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3 years ago

A sample of a substance that has a density of 0.844 g/mL has a mass of 0.549 g. Calculate the volume of the sample.

Chemistry

1 answer:

Virty [35]3 years ago

7 0

D = m / V

0.844 = 0.549 / V

V = 0.549 / 0.844

V = 0.650 mL

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Question 4 need this answered asap (chemistry)

aleksandrvk [35]

Answer:

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3 years ago

A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o

HACTEHA [7]

Answer:

1.099 gmL¯¹ ≈ 1.1 gmL¯¹

Explanation:

From the question given above, the following were obtained:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

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Concentration of solution =?

Next, we shall determine the mass of sodium hydroxide, NaOH. This can be obtained as as illustrated below:

Mass of empty cylinder = 9.5 g

Mass Cylinder + NaOH = 31.92 g

Mass of NaOH =?

Mass of NaOH = (Mass Cylinder + NaOH) – (Mass of empty cylinder)

Mass of NaOH = 31.92 – 9.5

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Finally, we shall determine concentration of the solution as follow:

Mass of NaOH = 22.42 g

Volume of solution = 20.4 mL

Concentration of solution =?

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Concentration of solution = 22.42 / 20.4

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4 years ago

An asteroid in space has traveled 4,500 km in 60 s, what is the average speed of the asteroid?

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3 years ago

400. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution

ch4aika [34]

<u>Answer:</u> The molecular weight of protein is $1.14\times 10^2g/mol$

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

or,

$\pi=i\times \frac{m_{solute}\times 1000}{M_{solute}\times V_{solution}\text{ (in mL)}}}\times RT$

where,

$\pi$ = Osmotic pressure of the solution = 0.0861 atm

i = Van't hoff factor = 1 (for non-electrolytes)

$m_{solute}$ = mass of protein = 400 mg = 0.4 g (Conversion factor: 1 g = 1000 mg)

$M_{solute}$ = molar mass of protein = ?

$V_{solution}$ = Volume of solution = 5.00 mL

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[25+273]K=298K$

Putting values in above equation, we get:

$0.0861atm=1\times \frac{0.4g\times 1000}{M\times 100}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\M=1136.62g/mol=1.14\times 10^2g/mol$

Hence, the molecular weight of protein is $1.14\times 10^2g/mol$

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3 years ago