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aniked [119]
2 years ago
8

Please help I will mark brainliest

Chemistry
1 answer:
Lubov Fominskaja [6]2 years ago
8 0

Explanation:

light blue is the answer second one

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Which of these salts is insoluble in water?
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An oxidation-reduction reaction that is also a synthesis reaction
Aleonysh [2.5K]

The example of a chemical combination that can also be said to be a synthesis reaction is; 2Mg(s) +O_{2} (g) -- > 2 MgO(s)

<h3>What is a synthesis reaction?</h3>

We know that the meaning of the synthesis reaction is quite central as we attempt to answer the question that is lying right before us here. If we are talking about the synthesis reaction then our mind must have to go to the kind of reaction in which there is the combination of two or more so as to produce a single compound.

As we are talking about the synthesis reaction, we would have to look at the reaction that would give rise to a single product as it has been written. Let us also recall that we have been told that the reaction is a  oxidation-reduction reaction thus there is a change in oxidation number in the process of the reaction as shown.

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7 0
8 months ago
If you have 3 g of each substance, which would take up the smallest volume?
VashaNatasha [74]

Answer:

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f the mass of an object is 35 grams and it takes up 7 cm3 of space, calculate the density.

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8 0
2 years ago
Chamber 1 and Chamber 2 have equal volumes of 1.0L and are assumed to be rigid containers. The chambers are connected by a valve
vitfil [10]
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas  than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)

=> p1 = 2 p2

Which is easy to demonstrate using ideal gas equation:

p1 = nRT/V = 2.0 mol * RT / 1 liter

p2 = nRT/V = 1.0 mol * RT / 1 liter

=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2

2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.

So, the pressure in both chambers (which form one same vessel) is:

p = nRT/V = 3.0 mol * RT / 2liter

which compared to the initial pressure in chamber 1, p1, is:

p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1

So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.

You can also see how the pressure in chamber 2 changes:

p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
5 0
3 years ago
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