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Alex
2 years ago
15

When carbon-containing compounds are burned in a limited amount of air, some CO(g) as well as CO₂(g) is produced. A gaseous prod

uct mixture is 35.0 mass % CO and 65.0 mass % CO₂. What is the mass % of C in the mixture?
Chemistry
1 answer:
solong [7]2 years ago
3 0

Here, we are going to calculate the mass % of C in the mixture.

What is a Mixture?

A mixture is composed of one or more pure substances in varying composition. There are two types of mixtures: heterogeneous and homogeneous. Heterogeneous mixtures have visually distinguishable components, while homogeneous mixtures appear uniform throughout.

Given that,

The mass % of CO =35.0% =35.0 g in 100 g mixture

The mass % of CO2 = 65% =65 g in 100 g mixture

Therefore,

The mass of C from CO = 15.007 g C

Similarly,

The mass of C from CO2 = 17.738 g C

Thus, the total mass of C = 15.007 g+17.738 g =32.745 g

Therefore,

The mass % of C= 32.745% =32.7%

Thus, the mass % of C in the mixture is 32.7%

To learn more about carbon-containing compounds click on the link below:

brainly.com/question/13381262

#SPJ4

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Answer:

Acid-base indicators are chemicals used to determine whether an aqueous solution is acidic, neutral, or alkaline. Because acidity and alkalinity relate to pH, they may also be known as pH indicators. Examples of acid-base indicators include litmus paper, phenolphthalein, and red cabbage juice.

Explanation:

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HeLp!!!
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Answer:

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  2. Submarines can be detected by using MADs
  3. The history of MAD development

In 1917 the interest in the detection of submarines started with the study hydrophones. Then in 1918, the U.S. considered to use the magnetism in this area but it did not result to be practical as it had a limited detection range.

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Due to the limited range and its lack of ability to detect the magnetic variance from different sources, MAD started to be used in combination with sonobuoys. This combination allowed an aircraft to localize submarines with the confirmation of sonobuoys.

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4 years ago
Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH. When the two solutions are
kolbaska11 [484]

Explanation:

Molarity of copper sulfate solution = 1.00 M

Volume of the copper sulfate solution  = 50.0 mL = 0.050 L

Moles of copper sulfate = n

1.00M=\frac{n}{0.050 L}

n = 0.050 L × 1.00 M= 0.050 mol

1 mol of copper sulfate has 1 mol of copper . Then 0.050 mol of copper sulfate has :

1\times 0.050 mol=0.050 mol of copper

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b) The identity of the compound which formed after the reaction is copper hydroxide.

c) The complete equation for the reaction that occurs when copper sulfate and potassium hydroxide are mixed:

CuSO_4(aq)+2KOH (aq)\rightarrow Cu(OH)_2(s)+K_2SO_4 (aq)

d) CuSO_4(aq)\rightarrow Cu^{2+}(aq) +SO_{4}^{2-}(aq)..[1]

KOH (aq)\rightarrow 2K^+(aq) +OH^-(aq)..[2]

Cu^{2+}(aq) +SO_{4}^{2-}(aq)+2K^+(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)+SO_{4}^{2-}(aq)+2K^+(aq)

Common ion both sides are removed. The net ionic equation is given as:

Cu^{2+}(aq) +2OH^-(aq)\rightarrow Cu(OH)_2(s)(aq)

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Mass of final solution ,m= 1 mL

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m=d\time V=1 g/ml\times 100 mL = 100 g

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Change in temperature of the solution,ΔT = 27.7 °C- 21.5 °C=6.2°C

Q=mc\Delta T

Q=100 g\times 4.186 J/g ^oC\times 6.2^oC=2595.32 J=2.595 kJ

Enthalpy of the reaction = ΔH = \frac{Q}{\text{Moles of copper}}

ΔH = \frac{2.595 kJ}{0.050 mol}=51.90 kJ/mol

The ΔH for the reaction that occurs on mixing is 51.90 kJ/mol.

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Filling in the gaps in the excerpt below

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Hence we can conclude that the answers to your questions are as listed above.

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