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yanalaym [24]
3 years ago
10

1. Write the equilibrium constant expression for the following:

Chemistry
1 answer:
melisa1 [442]3 years ago
3 0

Answer:

c

Explanation:

it could honestly be wrong but I'm not sure

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Give two test substances that would react with each other to produce salt and water​
Mademuasel [1]

Answer:

NaOH +HCl==>Nacl+H2O

KOH+HCl==>KOH+H2O

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3 years ago
Calculate the density of carbon dioxide at STP
Artist 52 [7]

Answer:

Density = mass/volume

= 44/22.4

= 1.96 gram/liter

The density of the Carbon Dioxide at S.T.P. (Standard Temperature and Volume) is 1.96 gram/liter.

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3 years ago
2HgO(s)---->2Hg(I)+O2(g) which type of reaction occurs
kaheart [24]
Decomposition,because 1 breaks down into 2
8 0
3 years ago
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You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

3 0
3 years ago
How do I identify polar bonds?
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Identify<span> each </span>bond<span> as either </span>polar<span> or nonpolar.</span>
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4 years ago
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