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gavmur [86]
2 years ago
7

Part B

Chemistry
1 answer:
iragen [17]2 years ago
8 0

Answer:

experiment can show the movement of pollutants through the groundwater!

Explanation:

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What mass of mgf2 is contained in 80.85 g of a 22.4% by mass solution of mgf2 in water?
Aleks [24]
Formula: % by mass = (mass of solute / mass of solution] *100

Data:

mass of solution = 80.85 g

% by mass = 22.4%

Unknown = mass of solute

Solution

% by mass = (mass of solute / mass of solution] *100 = >

mass of solute = % by mass * mass of solution / 100

mass of solute = 22.4 * 80.85 / 100 = 18.11 g

Answer: 18.11 g
4 0
3 years ago
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Differentiate the types of intermolecular forces
marshall27 [118]

Answer:

  1. Dipole interactions
  2. London dispersion forces
  3. Hydrogen bonds

Credit goes to: chem.libretexts.org

6 0
3 years ago
A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out
scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol

  • The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = \frac{2}{1}\times 0.0021=0.0042mol of potassium bromide

  • Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g

  • To calculate the percentage composition of KBr in the mixture, we use the equation:

\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0
3 years ago
How has the Chesapeake Bay watershed changed over time?
Alisiya [41]

Answer:

The Chesapeake Bay watershed continues to develop as population in the region grows. Development and urbanization at the cost of natural landscapes can lead to increased pollution of nutrients and sediment to the Bay, especially from stormwater runoff.

Explanation:

5 0
2 years ago
An unsaturated solution is formed when 80 grams of a salt is dissolved in 100 grams of water at 40. This salt could be ?
LekaFEV [45]
A simple way to go about this is that we look at the solubility curve, on the x axis we first look at the temperature and then the corresponding value of solute/100g H2O on the y axis, from the 4 curves above only NaNO3 has a curve that can accommodate  80g of salt at 40 without being Saturated since at 40 degrees it can accommodate 105g of salt to become completely Saturated.
7 0
2 years ago
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