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3 years ago

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A 0.750-g sample of a compound that might be iron(II) chloride, nickel(II) chloride, or zinc chloride is dissolved in water, and

22.40 mL of 0.515 M AgNO3 is required to completely precipitate all the chloride ion as AgCl. Name the compound, and write its formula.

1 answer:

DedPeter [7]3 years ago

4 0

Answer: The compound is nickel (II) chloride.

NiCl2

Explanation:

Let's assume that the compound is X.

Then,

XCl2 + 2AgNO3 ---> X(NO3)2 + 2AgCl

Given,

Volume of AgNO3 = 22.4ml = 0.0224L

Concentration of AgNO3 = 0.515M

no of moles = concentration x volume

:• n(AgNO3) = 0.515 x 0.0224

n(AgNO3) = 0.011536 mol

Therefore, no of moles of AgNO3 is 0.0011536mol

From the equation above, we can deduce that;

n(XCl2) = 1/2n(AgNO3)

:• n(XCl2) = 1/2 × 0.011536

n(XCl2) = 0.005768

Therefore number of moles of XCl2 is 0.005768 mol.

Mass, m, of XCl2 = 0.75g (Given)

Molar mass = mass/ no of moles

Molar mass of XCl2 = 0.75/0.005768

:• molar mass of XCl2= 130.03g/mol

Molar mass of Cl = 35.453

Cl2= 2 x 35.453 = 70.906

Molar mass of XCl2 = M.mass of X +

M.mass of Cl2

:• 130.03 = M.mass of X + 70.906

Molar mass of X = 130.03 - 70.906

Molar mass of X = 59.1

By comparing to literature, we can conclude that X is Nickle ,Ni.

Therefore, the compound is NiCl2 , Nickle(II) Chloride.

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Determine the number of atoms in 41.0 grams of calcium, Ca. (The mass of one mole of calcium is 40.08 g.)

Dovator [93]

Hi there!

Let me know if you have questions about my answer:

6.15 × 10²⁴

Explanation:

To find the number of atoms, divide the mass of calcium (m) by its molar mass (MM). This will give you the number of moles of calcium (n).

$n = m/MM$ Start with a formula

$=\frac{41.0g}{40.08g/mol}$ Substitute values from the question

$= \frac{41.0g*1mol}{40.08g}$ Simplify

$= \frac{41.0mol}{40.08}$ Cancel out the units "g"

$= 1.022...mol$ Number of moles of calcium, with 4 significant figures

Then, multiply the number of moles of calcium (n) by Avogadro's number ( $N_{A}$ ) to find the number of calcium atoms.

$Number.of.Ca=n*N_{A}$ Start with a formula

$= 1.022mol*\frac{6.022*10^{23}Ca}{1mol}$ Substitute values from the question

$= \frac{1.022mol*6.022*10^{23}Ca}{1mol}$ Simplify

$= 1.022*(6.022*10^{23}Ca)$ Cancel out the units "mol"

$= 6.154*10^{24} Ca$ Multiply, leave one extra digit to help round. "4" tells you to round down.

$= 6.15*10^{24} Ca$ Final answer to 3 significant figures*

*When multiplying and dividing numbers, your final answer will have the same number of significant figures as the number with the least number of significant figures in the question.

40.08 has 4 significant figures. 41.0 has 3 significant figures. So, you want 3 significant figures in your final answer.

I hope this helps!

Learn more about Avogadro's number here:

brainly.com/question/1445383

Check out another answer that uses significant figures:

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jeka94

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$T_f$ = 25.05°C

Explanation:

Given:

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