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Rzqust [24]
3 years ago
13

How many moles are in 75.0 grams of K2S?

Chemistry
1 answer:
notka56 [123]3 years ago
3 0

Answer:

0.6802 mol

Explanation:

1 grams K2S  = 0.00907 mol

0.00907 x 75 = 0.6802 mol

You might be interested in
A compound contains 40 g of calcium, 12 g of carbon, and 48 g of oxygen. What is its empirical formula
Deffense [45]

Answer:

CaCO3

Explanation:

answer attached

hope this helps!

5 0
2 years ago
A person tries to pour exactly one kilogram of sand onto a scale on five
blsea [12.9K]

Answer:

A. While inconsistent, his average of 1.0 kg made his results quite

accurate

Explanation:

Accuracy is how close the value you gain from test or observation compared with the real value. In this test, all tests relatively close to 1kg and the average of the test is 1kg so this tool can be considered quite accurate.

The precision determines how close the value of multiple repetitions of the test. In this case, the value varies from 0.8kg to 1.2kg. The highest value is about 50% higher than the lowest value, so the range is pretty far and the test can be considered not precise.

8 0
3 years ago
Explain how the concentration of hydronium ions and hydroxide ions are related to pH.
Dominik [7]
Hello!

The concentration of hydronium ions is related to pH by the following simple equation:

pH=-log[H_3O^{+}]

From this equation, you can see that as the hydronium concentration is higher, the pH will be lower.

The concentration of the OH⁻ ions is related to the pH by the following set of equations

pOH=-log[OH^{-}] \\ pH=14-pOH

You can see that as the concentration of OH⁻ is higher, the pOH is lower and thus the pH is higher. 

When the pH of the solution is less than 7, the solution is acidic.

When the pH of the solution is higher than 7, the solution is basic.

Have a nice day!

8 0
3 years ago
The formation of tert-butanol is described by the following chemical equation: (CH3), CBr (aq) + OH(aq) → Br" (aq) +(CH), COH(aq
Dafna11 [192]

Answer:

(CH_3)_3С^+ (aq) + OH^- (aq)\rightarrow (CH_3)_3COH (aq)

Explanation:

There are two ways of looking at this problem. The first way, slightly more advanced, is to understand that the carbocation formed is an intermediate in this reaction: it is formed in one step and consumed in the subsequent step.

Secondly, we have hydroxide involved as our reactant, so it should be our second reactant in the second bimolecular step.

Thirdly, the product formed would be a combination of the anion and cation, one of our products, this means we have the following second step:

(CH_3)_3С^+ (aq) + OH^- (aq)\rightarrow (CH_3)_3COH (aq)

Another way is to verify this knowing that by adding all of the steps should yield a net equation, notice if we add the two steps together (reactants on one side and products on the other), we obtain:

(CH_3)_3С^+ (aq) + OH^- (aq) + (CH_3)_3CBr (aq)\rightarrow (CH_3)_3COH (aq) + (CH_3)_3C^+ (aq) + Br^- (aq)

Notice that the intermediate carbocation cancels out on both sides to yield the final net equation:

OH^- (aq) + (CH_3)_3CBr (aq)\rightarrow (CH_3)_3COH (aq) + Br^- (aq)

This means we have the correct second step.

6 0
3 years ago
What is true about a solution of 1.0 M HF? HF has a higher [OH-] than a solution of 1.0 M HCl. HF has a lower pH than a solution
Wittaler [7]
<span>HF has a higher [OH-] than a solution of 1.0 M HCl. is the answer</span>
6 0
3 years ago
Read 2 more answers
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