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Step2247 [10]
1 year ago
9

b → p experiment number [a] (m) [b] (m) initial rate (m/s) 1 0.273 0.763 2.83 2 0.273 1.526 2.83 3 0.819 0.763 25.47 22) the rat

e law for this reaction is rate
Chemistry
1 answer:
Mazyrski [523]1 year ago
8 0

The rate law for this reaction is [A]².

Balanced chemical reaction used in this experiment: A + B → P

The reaction rate is the speed at which reactants are converted into products.

Comparing first and second experiment, there is no change in initial rate. The concentration of reactant B is increased by double. Initial rate does not depands on concentration of reactant B.

Comparing first and third experiment, initial rate is nine times greater, while concentration of reactant A is three times greater. Conclusion is that concentration of reactant A is squared and the rate is [A]².

More info about rate law: brainly.com/question/16981791

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7 0
3 years ago
A substance with a pH of 10 is how much more basic than a substance with a pH of 2?
Kazeer [188]

Answer:

The same holds true for pH values above 7, each of which is ten times more alkaline (another way to say basic) than the next lower whole value. For example, pH 10 is ten times more alkaline than pH 9 and 100 times (10 times 10) more alkaline than pH 8.

Explanation:

7 0
3 years ago
Convert 3.52 × 10^22 atoms of gold to moles​
IrinaVladis [17]

Answer:

5.85 x 10⁻² mol Au

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Divide the atoms of gold with Avogadro's number to find the solution.

5 0
3 years ago
A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH
Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5    → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0
3 years ago
If you have 85g of sulfur dioxide at STP, how many moles do you have?
Step2247 [10]

Answer: 1.32

Explanation:

Given

85 gm of sulfur dioxide is present at STP

The molar mass of sulfur dioxide is 32+2\times 16=64\ g/mol

The number of moles of sulfur dioxide is

\Rightarrow n=\dfrac{85}{64}\\\\\Rightarrow n=1.32\ \text{mol}

3 0
3 years ago
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