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Galina-37 [17]
2 years ago
6

Tìm câu sai về độ pH?

Chemistry
1 answer:
galina1969 [7]2 years ago
5 0

Answer:

B

Mark as the Brainliest please

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When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to
Reika [66]

Answer:

2

0.4167\ \text{M}^{-1}\text{min}^{-1}

Explanation:

Half-life

{t_{1/2}}A=2\ \text{min}

{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}

So

\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

Comparing the exponents we get

1=n-1\\\Rightarrow n=2

The order of the reaction is 2.

t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}

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3 0
2 years ago
An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula
uysha [10]

Answer:

The answer is: <u>Al2O3</u>

Explanation:

The data they give us is:

  • 0.545 gr Al
  • 0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

  • 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
  • 0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

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  • 0.03 mol O / 0.02 = 1.5  O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

  • 1Al * 2 = 2Al
  • 1.5O * 2 = 3O

Now the answer is given correctly:

  • Al2O3

8 0
2 years ago
Which term names the concentration of salts dissolved in a liquid?
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