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Inessa05 [86]
3 years ago
7

what is the molarity of 20.0 ml of a KCl solution that reacts completely with 30.0 ml of a 0.400 M Pb(NO3)2 solution?

Chemistry
1 answer:
Alex17521 [72]3 years ago
7 0

Answer:

[KCl] = 1.2 M

Explanation:

We need to complete the reaction:

2KCl(aq) + Pb(NO₃)₂(aq)  → 2KNO₃(aq) + PbCl₂(s)↓

By stoichiomety we know that 1 mol of chloride needs 1 mol of nitrate to react:

Let's find out the moles of nitrate, we have:

Molarity = mol/volume(L)

We convert the volume → 30 mL . 1L/1000mL = 0.030L

Molarity . volume(L) = moles → 0.400 M . 0.030L = 0.012 moles

Therefore, we can make a rule of three.

1 mol of nitrate reacts with 2 moles of chloride

Then, 0.012 moles of nitrate must react with (0.012 . 2) / 1 = 0.024 moles of KCl

We convert the volume from mL to L → 20 mL . 1L /1000mL = 0.020L

Molarity = mol /volume(L) → 0.024 mol /0.020L = 1.2 M

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Explanation:

In order to justify Marian's statement we have to look at the '' Amount per serving calories'' ⇒

In food label A we can see that this value is 160 calories

In food label B we can see that this value is 50 calories

⇒ 160 calories is slightly more than three times 50 calories

Otherwise If we want to justify Johan statement we need to look at the '' serving size '' ⇒

In food label A we can see that the serving size is 1 cup (237 mL)

In food label B we can see that the serving size is \frac{1}{4} cup (56g)

Working with the information of the food label A we can write the following expression :

\frac{1Cup}{160calories}=\frac{\frac{1}{4}Cup}{x}  ⇒ Looking at the value of ''x'' ⇒

x=160calories(\frac{1}{4})=40calories

x=40calories

If we look at the same amount of portion volume :

In \frac{1}{4} cup of food A we have 40 calories

In \frac{1}{4} cup of food B we have 50 calories

We could conclude that Food B has more calories.

That's how both claims could both be justified.

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4 years ago
What is oxidation number of S in H2SO5 ???​
balandron [24]

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\large\underbrace\red{♡+6 :/is :/the :/oxidation:/ number}

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4 0
3 years ago
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Consider the reaction of KOH with H3PO4 to form K3PO4 and H2O. If 5.38 g H3PO4 is reacted with excess KOH and 7.97 g of K3PO4 is
balu736 [363]

Answer:75%

Explanation:

First, the balanced reaction equation must be written out clearly as a guide to solving the problem. The molar masses of H3PO4 and K3PO4 are then calculated as they will be consistently required in solving the problem. The theoretical yield is obtained from the amount of H3PO4 reacted. Since 1 mole of H3PO4 yields 1 mole of K3PO4, 0.05 moles of H3PO4 yields 0.05 moles of K3PO4. The mass of K3PO4 is produced is then the product of 0.05 and it molar mass hence the theoretical yield. The % yield is calculated as shown.

7 0
3 years ago
What is the abbreviation for the unit nanometers?
Mila [183]

Answer: nn

Explanation:

The nanometre (international spelling as used by the International Bureau of Weights and Measures; SI symbol: nm) or nanometer (US spelling) is a unit of length in the metric system, equal to one billionth (short scale) of a metre (0.000000001 m).

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3 years ago
8. What was the original concentration in the BHL sample, if the dilution is 1:500 and the concentration 0.07 mg/ml
Yakvenalex [24]

Answer:

The original concentration is "35 mg/ml".

Explanation:

According to the question,

The solution is diluted,

= 1:50

The initial volume,

V1 = 1 ml

Final concentration,

= 0.07 mg

then,

The final volume,

V2 = 500 ml

As we know,

⇒ V_1N_1=V_2N_2

or,

⇒ N_1=\frac{V_2N_2}{V_1}

On substituting the values, we get

⇒       =\frac{500\times 0.07}{1}

⇒       =\frac{35}{1}

⇒       =35 \ mg/ml

4 0
3 years ago
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