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3 years ago

14

Can someone answer this question? It's not that long

1 answer:

Kitty [74]3 years ago

6 0

Answer:

sections 2 and 3

Explanation:

logic

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What is the main greenhouse gas responsible for influencing the world climate

cluponka [151]

Answer:

carbon dioxide

Explanation:

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3 years ago

Read the following chemical equation.

Anton [14]

Answer:

Bromine gains an electron

Explanation:

According to oxidation and reduction

8 0

3 years ago

Read 2 more answers

Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.

goblinko [34]

Answer:

Vapour pressure of cyclohexane at 50°C is 490torr

Vapour pressure of benzene at 50°C is 90torr

Explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

$P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}$

In the first solution:

$X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625$

$X_{benzene}=\frac{1.5mol}{2.5mol+1.5mol} =0.375$

$340torr = 0.625P^0_{A}+0.375P^0_{B}$ <em>(1)</em>

For the second equation:

$X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700$

$X_{benzene}=\frac{1.5mol}{3.5mol+1.5mol} =0.300$

$370torr = 0.700P^0_{A}+0.300P^0_{B}$ <em>(2)</em>

Replacing (2) in (1):

$340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})$

$340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}$

-122.5torr = -0.250P°A

$P^0_{A} = 490 torr$

<em>Vapour pressure of cyclohexane at 50°C is 490torr</em>

And for benzene:

$370torr = 0.700*490torr+0.300P^0_{B}$

$P^0_{B}=90torr$

<em>Vapour pressure of benzene at 50°C is 90torr</em>

3 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

50 POINTS for a legitimate response. Please answer the question, chose the correct unit conversions on the right. Please read th

makkiz [27]

Answer:

0.121 millimeters

5 0

2 years ago