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fiasKO [112]
2 years ago
9

What volume of water would be added to 16.5 ml of a 0.0813 m solution of sodium borate in order to get a 0.0200 m s?

Chemistry
1 answer:
raketka [301]2 years ago
7 0
When diluting solutions from concentrated solutions the following formula can be used 
c1v1 = c2v2 
where c1 is concentration and v1 is volume of the concentrated solution 
and c2 is concentration and v2 is volume of the diluted solution to be prepared
substituting these values 
0.0813 M x 16.5 mL = 0.0200 M x V
V = 67.1 mL
the volume of the diluted solution prepared is 67.1 mL.
the volume of water that should be added to get a final volume of 67.1 mL is (67.1 - 16.5 ) = 50.6 mL
a volume fo 50.6 mL should be added 
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A propane stove burned 470 grams propane and produced 625 grams of water (this is the actual yield) C3H8 +5O2=3CO2+4H20. What wa
Liula [17]

Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

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3 years ago
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You have to figure out a way to write the two unknown abundances in terms of one variable.

The total abundance is 1 (or 100%). So if you say the abundance for the first one is X then the abundance for the second one has to be 1-X (where X is the decimal of the percentage so say 0.8 for 80%).

203(X) + 205(1-X) = 204.4

Then you just solve for X to get the percentage for TI-203.
And then solve for 1-X to get the percentage for TI-205.

After that the higher percentage would be the most abundant.

203x + 205 - 205x = 204.4
-2x + 205 = 204.4
-2x = -0.6

x = 0.3
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Then the TI-205 would have the highest percentage and would be the most abundant.
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