What is responsible for morphogenesis

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

ale4655 [162]

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

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.

6 0

3 years ago

Assume that the reaction of aqueous hydrobromic acid solution and potassium hydroxide base undergoes a complete neutralization r

Tamiku [17]

Answer:

a.

$HBr + KOH → KBr + H _{2} O$

b.

$1000 \: ml \: contains \: 0.685 \: moles \\ 55.4 \: ml \: contains \: ( \frac{55.4 \times 0.685}{1000} ) \\ = 0.038 \:moles \\ 1 \: mole \: of \: hydrobromic \: acid \: produces \: 1 \: mole \: of \: water \\ 0.038 \: moles \: produce \: (0.038 \times 1) \\ = 0.038 \: moles \\ 1 \: mole \: of \: water \: weighs \: 18 \: g \\ 0.038 \: moles \: weighs \: (0.038 \times 18) \: g \\ = 0.684 \: g$

c.

0.042 M

d.

1.4

5 0

3 years ago

A chemist needs 0.550 mol selenium for a reaction. What mass of selenium should a chemist use?

-BARSIC- [3]

Answer:

The chemist would require to use 43.43 grams.

Explanation:

In order to solve this problem we need to know<u> how much do 0.550 moles of selenium weigh</u>. To do that we use selenium's<em> molar mass </em>and multiply it by the given number of moles:

0.550 mol * 78.96 g/mol = 43.43 g

The chemist would require to use 43.43 grams.

4 0

3 years ago

What are the 3 principles of the Kinetic molecular theory?

Sever21 [200]

Answer:

u already know the answer

Explanation:answer

6 0

3 years ago

Read 2 more answers

The increasing concentration of a chemical as it moves through the links in a food chain is called ________.

algol13

Answer: Biological Magnification

Explanation:

Organisms acquire toxic substance from the environment along with nutrients and water. Some of the toxins are metabolized and excreted, but others accumulate in specific tissues, especially fat. One of the reasons accumulated toxins are particularly harmful is that the become more concentrated in successive trophic level of the food web, this is the process of biological magnification.

Magnification occurs because the biomass at any given level is produced from a must larger biomass ingested from the level below. Thus the top-level carnivores tend to be the organism most severely affected by toxic compounds in the environment.

Examples of toxins that demonstrate biology magnification are chlorinated hydrocarbons, and many pesticides.

4 0

3 years ago