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3 years ago

13

Can somobody help me??????

1 answer:

Katen [24]3 years ago

5 0

Answer:

Interphase, mitosis, and cytokenesis

Explanation:

Interphase is the phase in which the cell is growing, and mitosis and cytokenesis make up the phase in which daughter cells are formed

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Which compounds are lewis acids? select all that apply. (ch3ch2)2hc+ cl3c− ch3nh2 bbr3?

alexandr402 [8]

The Lewis acids are electron pair acceptors due to the availability of vacant orbitals they accept electrons from the donor compounds.

$(CH_3CH_2)_2CH^{+}$ is a Lewis acid due to the unfilled valence shell on carbon atom (possessing positive charge) it accepts electron readily.
$CCl_{3}^{-}$ is not a Lewis acid due to the presence of negative charge which represents that there are electrons available for donation.
$CH_3NH_2$ is not a Lewis acid due to the presence of lone pair on nitrogen which is easily donated to the acceptors.
$BBr_3$ is a Lewis acid due to the availability of vacant p-orbital on boron atom.

Hence, $(CH_3CH_2)_2CH^{+}$ and $BBr_3$ are Lewis acids.

3 0

3 years ago

What kind of intermolecular force is between hydrogen sulfide and hydrogen iodide?

RUDIKE [14]

I will have to go with carbon monoxide

8 0

3 years ago

Atoms of Ca and F would be expected to react in a ratio of

Gemiola [76]

Answer:

A 1:1

Explanation:

i thinks it's right

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3 years ago

What branch of science is, matter, energy, motion, forces, time

serious [3.7K]

These are terms that we use in physics

6 0

3 years ago

A solution is made by mixing 50 ml of 2.0m k2hpo4 and 25 ml of 2.0m kh2po4. the solution is diluted to a final volume of 200 ml.

Volgvan

50 mL of 2.0 M of $K_2HPO_4$ and 25 mL of 2.0 M of $KH_2PO_4$ were mixed to make a solution

Final volume of the solution after dilution = 200 mL (given)

Final concentration of $K_2HPO_4$ , [ $K_2HPO_4$ ] = $\frac{50 mL\times 2 M}{200 mL} = 0.5 M$

Final concentration of $KH_2PO_4$ , [ $KH_2PO_4$ ] = $\frac{25 mL\times 2 M}{200 mL} = 0.25 M$

Using Hasselbach- Henderson equation:

$pH = pK_a+ log \frac{[salt]}{[acid]}$

$pka of KH_2PO_4 = 6.85$

Substituting the values:

$pH = 6.85+ log \frac{0.5}{0.25}$

$pH = 6.85+ log 2$

$pH = 6.85+ 0.3 = 7.15$

Hence, the $pH$ of the final solution is 7.15.

8 0

3 years ago

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