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2 years ago

What is the final volume?

Chemistry

1 answer:

zaharov [31]2 years ago

6 0

Answer:

Option A. 9.4 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 8 L

Initial temperature (T₁) = 293 K

Final temperature (T₂) = 343 K

Final volume (V₂) =?

V₁ / T₁ = V₂ / T₂

8 / 293 = V₂ / 343

Cross multiply

293 × V₂ = 8 × 343

293 × V₂ = 2744

Divide both side by 293

V₂ = 2744 / 293

V₂ = 9.4 L

Therefore, the final volume of the gas is 9.4 L

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Choose the reaction that represents the combustion of C6H12O2.

Leona [35]

Answer:

The answer to your question is: letter D

Explanation:

In a combustion reaction, the reactants are always a molecule with Carbon that reacts with oxygen and the products are carbon dioxide and water.

According to the explanation, the only possible solution is:

a) C₆H₁₂O₂(l) ⇒ 6 C(s) + 6 H₂(g) + O₂(g)

b) Mg(s) + C₆H₁₂O₂(l) ⇒ MgC₆H₁₂O₂(aq)

c) 6 C(s) + 6 H₂(g) + O₂(g) ⇒ C₆H₁₂O₂(l)

d) C₆H₁₂O₂(l) + 8 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(g)

e) None of the above represent the combustion of C₆H₁₂O₂.

4 0

2 years ago

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What are the radicals?Write their types with four examples.

polet [3.4K]

Answer:

<h3>A group of atoms behaving as a unit in a number of compounds. </h3>

Explanation:

<h2>Sorry for giving half answer only .I known this much only.</h2><h2>I hope U like my answer and my answer help u.</h2>

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3 years ago

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Plz help me solve this question is it A,B,C or D

Rashid [163]

Answer:

B - To increase the rate of the reaction

Explanation:

Catalysts speed up the reaction without being reactants or products, so aren't used up in the reaction.

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2 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

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3 years ago

Look at the rock sitting on the hill in the picture above. Gravity should make the rock slide down the hill. What force is actin

nexus9112 [7]

I believe the answer you are looking for is Static Friction. Static Friction is the force that holds an object in place until it starts to move. Then it switches to rolling friction.

For example, if you have a 1/2 ton truck sitting in front of you and the truck is in neutral. (meaning it can roll if pushed). The truck is extremely hard to move at first. That is because static friction is holding it in place until the amount of force exceeds the limit of static friction.

So if we continue to push at the truck and you feel it starting to move, then once it starts moving it is much easier to push, that is because we moved past static friction to rolling friction. Rolling friction is what helps slow things down. If you roll a ball across a carpet floor it eventually comes to a stop.

5 0

3 years ago

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