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kakasveta [241]
3 years ago
12

What can a drop of liquid mercury be described as?

Chemistry
2 answers:
Tomtit [17]3 years ago
8 0
A drop of iquid mercury is D. <span> a pure substance and an element. This is because mercury is an element which exists as liquid in the normal or standard conditions. It is a metal that is toxic to all environments. It is pure because it is composed only of one kind of substance.</span>
mezya [45]3 years ago
6 0

Answer: D) a pure substance and an element

Explanation:

Mercury is an element which is composed of similar atoms and is a pure substance.

Element is a pure substance which is composed of atoms of similar elements.

Compound is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.

Mixtures are impure substances and are heterogeneous materials as they do not have uniform composition and the components are not evenly distributed throughout the material.

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At lunch, the babysitter gives Tanya and Damien each a chocolate candy. Tanya puts her candy on her plate. Damien saves his cand
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3 years ago
Problem Page A chemist measures the amount of bromine liquid produced during an experiment. He finds that of bromine liquid is p
masha68 [24]

The question is incomplete, here is the complete question:

A chemist measures the amount of bromine liquid produced during an experiment. She finds that 766.g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.

<u>Answer:</u> The amount of liquid bromine produced is 4.79 moles.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Given mass of liquid bromine = 766. g

Molar mass of liquid bromine, (Br_2) = 159.8 g/mol

Putting values in above equation, we get:

\text{Moles of liquid bromine}=\frac{766.g}{159.8g/mol}=4.79mol

Hence, the amount of liquid bromine produced is 4.79 moles.

6 0
3 years ago
The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.
madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

H_3BO_3\rightarrow H^+H_2BO_3^-

at t=0  cM              0             0

at eqm c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 4.4 M and \alpha = ?

K_a=5.8\times 10^{-10}

Putting in the values we get:

5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}

(\alpha)=0.000011

[H^+]=c\times \alpha

[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M

Also pH=-log[H^+]

pH=-log[4.8\times 10^{-5}]=4.3

Thus pH of a 4.4 M H_3BO_3 solution is 4.3

3 0
3 years ago
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