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3 years ago

A 4,000 mL solution of AgNO3 contains 17.00 g of solute in water. Calculate the molar concentration of the solution.

Chemistry

2 answers:

MariettaO [177]3 years ago

6 0

The answer is 0.03 M

blsea [12.9K]3 years ago

3 0

The molarity is moles/liters.

First, convert 4,000 mL to L:

4000 mL --> 4 L

Now, you must convert the 17 g of solute to moles by dividing the number of grams by the molar mass. The molar mass of AgNO3 is 169.87 g/mol:

17 / 169.87 = .1

Now that you have both the number of moles and the liters, plug them into the initial equation of moles/liters:

.1/4 = .025

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Explanation:

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Answer:

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(9) Calcium is carrying +2 charge called as $Ca^{+2}$ cation and chloride $Cl^{-1}$ is an anion carrying -1 charge. They form $CaCl_2$ .

(10) Calcium is carrying +2 charge called as $Ca^{+2}$ cation and phosphate $PO_4^{-3}$ is an anion carrying -3 charge. They form $Ca_3(PO_4)_2$ .

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(14) Ammonium ion is carrying +1 charge called as $NH_4^{+1}$ cation and phosphate $PO_4^{-3}$ is an anion carrying -3 charge. They form $NH_4_3PO_4$ .

(15) Ammonium ion is carrying +1 charge called as $NH_4^{+1}$ cation and sulfate $SO_4^{-2}$ is an anion carrying -2 charge. They form $NH_4_2SO_4$ .

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3 years ago

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Answer:

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Explanation:

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D is mercury, a metal reacting with oxygen. Two elements reacting. Neither are an acid or a base

E. is an acid reacting with a metal to liberate hydrogen. There is no base

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3 years ago

A compound is 52.0% zinc 9.6% carbon and 38.4% oxygen. Calculate the empirical formula of the compound.

skad [1K]

Answer:

So a compound is 52% Zinc(Zn), 9.6% Carbon(C), and 38.4% Oxygen (O). Let’s first start off by assuming that we have 100 g of this compound. This means that we have 52 g of Zinc, 9.6 g of Carbon, and 38.4 g of Oxygen.Zinc = 65.38 g/molCarbon = 12 g/molOxygen = 16 g/molThis means we have:52 g of Zn(1 mol Zn/65.38 g of Zn) ≈0.8 mol of Zn.9.6 g of C(1 mol C/12 g of C) = 0.8 mol of C38.4 g of O(1 mol of O/16 g of O) = 2.4 mol of O.

Explanation:

What we want to do next is divide each element by the common factor of all of them, which is 0.8. In most cases, you divide each element by the element with the least amount of moles. After we divide each by 0.8, you’ll notice you have 1 Zn, 1 C, and 3 O. This gives you the empirical formula of ZnCO3, or Zinc Carbonate.

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1 year ago

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andriy [413]

Answer:

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Explanation:

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8 0

2 years ago