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olga55 [171]
3 years ago
8

How many half-lives are required for the concentration of reactant to decrease to 12. 5% of its original value?.

Chemistry
1 answer:
Tcecarenko [31]3 years ago
5 0

Answer:

Three half-lives

Explanation:

12.5% is one-eighth. One eighth is the cube of one half, so it would take three half-lives to reduce a reactant's concentration to 12.5%

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What is the perecentage yield of a reaction in which 200g PCl3 reacts with excess water to form 128g HCl according to the f.F re
castortr0y [4]

Answer:

The percentage yield is 80.36% (see calculations in attachment).

Explanation:

The theoretical  yield of the reaction is the <u>amount of product that would result if all the  limiting reagent reacted.</u> The theoretical yield is calculated using the balanced equation.

In practice, the actual yield, or the <u>amount of  product actually obtained from a reaction</u>, is almost always less than the theoretical  yield.

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%yield = actual yield ÷ theoretical yield × 100%

First we need to make sure that the equation is properly balanced. In the question they provide the balanced equation.

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3 years ago
Brainliest if answered correctly <br>How many Calcium (CA) atoms are in carbon tetrachloride​
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Answer:

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3 years ago
Read 2 more answers
0.475 g H, 7.557 gS, 15.107 g O. Express your answer as a chemical formula.
max2010maxim [7]

Answer:

H₂SO₄

Explanation:

We have a compound formed by 0.475 g H, 7.557 g S, 15.107 g O. In order to determine the empirical formula, we have to follow a series of steps.

Step 1: Calculate the total mass of the compound

Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g

Total mass = 23.139 g

Step 2: Determine the percent composition.

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S: (7.557g/23.139g) × 100% = 32.66%

O: (15.107g/23.139g) × 100% = 65.29%

Step 3: Divide each percentage by the atomic mass of the element

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S: 32.66/32.07 = 1.018

O: 65.29/16.00 = 4.081

Step 4: Divide all the numbers by the smallest one

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S: 1.018/1.018 = 1

O: 4.081/1.018 ≈ 4

The empirical formula of the compound is H₂SO₄.

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