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Greeley [361]
2 years ago
13

g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol

ution to yield 0.904 g of precipitate, what is the molarity of lead(II) ion in the original solution
Chemistry
1 answer:
Ronch [10]2 years ago
3 0

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

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If evaporation occurs what will change 0.70 molar solution of CuSO4
faust18 [17]

Answer:

A. Molarity will increase .

Explanation:

Molarity = moles of solute per litre of solution

= moles of solute / volume of solution

If evaporation occurs , volume of solution decreases and moles of solute remains constant . Hence denominator decreases and numerator remains constant .

Hence the molarity increases .

3 0
3 years ago
In another experiment, if 80 xo3 molecules react with 104 brz3 molecules how many br2 molecules will be produced which reactant
BaLLatris [955]

This is an incomplete question, here is a complete question.

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

In another experiment, if 80 XO_3 molecules react with 104 BrZ_3 molecules. How many Br_2 molecules will be produced which reactant will be used up in the reaction.

Answer : The number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

Explanation :

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

First we have to determine the limiting reagent.

From the balanced reaction we conclude that,

As, 8 molecules of BrZ_3 react with 6 molecule of XO_3

So, 104 molecules of BrZ_3 react with \frac{104}{8}\times 6=78 molecule of XO_3

From this we conclude that, XO_3 is an excess reagent because the given moles are greater than the required moles and BrZ_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the molecules of Br_2

From the reaction, we conclude that

As, 8 molecules of BrZ_3 react to give 4 molecules of Br_2

So, 104 molecules of BrZ_3 react to give \frac{104}{8}\times 4=52 molecules of Br_2

Hence, the number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

4 0
3 years ago
How can you determine the frequency of a wave when the wavelength is known?
Korvikt [17]
Frequency = velocity (aka speed) / wave length
3 0
3 years ago
What is the volume of 4.78g of O2 gas at STP?
My name is Ann [436]

Answer:

Explanation:

The trick here is to realize that if you know the volume of a gas at STP, you can use the fact that

1

mole of any ideal gas occupies

22.7 L

under STP conditions to calculate how many moles of gas you have in your sample.

Under STP conditions:

1 mole of an ideal gas = 22.7 L

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

In your case, you know that your sample of gas occupies

2.28 L

under STP conditions, which are currently defined as a pressure of

100 kPa

and a temperature of

0

∘

C

.

This means that your sample will contain

2.28

L

⋅

molar volume of a gas at STP



1 mole gas

22.7

L

=

0.10044 moles gas

Now, the molar mass of the gas is the mass of exactly

1

mole of the gas. In your case, you know that you get

3.78 g

for every

0.10044

moles, which means that you have

1

mole

⋅

3.78 g

0.10044

moles

=

37.6 g

Since this is the mass of

1

mole of gas, you can say that the molar mass of the gas is

molar mass = 37.6 g mol

−

1

−−−−−−−−−−−−−−−−−−−−−−−

8 0
2 years ago
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