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STALIN [3.7K]
3 years ago
15

A student combines a solution of NaCl(aq) with a solution of AgNO3(aq), and a precipitate forms. Assume that 50.0mL of 1.0MNaCl(

aq) and 50.0mL of 1.0MAgNO3(aq) were combined. According to the balanced equation, if 50.0mL of 2.0MNaCl(aq) and 50.0mL of 1.0MAgNO3(aq) were combined, the amount of precipitate formed would:_________
Chemistry
1 answer:
gtnhenbr [62]3 years ago
8 0

Answer:

Based on the given information, the balanced equation is:

NaCl + AgNO3 = AgCl + NaNO3

Now the moles present in 50 ml of 1M NaCl is,

= 50 * 1/1000 mole = 0.05 mole

And the moles present in 50 ml of 1 M AgNO3 is,

= 50*1/1000 mole = 0.05 mole

Therefore, 0.05 mole of NaCl combines with 0.05 mole of AgNO3 to precipitate 0.05 mole of AgCl.

Now in the second case, to balance the chemical equation, 50 ml of 2 M NaCl combines with 50 ml of 1 M AgNO3. So, the moles present in 50 ml of 2 M NaCl will be,

= 50*2/1000 mole = 0.1 mole

However, the amount of AgNO3 in the second case is not changing, therefore, the amount of AgCl precipitated will be,

= 0.1 - 0.05 = 0.05 mole

Therefore, the amount of precipitate would not change as there is no change in the amount of AgNO3.

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2 years ago
Two descriptions for a sub-atomic particle are listed below:
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Can not be D or B and A is not enough so C is right.
4 0
3 years ago
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What would you observe in a solution of 30 g of KC1O3 in 100 g of water at 10 c?
Mrac [35]

Answer:

Option A is correct. About 5 g of the KClO3 is dissolved

Explanation:

KClO3 is not very good soluble in water.

So, Option C is impossible, because KClO3 is poorly soluble in water.

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7 0
3 years ago
A student has 100. mL of 0.400 M CuSO4 (aq) and is asked to make 100. mL of 0.150 M CuSO4 (aq) for a spectrophotometry experimen
GenaCL600 [577]

Answer:

We have to take 37.5 mL of a 0.400 M solution

Explanation:

Step 1: Data given

Stock volume = 100 mL  = 0.100L

Stock concentration 0.400 M

Volume of solution he wants to make = 100 mL = 0.100L

Concentration of solution he wants to make = 0.150 M

Step 2: Calculate the volume of 0.400 M CuSO4 needed

C1*V1 = C2*V2

⇒with C1 = the stock concentration = 0.400M

⇒with V1 = the volume of the stock = TO BE DETERMINED

⇒with C2 = the concentration of the solution he wants to make = 0.150 M

⇒with V2 = the volume of the solution made = 0.100 L

0.400 M * V1 = 0.150M * 0.100L

V1 = (0.150M*0.100L) / 0.400 M

V1 = 0.0375 L = 37.5 mL

We have to take 37.5 mL of a 0.400 M solution

5 0
3 years ago
Read 2 more answers
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