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Mumz [18]
3 years ago
9

Air can be considered a ____________ of gaseous elements and compounds.

Chemistry
1 answer:
vodomira [7]3 years ago
5 0
The answer would be B - homogeneous mixture
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5. Why are lonic compounds good conductors of electricity?
ziro4ka [17]

Answer:

The options are unclear, however, the correct option is:

Aqueous solutions of ionic compounds cause to dissociate, hence, ions are free to conduct electricity

Explanation:

Ionic compounds are compounds formed from ions (charged atoms). For example, NaCl is an ionic compound from the following ions; Na+ (cation) and Cl- (anion). One characteristics of ionic compounds is their ability to dissociate into the ions that form them when in an aqueous solution i.e. NaCl will dissociate into Na+ and Cl- when in an aqueous solution.

These disssociated ions are free to conduct electricity, hence, making ionic compounds good conductors of electricity.

6 0
2 years ago
At which instance contain 1 molecule of substances from following?
Black_prince [1.1K]

Answer:

can I have brainleis answer pls I'm new

6 0
2 years ago
The diagram shows a model of the nitrogen cycle. which role do plants play in the nitrogen cycle?
kenny6666 [7]

Answer

D

Explanation:

They take up usable forms of nitrogen found in soil

3 0
3 years ago
Read 2 more answers
Please help! Chemistry 113 smartworks question. Thanks !!!
Dominik [7]
Kc' =Kc^1/3
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=0.182716013
3 0
3 years ago
Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne
777dan777 [17]

Answer:

Explanation:

From the information given:

CaF_2 \to Ca^{2+} + 2F^-

Ksp = 3.2 \times 10^{-11}

no of moles of Ca^{2+} = 0.01 L × 0.0010 mol/L

no of moles of Ca^{2+} = 1 \times 10^{-5} \ mol

no of moles of F^- = 0.01 L × 0.00010 mol/L

no of moles of F^- = 1 \times 10^{-6}\ mol

Total volume = 0.02 L

[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\  \\  \[[Ca^{2+}}] = 0.0005 \ mol/L

[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}

[F^{-}] = 5 \times 10^{-5}  \ mol/L

Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}

Since Q<ksp, then there will no be any precipitation of CaF2

3 0
3 years ago
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