14)

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

Warm air masses tend to originate from:

Readme [11.4K]

Answer:

area where it's Summertime and tropical areas

Explanation:

8 0

3 years ago

Read 2 more answers

Which order shows the size of the inner planets from smallest to largest

elixir [45]

Answer:

PLUTO

MERCURY

MARS

VENUS

EARTH

NEPTUNE

URANUS

SATURN

JUPITER

Explanation:

4 0

3 years ago

Read 2 more answers

Calculate the freezing point and boiling point of a solution containing 14.2 gg of naphthalene (C10H8)(C10H8) in 111.0 mLmL of b

krek1111 [17]

Answer:

The freezing point of solution = -0.34 °C

The boiling point of solution = 82.98 °C

Explanation:

Step 1: Data given

Mass of naphthalene = 14.2 grams

Molar mass naphthalne = 128.17 g/mol

Volume of benzene = 111.0 mL

Density of benzene = 0.877 g/mL

Kf(benzene)=5.12°C/m

Freezing point benzene = 5.5 °C

Kb(benzene)=2.53°C/m

Boiling point benzene = 80.1 °C

Step 2: Calculate mass of benzene

Mass benzene = density * volume

Mass benzene = 0.877 g/mL * 111.0 mL

Mass benzene = 97.3 grams

Step 3: Calculate moles naphthalene

Moles naphthalene = mass naphthalene / molar mass napthalene

Moles napthalene = 14.2 grams / 128.17 g/mol

Moles naphthalene = 0.111 moles

Step 4: Calculate molality

Molality = moles naphthalene / mass benzene

Molality = 0.111 moles / 0.0973 kg

Molality = 1.14 molal

Step 5: Calculate freezing point of a solution

ΔT = i*kf*m

ΔT = 1 * 5.12 °C/m * 1.14 m

ΔT = 5.84 °C

ΔT = T(pure solvent) − T(solution)

The freezing point of solution = T pure - ΔT

5.5 - 5.84 = -0.34 °C

The freezing point is -0.34 °C

Step 6: Calculate boiling point of a solution

ΔT = i*kb*m

ΔT = 1 * 2.53 °C/m * 1.14 m

ΔT = 2.88 °C

ΔT = Tb (solution) - Tb (pure solvent)

The boiling point of solution = T pure + ΔT

The boiling point of solution = 80.1 °C + 2.88

The boiling point of solution = 82.98 °C

6 0

3 years ago

Any two substances which are odourless

STatiana [176]

Answer:

This question perplexes me somewhat. If we as humans are not able to detect a scent or odor from certain substances, that does not of necessity mean those substances are not giving off an odor, and are therefore not “truly” odorless, just odorless to the less sensitive noses of humans.

Water is supposedly odorless, but I know quite well of a test where subjects did a blind tasting, only instead of wine these people were tasting Australian waters, sourced from different locations around the country.

All of these tasters (professionals) were able to differentiate the waters and from what I recall, were all accurate. These wine buffs use their palates as well as their taste-buds, so, were there no odor, their task would have had a greater degree of difficulty.

My own baby sister has such a sensitive palate that she can tell you if you serve her a glass of water from the fridge whether that water was fresh, straight from the tap (unfiltered), filtered or had been boiled before and whether it had been boiled for more than five minutes or not. Truly.

So I would find it difficult to answer your question, “What substances are truly odorless?” , because I am not sure of the criteria you mean by “truly” odorless. In other words, If you are talking about zero degrees, are you talking of the freezing point of water, absolute zero or zero degrees Kelvin?

I’m sure there are substances that are odorless, to humans, but are those same substances “truly” odorless, or only to our perception of them?

Hope this helps, have a wonderful day/night, and stay safe!

8 0

3 years ago