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2 years ago

As shown below, a hammer was placed into a graduated cylinder. What is the volume displacement of the hammer?

Chemistry

1 answer:

Rufina [12.5K]2 years ago

4 0

Answer:

B. 4 ml

Explanation:

Volume displacement of the hammer = volume volume of water when the hammer was placed into the cylinder - volume of water only before the hammer was placed into the cylinder.

Volume volume of water when the hammer was placed into the cylinder = 69 ml

Volume of water only before the hammer was placed into the cylinder = 65 ml

Volume displacement of the hammer = 69 ml - 65 ml = 4 ml

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a chemist dissolves 0.564 moles of manganese (IV) oxide (MnO2) in water, and adds enough water to make 0.510 L of solution. Calc

ladessa [460]

Answer:

The molarity of the solution is 1.1 $\frac{moles}{liter}$

Explanation:

Molarity is a measure of the concentration of that substance that is defined as the number of moles of solute divided by the volume of the solution.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

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

Molarity is expressed in units $\frac{moles}{liter}$

In this case

number of moles of solute= 0.564 moles
volume= 0.510 L

Replacing:

$molarity=\frac{0.564 moles}{0.510 L}$

Solving:

molarity= 1.1 $\frac{moles}{liter}$

The molarity of the solution is 1.1 $\frac{moles}{liter}$

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2 years ago

Andre is packing for a move to Kentucky. He needs to pack 20 small paperback books whose total mass is 5 kg and some pillows who

storchak [24]

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hope I helped

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3 years ago

What is the molality of a solution of water and KCl if the freezing point of the solution is –3°C? (Kf = 1.86°C/m; molar mass of

Elina [12.6K]

Some of the solutions exhibit colligative properties. These properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering. Calculations are as follows:

ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg

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3 years ago

Read 2 more answers

Ethylene glycol (antifreeze) has a density of 1.11 g/cm3. What is the mass in grams of 387 mL of ethylene glycol?

harkovskaia [24]

To start, 1 cubic centimeter = 1 milliliter, so now you have 1.11g/mL.

Now multiply 1.11 by 387 to get the mass of antifreeze in grams, since the mL is canceled out.

387 mL x 1.11g/mL = 429.57 g

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2 years ago

Cuántos moles y gramos de Hcl pueden producirse a partir de 945g de ácido sulfúrico.?

Xelga [282]

I’m confused and that comment above is probably a trafficker

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2 years ago