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kap26 [50]
3 years ago
12

Identify the following silver salts as soluble or insoluble in water. A g N O 3 AgNOX3 Choose... A g C l AgCl Choose... A g I Ag

I Choose...
Chemistry
1 answer:
maxonik [38]3 years ago
5 0

Answer:

AgNO3 -soluble in water

AgCl- insoluble in water

AgI- Insoluble in water

Explanation:

The solubility of chemical compounds in water is easily predicted by a set of rules generally referred to as the solubility rules. These rules are usually based on experimental observation of diverse groups of compounds.

According to the solubility rules, nitrates are soluble in water including the nitrates of silver. However, halides are soluble in water except those of silver, lead and mercury.

You might be interested in
(will give brainliest) show your work. How many grams of Copper(I) nitrate, CuNO3 are required to produce 88.0 grams of aluminum
ValentinkaMS [17]

Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

<h3>What is stoichiometry of a reaction?</h3>

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

Learn more about stoichiometry at: brainly.com/question/16060223

Therefore, 156.114 g of CuNO3

4 0
1 year ago
Im hvaing a hard time getting the right answer
Tanya [424]

Answer:

V=23.9mL

Explanation:

Hello!

In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:

n=2.49g*\frac{1mol}{125.55 g}=0.0198mol

Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:

M=\frac{n}{V}\\\\V=\frac{n}{M}

By plugging in the moles and molarity, we obtain:

V=\frac{0.0198mol}{0.830mol/L}=0.0239L

Which in mL is:

V=0.0239L*\frac{1000mL}{1L}\\\\V=23.9mL

Best regards!

6 0
3 years ago
A 5 gram round ball has a density of 1.25 grams/milliliter. What is the volume of the round ball?
Arlecino [84]

Explanation:

m=5g

density=1.25g/ml

density=m/v

v=m/density=5/1.25

4 0
3 years ago
the half life for strontium-90 is 29 years. how many half-lives did the sample go through at the end of 87 year​
Nesterboy [21]
90 divided by 29 x 87 = 290
4 0
2 years ago
Read 2 more answers
Why is it important to have the International System of Units?
omeli [17]
<span>It can be used by scientists everywhere its important to have the International System of Units.</span>
4 0
3 years ago
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