^-7

k = 4.7 * 10^2 mmol-1 L s-1

e) The reaction occurs in one step because;

1) The rate law agrees with the experimental data.

2) The sum of the order of reaction of each specie in the rate law gives the overall order of reaction.

8 0
3 years ago
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