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rosijanka [135]
2 years ago
8

A student used 1.2 g of hydrate in the first trial and 1.6 g in the second trial. Explain fully why her mass percent of water wi

ll be the same for both trials even though she used more hydrate in the second trial. (Do not simply state that the mass percent is an intensive property ).
Chemistry
1 answer:
Pachacha [2.7K]2 years ago
8 0

Both trials of 1.2 g and 1.6 g will have the same mass percent of water because the ratio of the salt to the water of hydration is always constant for any hydrated salt.

<h3>Water of hydration</h3>

For every hydrated salt, the ratio of the salt to the water of hydration remains constant irrespective of the amount of salt taken for experimental analysis.

For example, assuming the mass percent of water in 10g of a hydrated salt is 40%, if 100g of the same salt is taken, the mass percent will remain 40%.

More on water of hydration can be found here: brainly.com/question/11202174

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Phantasy [73]
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Hope that helps!!!!
8 0
3 years ago
Question 14 A student dissolves 1.5g of styrene C8H8 in 225.mL of a solvent with a density of 1.02/gmL . The student notices tha
Yuki888 [10]

We must to know:

Cm = molarity = niu / Vs, when the niu = no. of moles and Vs = Volume of solution

the no. niu = mass / molecular mass of substance

molecular mass of C8H8 = 12x8+8x1 = 104 g/mol

=> niu = 1,5 / 104 = 0,0144 moles C8H8

=> Cm = 0,0144/0,225 = 0,06 mol/L

Cmm = molality = niu (C8H8) / mass of solvent (kg)

=> p = mass / V => mass (solvent) = p x V

=> 225 x 1,02 = 229,5 g solvent = 0,2295 kg solvent

=> Cmm = 0,0144 / 0,229,5 = 0,063

8 0
3 years ago
Read 2 more answers
How many miles can you drive, if your car gets 21.3 miles per gallon and you have 12.0 gallons of gas?
Travka [436]

Answer:

255.6

Explanation:

If you have 12 gallons and get 21.3mpg,

-Multiply 21.3 by 12

-you can travel 255.6 miles before running out of gas.

-If you need to estimate, round up to 256 miles.

6 0
4 years ago
A Wisconsin farmer taps a maple tree and collects 200 liters of sap. He determines the sucrose (C12H2011) concentration to be 0.
Minchanka [31]

Answer:

6.0 L

Explanation:

Use the dilution equation M1V1 = M2V2

M1 = 0.075 M

V1 = 200 L

M2 = 2.5 M

V2 = ?

Solve for V2 --> V2 = M1V1/M2

V2 = (0.075 M)(200 L) / (2.5 M) = 6.0 L

4 0
3 years ago
What is the percent composition by mass of hydrogen in nh4hco3
Simora [160]

<u>Answer:</u> The percent composition by mass of hydrogen in given compound is 6.33 %

<u>Explanation:</u>

We are given:

A chemical compound having chemical formula of NH_4HCO_3

It is made up by the combination of 1 nitrogen atom, 5 hydrogen atoms, 1 carbon atom and 3 oxygen atoms

To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:

\%\text{ composition of Hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of compound}}\times 100

Mass of compound = [(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol

Mass of hydrogen = (5\times 1)=5g/mol

Putting values in above equation, we get:

\%\text{ composition of Hydrogen}=\frac{5g/mol}{79g/mol}\times 100=6.33\%

Hence, the percent composition by mass of hydrogen in given compound is 6.33 %

6 0
3 years ago
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