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Ronch [10]
3 years ago
9

1. Which of the following describes a biotic influence on population size?

Chemistry
1 answer:
Alborosie3 years ago
5 0

Answer:

A

Explanation:

decreased rainfall leads to less space for fish in a pond to live.

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How many atoms are in .800 g of Ca
Goshia [24]
The answer for this question is 0.8
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Strontium sulfate becomes less soluble in an aqueous solution when sodium sulfator is added because
horsena [70]

Answer:

The addition of sulfate ions shifts equilibrium to the left.

Explanation:

Hello!

In this case, according to the following ionization of strontium sulfate:

SrSO_3(s)\rightleftharpoons Sr^{2+}+SO_4^{2-}

It is evidenced that when sodium sulfate is added, sulfate, SO_4^{2+} is actually added in to the solution, which causes the equilibrium to shift leftwards according to the Le Ch athelier's principle. Thus, the answer in this case would be:

The addition of sulfate ions shifts equilibrium to the left.

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3 years ago
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Which is the smallest particle into which water (H2O) in a glass can be broken down and still remain water? A. a water molecule
Gnesinka [82]

Answer:-

A. A water molecule

Explanation:-

A molecule is the smallest particle of a compound that retains all it's chemical properties.

Here H2O is a compound. So the smallest particle that will retain all the chemical properties and still remain water is water molecule.

Atoms Hydrogen and Oxygen both have different chemical properties from water H2O and are thus different.

Hydrogen peroxide is different molecule from water. So it is also not water.

6 0
3 years ago
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What mass of sodium hydroxide must be heated in order to produce 4.5 g of water?
inysia [295]

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If 9.85 grams of copper metal react with 31.0 grams of silver nitrate, how many grams of copper nitrate can be formed and how ma
goldfiish [28.3K]
The balanced chemical reaction:

<span>Cu + 2AgNO3 = Cu(NO3)2 + 2Ag
</span>
We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.

9.85 g Cu ( 1 mol Cu / 63.55 g Cu ) = 0.15 mol Cu
31.0 g AgNO3 ( 1 mol AgNO3 / 169.87 g AgNO3 ) = 0.18 mol AgNO3

The limiting reactant is AgNO3.

0.18 mol AgNO3 ( 1 mol Cu(NO3)2 / 2 mol AgNO3 ) (187.56 g / 1 mol) =16.88 g Cu(NO3)2

0.15 mol Cu - 0.18 mol AgNO3 ( 1 mol Cu / 2 mol AgNo3) = 0.06 mol Cu excess

<span>0.06 mol Cu ( 63.55 g Cu / 1 mol Cu ) = 3.81 g Cu excess</span>
3 0
3 years ago
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