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lesya692 [45]
4 years ago
15

An object has a density of 1.75 g/mL. Will it float on water? Why?

Chemistry
1 answer:
Advocard [28]4 years ago
4 0

Answer:

The object will not float in water because the density of the object is greater than that of water.

Explanation:

Density of object = 1.75 g/mL

Density of water = 1 g/mL

The Density of the object is 1.75 g/mL and that of water is 1 g/mL. This implies that the object is denser (i.e heavier) than water. Therefore, the object will not float in water, rather it will sink in water since it's density is greater than that of water.

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Indicate whether aqueous solutions of each of the following solutes will contain only ions, only molecules, or mostly molecules
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Answer:

Following are the solution to the given points:

Explanation:

In point a, the answer is only ions because of NH_4Cl has a strong electrolyte.

In point b, the answer is the only molecules because of ethanol CH_3CH_2OH, it has a nonelectrolyte.

In point c, the answer is the few ions because of hydrocyanic acid HCN, which has a weak electrolyte.

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If 8.74 g of CuNO3 is dissolved in water to make a 0.700 M solution, what is the volume of the solution
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Where do reactions in a solid generally take place?
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Answer:

It's D.  On the surface of the solid.

Explanation:

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what mass of sulfur dioxide would be produced if 64 tonnes of sulfur was completely reacted with oxygen gas?
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7 0
4 years ago
What mass of NaNO3 must be dissolved to make 838mL of a 1.25 M solution
denis-greek [22]

Answer:

89.04 g of NaNO₃.

Explanation:

We'll begin by converting 838 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

838 mL = 838 mL × 1 L / 1000 mL

838 mL = 0.838 L

Next, we shall determine the number of mole of NaNO₃ in the solution. This can be obtained as follow:

Volume = 0.838 L

Molarity = 1.25 M

Mole of NaNO₃ =?

Mole = Molarity × volume

Mole of NaNO₃ = 1.25 × 0.838

Mole of NaNO₃ = 1.0475 mole

Finally, we shall determine the mass of NaNO₃ needed to prepare the solution. This can be obtained as follow:

Mole of NaNO₃ = 1.0475 mole

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Mass of NaNO₃ =?

Mass = mole × molar mass

Mass of NaNO₃ = 1.0475 × 85

Mass of NaNO₃ = 89.04 g

Therefore, 89.04 g of NaNO₃ is needed to prepare the solution.

4 0
3 years ago
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