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sleet_krkn [62]
3 years ago
12

The table shows the chemical formulas for four substances which substances have the same number of carbon atoms

Chemistry
1 answer:
marissa [1.9K]3 years ago
7 0

Answer:

Substance 1 and 3.

Explanation:

The following data were obtained from the question:

Substance >>>>> chemical formula

1 >>>>>>>>>>>>>> C2H6O

2 >>>>>>>>>>>>>> C8H18

3 >>>>>>>>>>>>>> CH3CH2Br

4 >>>>>>>>>>>>>> C4H10O

Next, we shall determine the number of carbon in each substance as follow:

Substance 1 (C2H6O):

Number of carbon atom = 2

Substance 2 (C8H18)

Number of carbon atom = 8

Substance 3 (CH3CH2Br)

Number of carbon atom = 1 + 1 = 2

Substance 4 (C4H10O)

Number of carbon atom = 4

From the above illustrations,

Substance 1 and 3 has the same number of carbon atom

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A chemist dissolves 484 .mg of pure perchloric acid in enough water to make up 240.mL of solution. Calculate the pH of the solut
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Answer:

1.70

Explanation:

The molar mass of perchloric acid is 100.46 g/mol. The moles corresponding to 484 mg (0.484 g) are:

0.484 g × (1 mol/100.46 g) = 4.82 × 10⁻³ mol

4.82 × 10⁻³ moles are dissolved in 240 mL (0.240 L) of solution. The molar concentration of perchloric acid is:

4.82 × 10⁻³ mol/0.240 L = 0.0201 M

Perchloric acid is a strong monoprotic acid, that is, it dissociates completely, so [H⁺] = 0.0201 M.

The pH is:

pH = -log [H⁺] = -log 0.0201 = 1.70

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3 years ago
Calculate the molar mass of butane. Butane's rate of diffusion is 3.8 times slower than that of hleium.
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Answer:

58 g/mol

Explanation:

According to Graham's law, the rate of diffusion of a gas (r) is inversely proportional to the square root of its molar mass (M). Butane's rate of diffusion is 3.8 times slower than that of helium, that is, rButane = rHe/3.8, or rHe/rButane = 3.8. Then,

\frac{rHe}{rButane} =3.8=\sqrt{\frac{M(Butane)}{M(He)} } =\sqrt{\frac{M(Butane)}{4.00g/mol} } \\M(Butane)=3.8^{2} \times 4.00g/mol\\M(Butane)=58g/mol

8 0
3 years ago
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Describe the electrolytic process for refining copper.
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Explanation:

As the anodes dissolve away, the cathodes on which the pure metal is deposited grow in size. The impurities in the copper anode include lead, zinc, nickel, arsenic, selenium, and several precious metal including gold and silver.

3 0
3 years ago
When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant so
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Answer: 54.4 kJ/mol

Explanation:

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.0M\times 0.05=0.05mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.0\times 0.05L=0.05mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 mole

Now we have to calculate the mass of water:

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 50ml+50ml=100ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 100ml=100g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 100 g

T_{final} = final temperature of water = 27.5^0C

T_{initial} = initial temperature of metal = 21.0^0C

Now put all the given values in the above formula, we get:

q=100g\times 4.18J/g^oC\times (27.5-21.0)^0C

q=2719.6J=2.72kJ

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of HCl:

0.05 moles of HCl releases heat = 2.72 KJ

1 mole of HCl releases heat =\frac{2.72}{0.05}\times 1=54.4KJ

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ

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