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3 years ago

7

An experiment produced 0.10 g CO2 with a volume of 0.056 L at STP. If the accepted density of CO2 at STP is 1.96 g/L, what is th

e approximate percent error?

1 answer:

Mandarinka [93]3 years ago

3 0

The experimental density of CO2 at STP is 0.10/0.056=1.78 g/L. The percent error equals to (1.96-1.78)/1.96*100%=9.18%. So the answer is 9.18%.

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According to the law of constant composition, how many grams of oxygen does this isolated sample contain?

KatRina [158]

The Mass of oxygen in isolated sample is 8.6 g

<h3>What is the Law of Constant composition?</h3>

The law of constant composition states that pure samples of the same compound contain the same element in the same ratio by mass irrespective of the source from which the compound is obtained.

Considering the given ascorbic acid samples:

Laboratory sample contains 1.50 gg of carbon and 2.00 gg of oxygen

mass ratio of oxygen to carbon is 2 : 1.5

Isolated sample will contain 2/1.5 * 6.45 g of oxygen.

Mass of oxygen in isolated sample = 8.6 g

In conclusion, the mass of oxygen is determined from the mass ratio of oxygen and carbon in the compound.

Learn more about the Law of Constant composition at: brainly.com/question/1557481

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Note that the complete question is given below:

A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains 1.50 g of carbon and 2.00 g of oxygen. Another sample of ascorbic acid isolated from citrus fruits contains 6.45 gg of carbon. According to the law of constant composition, how many grams of oxygen does this isolated sample contain?

Express the answer in grams to three significant figures.

8.47 g

8 0

2 years ago

0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati

erastova [34]

Answer:

$[Ag^{+}]=4.2\times 10^{-2}M$

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

$BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ba^{2+}][CrO_{4}^{2-}]$

$1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]$

$[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}$

Now,

$Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]$

$1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})$

$[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}$

$[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M$

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

7 0

3 years ago

The pOH of a solution is 9.21. Calculate the hydrogen ion concentration of the solution. Be sure to report your answer to the co

11Alexandr11 [23.1K]

Answer:

[OH-] = 6.17 *10^-10

Explanation:

Step 1: Data given

pOH = 9.21

Step 2: Calculate [OH-]

pOH = -log [OH-] = 9.21

[OH-] = 10^-9.21

[OH-] = 6.17 *10^-10

Step 3: Check if it's correct

pOH + pH = 14

[H+]*[OH-] = 10^-14

pH = 14 - 9.21 = 4.79

[H+] = 10^-4.79

[H+] = 1.62 *10^-5

6.17 * 10^-10 * 1.62 * 10^-5 = 1* 10^-14

3 0

3 years ago