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AnnyKZ [126]
2 years ago
7

A student dissolves some solid NaNO3 in a beaker of water. Which are the solute and the solvent? Choose all answers that are cor

rect. A. solute: water B. solvent: water C. solvent: NaNO3 D. solute: NaNO3
Chemistry
1 answer:
Paul [167]2 years ago
3 0
The correct answer is D. The solute in this solution is the solid sodium nitrate (NaNO3) which is dissolved in the solvent, the water. Solute is the minor component in a solution whereas the solvent is the major component in the solution.
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The density of gas particles in a section of Earth’s atmosphere decreases. Which of the following is the most likely explanation
Bad White [126]
<span>This is due to the fact that the air pressure in that certain section of Earth’s atmosphere decreased. As density of gas particles decreases as air pressure decreases. Therefore, density of gas particles and air pressure have a direct relationship. An increase in air pressure would then effect to an increase in gas particles. </span>
6 0
1 year ago
explain what you would do expect caesium astatide to look like .will it be soluble in water ?explain your reasoning​
son4ous [18]

Answer:

it will not be soluble in water Becoz it can only be

separated by passing it through silver nitrate solution

Explanation:

i hope you understand

3 0
1 year ago
A 1.0-gram sample of solid iodine is placed in a tube and the tube is sealed after all of the air is removed. The tube and the s
yKpoI14uk [10]

Answer:

27.0

Explanation:

Because Mass can neither be created nor be destroyed hence total mass of sample of iodine and tube remain equal as it is sealed.

7 0
2 years ago
The factor 0.01 corresponds to which prefix? A) milli B) deci C) deka D) centi
BARSIC [14]
The answer is: D) centi
5 0
2 years ago
Use the following half-reactions to design a voltaic cell: Sn4+(aq) + 2 e− → Sn2+(aq) Eo = 0.15 V Ag+(aq) + e−→ Ag(s) Eo = 0.80
AVprozaik [17]

Answer:

E° = 0.65 V

Explanation:

Let's consider the following reductions and their respective standard reduction potentials.

Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V

Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:

Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V

The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V

4 0
2 years ago
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