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3 years ago

Which tool was most likely used in a procedure if the lab report shows that approximately 300 mL of water was used ?

Chemistry

2 answers:

pogonyaev3 years ago

6 0

100 mL graduated cylinders

Travka [436]3 years ago

4 0

Graduated Cylinder is the answer to this.

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Formed when chemicals in the air get into rain and up the acidity levels

snow_lady [41]

Acid Rain is formed when chemicals in the air get into rain and up the acidity levels!!!!!!

Hope this helps guys!

4 0

3 years ago

What is the mass of solute in a 500 mL solution of 0.200 M

Fofino [41]

16.4 grams is the mass of solute in a 500 mL solution of 0.200 M .

sodium phosphate

Explanation:

Given data about sodium phosphate

atomic mass of Na3PO4 = 164 grams/mole

volume of the solution = 500 ml or 0.5 litres

molarity of sodium phosphate solution = 0.200 M

The formula for molarity will be used here to know the mass dissolved in the given volume of the solution:

The formula is

molarity = $\frac{number of moles of solute}{volume in litres}$

putting the values in the equation, we get

molarity x volume = number of moles

0.200 X 0.5= number of moles

number of moles = 0.1 moles

Atomic mass x number of moles = mass

putting the values in the above equation

164 x 0.1 = 16.4 grams

16.4 grams of sodium phosphate is present in 0.5 L of the solution to make a 0.2 M solution.

8 0

3 years ago

What is the mass in grams of 4.91 moles of ammonium sulfate?

Montano1993 [528]

1 mole is equal to 1 moles Ammonium Sulfate, or 132.13952 grams

4 0

3 years ago

Read 2 more answers

Calculate the freezing point and boiling point of a solution containing 8.15 g of ethylene glycol (C2H6O2) in 96.3 mL of ethanol

pishuonlain [190]

Answer: The freezing point of solution is -117.54°C and the boiling point of solution is 80.48°C

Explanation:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.789 g/mL

Volume of ethanol = 96.3 mL

Putting values in above equation, we get:

$0.789g/mL=\frac{\text{Mass of ethanol}}{96.3mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 96.3mL)=75.98g$

Calculating the freezing point:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = -114.1 °C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.99°C/m

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$-114.1-\text{Freezing point of solution}=1\times 1.99^oC/m\times \frac{8.15\times 1000}{62g/mol\times 75.98}\\\\\text{Freezing point of solution}=-117.54^oC$

Hence, the freezing point of solution is -117.54°C

Calculating the boiling point:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

Boiling point of pure solution = 78.4°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_b$ = molal boiling point elevation constant = 1.20°C/m.g

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-78.4=1\times 1.20^oC/m\times \frac{8.15\times 1000}{62\times 75.98}\\\\\text{Boiling point of solution}=80.48^oC$

Hence, the boiling point of solution is 80.48°C

3 0

3 years ago

My homework said"Solve using the standard algorithm. Check your quotient and remainder by using multiplication and addition."

Harman [31]

84 divided by 2 = 42.

2 goes into 8 four times, so there would be your 4.
Then 2 goes into 2 one time, and there would be your 2.

You check your answer by multiplying 42 by 2, which would give you, 84.

5 0

3 years ago

Read 2 more answers