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dlinn [17]
3 years ago
10

I don’t know how to answer these, can anyone answer any of them?

Chemistry
1 answer:
vlabodo [156]3 years ago
3 0

Answer:

1.) To convert between grams and moles, you would use the substance's molar mass. To go from grams to moles, divide the grams by the molar mass. 600 g58.443 g/mol = 10.27 mol of NaCl. It has been found that 1 mol of any gas at STP (Standard Temperature and Pressure = 0 °C and 1 atm) occupies 22.4 L

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Which of these phrases accurately describe disproportionation? Check all that apply.
Olin [163]

Answer:

A single compound is simultaneously oxidized and reduced.

Explanation:

In chemistry, disproportionation is a simultaneous oxidation and reduction of a single chemical specie.

What this means is that; in a disproportionation reaction, only one compound is both oxidized and reduced. This implies that two products are formed during disproportionation. One is the oxidized product while the other is the reduced product.

Consider the disproportionation of CuCl shown below;

2CuCl -----> CuCl2 + Cu

Here, CuCl2 is the oxidized product while Cu is the reduced product.

7 0
3 years ago
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The theoretical yield for a reaction is 55.9 g LiCl. The actual yield is 24.6 g LiCl. What is the percent yield of the reaction?
xenn [34]
In chemical reactions, the actual yield is not the same as the expected yield . Actual yield is lower than the theoretical yield . Then we have to find the yield percentage. To see what percentage of the theoretical yield is the actual yield.
Percent yield = actual yield / theoretical yield x 100%
Percent yield = 24.6/55.9 x100%
Percent yield = 44%
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3 years ago
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photoshop1234 [79]
Number 4 is
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Answer:

B

Explanation:

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What is the value of the activation energy of the uncatalyzed reaction?
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The value of the activation energy of an uncatalyzed reaction is greater than that of a catalyzed reaction. As we know, a catalyst provides an alternative path for the reaction to happen at a faster rate. So, for a catalyzed reactio, activation energy is lesser than the original path.
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