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NARA [144]
3 years ago
6

Which of the following lists the correct symbol for the ion that tin forms when it loses 2 electrons, the correct classification

of the ion as an anion or a cation, and the name of the ion?
A) Sn; cation; tin(II) ion
B) Sn; anion; tin(II) ion
C) Sn; anion; tin ion
D) Sn; cation; tin ion
Chemistry
2 answers:
ser-zykov [4K]3 years ago
6 0

Answer:

\huge\boxed{\sf Option \ A}

Explanation:

The symbol for TIN is Sn.

When Sn loses 2 electrons, it gets a double positive charge ( +2 ) and becomes \sf Sn ^{+2}.

It becomes a cation.

The name of Ion is Tin ( II ) Ion.

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
AysviL [449]3 years ago
3 0

Answer:

option a

Explanation:

gradpoint

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For atm to kPa conversion:

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What is the partial pressure of carbon dioxide in a container that contains 3.63 mol of oxygen, 1.49 mol of nitrogen, and 4.49 m
lana66690 [7]

Answer:

Partial pressure of CO₂ is 406.9 mmHg

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6 0
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The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa
Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of MgCl_2 (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of MgCl_2 (solute) = 3.37 g

First we have to calculate the mass of solute.

\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2

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Now we have to calculate the mass of solution.

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Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg

The molality of the solution is, 0.0381 mole/Kg.

Now we have to calculate the mass/mass percent.

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The mass/mass percent is, 25.67 %

Now we have to calculate the mass/volume percent.

\text{Mass by volume percent}=\frac{\text{Mass of solute}}{\text{Volume of solution}}\times 100=\frac{320.86}{1000}\times 100=32.086\%

The mass/volume percent is, 32.086 %

Therefore, the molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

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