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4 years ago

A cubic centimeter is equal in volume to a _____.

Chemistry

2 answers:

mafiozo [28]4 years ago

8 0

A cubic centimeter is equal in volume to one milliliter

dimulka [17.4K]4 years ago

5 0

Answer: ml

Explanation:-

S.I units of volume are $m^3$ and C.G.S unit of volume are $cm^3$

Units of ml are also used to measure volume of a substance and corresponds to $1cm\times 1cm\times 1cm$

A liquid which occupies a space of $1cm^3$ has volume of 1 ml.

$1cm^3=1ml$

Thus 1 $cm^3$ is equal to $1ml$

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Compared with the freezing-point depression of a 0.01 m c6h12o6 solution, the freezing-point depression of a 0.01 m nacl solutio

agasfer [191]

Answer:

Twice as much.

Explanation:

That's because the freezing point depression depends on the total number of solute particles.

C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)

0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.

NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)

1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).

That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.

6 0

4 years ago

Which of the following objects have kinetic energy?

jasenka [17]

A bee a bike a water fall

4 0

3 years ago

Which of the following is an element? A. MgBr B. HCl C.O .D. Cu(SO4)3

DanielleElmas [232]

The answer is c. the others are compounds.

5 0

3 years ago

Read 2 more answers

What is the concentration in mass percent of a solution prepared from 50.0g nacl and 150.0g of water?

otez555 [7]

M(NaCl)=50.0 g
m(H₂O)=150.0 g

m(solution)=m(NaCl)+m(H₂O)

w(NaCl)=100m(NaCl)/m(solution)=100m(NaCl)/{m(NaCl)+m(H₂O)}

w(NaCl)=100*50.0/(50.0+150.0)=25%

3 0

4 years ago

Noble gas compounds like KrF, XeCl, and XeBr are used in excimer lasers. Draw an approximate molecular orbital diagram appropria

Aneli [31]

Answer:

Here's what I get.

Explanation:

The MO diagrams of KrBr, XeCl, and XeBr are shown below.

They are similar, except for the numbering of the valence shell orbitals.

Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.

However, the MO diagrams are approximately correct.

The ground state electron configuration of KrF is

$(1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \, (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}$

KrF⁺ will have one less electron than KrF.

You remove the antibonding electron from the highest energy orbital, so the bond order increases.

The KrF bond will be stronger.

6 0

4 years ago