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asambeis [7]
3 years ago
12

In the combustion of gasoline, if it is calculated that 250.15 g of CO2 will be produced, but only 212.48 g are collected, what

is the reaction's efficiency?
Chemistry
1 answer:
S_A_V [24]3 years ago
8 0

Answer:

To calculate the Carbon Dioxide - CO 2 - emission from a fuel, the carbon content of the fuel must be multiplied with the ratio of molecular weight of CO 2 (44) to the molecular weight of Carbon (12) -> 44 / 12 = 3.7 Emission of CO 2 from combustion of some common fuels are indicated in the table

Explanation:

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Hey can someone help me and show work please
Ivahew [28]

Answer: 2.54g

Explanation:

Molar Mass of H2O2 = (2x1) + (2x16) = 34g/mol

1mole (34g) of H2O2 contains 6.02x10^23 molecules

Therefore Xg of H2O2 will contain 4.5x10^22 molecules i.e

Xg of H2O2 = (34x4.5x10^22)/6.02x10^23 = 2.54g

5 0
3 years ago
For each of the reactions provided, write a balanced equation. Then, using your knowledge of reaction types and the activity ser
pshichka [43]

Answer:

a.monadioxide

Explanation:

for maintaince of different gases in the air to separate reactions of chemicals

5 0
3 years ago
As a liquid is added to a beaker, the pressure exerted by the liquid on the bottom of the beaker does what
ololo11 [35]
So if you put a holo ball of paper in the water it sinks and as it sinks it gets smaller cuse pressure so it cuseus things to shrink cuse of pressure
5 0
3 years ago
(1 pt) You have 250.0 mL of 0.136 M HCl. Using a volumetric pipet, you take 25.00 mL of that solution, place it in a 100.00 ml v
alexdok [17]

Answer:

0.034M HCl is the concentration of the diluted solution

Explanation:

You take, initially, 25.00mL of the 0.136M HCl. Then, you dilute the solution to 100.00mL. The solution is diluted:

100.00mL / 25.00mL = 4. The solution was diluted 4 times.

That means the concentration of the diluted solution is:

0.136M / 4 =

<h3>0.034M HCl is the concentration of the diluted solution</h3>
7 0
3 years ago
Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

3 0
3 years ago
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