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inysia [295]
3 years ago
10

Suppose a solution was too concentrated for an accurate reading with the spectrophotometer. The concentrated solution was dilute

d by placing 1.00 mL of the concentrated solution in 4.00 mL of water. The solution was then placed in the spectrophotometer and an absorbance was obtained and after a few calculations the molar concentration was calculated to be 3.5 × 10–6 M. what was the concentraion of the original solution
Chemistry
1 answer:
yan [13]3 years ago
5 0
Using the law of <span>dilution:

</span>initial Molarity = 3.5x10⁻⁶ M

<span>Initial volume = 4.00 mL
</span>
final Molarity = ??

final volume = 1.00 mL

Therefore:

Mi x Vi = Mf x Vf

(3.5x10⁻⁶) x 4.00 = Mf x 1.00

1.4x10⁻⁵ = Mf x 1.00

Mf = 1.4x10⁻⁵ / 1.00 = 

 1.4x10⁻⁵ M 


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Answer: sulfer

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Science has had no negative impacts on society. Please select the best answer from the choices provided T F
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The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10
LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

(4) The number of moles of CO_2 produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of CO_2 is C_2H_6

Explanation :

<u>Part 1 :</u>

First we have to calculate the number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

As, 52 kJ energy required amount of C_2H_6 = 1 g

So, 1000 kJ energy required amount of C_2H_6 = \frac{1000}{52}=19.23g

and,

As, 49 kJ energy required amount of C_4H_{10} = 1 g

So, 1000 kJ energy required amount of C_4H_{10} = \frac{1000}{49}=20.41g

<u>Part 2 :</u>

Now we have to calculate the number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

Molar mass of C_2H_6 = 30 g/mole

Molar mass of C_4H_{10} = 58 g/mole

\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles

and,

\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles

<u>Part 3 :</u>

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of C_2H_6 is:

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

and,

The balanced chemical reaction for combustion of C_4H_{10} is:

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

<u>Part 4 :</u>

Now we have to calculate the number of moles of CO_2 produced by burning each fuel to produce 1000 kJ.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

From this we conclude that,

As, 1 mole of C_2H_6 react to produce 2 moles of CO_2

As, 0.641 mole of C_2H_6 react to produce 0.641\times 2=1.28 moles of CO_2

and,

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

From this we conclude that,

As, 1 mole of C_4H_{10} react to produce 4 moles of CO_2

As, 0.352 mole of C_4H_{10} react to produce 0.352\times 4=1.41 moles of CO_2

So, the fuel that emitting least amount of CO_2 is C_2H_6

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3 years ago
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Answer:

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Explanation:

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From the options, the only compound in which the central atom does not meet the octet rules is BH₃. The central atom is boron (B), which has 3 electrons in its valence shell. When B is combined with hydrogen (H), 3 electrons from the 3 atoms of H are added. The total amount of electrons is 6, fewer than 8 electrons needed to meet the rule.

hope this helps

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2 years ago
A 350.0 ml sample of gas has a temperature of 22.7 C and a pressure of 0.914 atm. What temperature would be needed for the same
Rus_ich [418]

Answer:

Explanation:

We shall apply gas law formula

P₁ V₁ / T₁ = P₂V₂ / T₂

.914  x 350 / ( 273 + 22.7 ) = 1 x 220 / T₂

1.0818 = 220 / T₂

T₂ = 203.36 K

= - 69.64 ⁰ C

7 0
3 years ago
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