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andreyandreev [35.5K]
4 years ago
11

Science question down below

Chemistry
1 answer:
Sergio039 [100]4 years ago
8 0
It's the last one the answer is the last one
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Which of the following bases are strong enough to deprotonate CH3CH2CH2C≡CH (pKa = 25), so that equilibrium favors the products?
seraphim [82]

Answer:

a, and f.

Explanation:

To be deprotonated, the conjugate acid of the base must be weaker than the acid that will react, because the reactions favor the formation of the weakest acid. The pKa value measures the strength of the acid. As higher is the pKa value, as weak is the acid. So, let's identify the conjugate acid and their pKas:

a. NaNH2 will dissociate, and NH2 will gain the proton and forms NH3 as conjugate acid. pKa = 38.0, so it happens.

b. NaOH will dissociate, and OH will gain the proton and forms H2O as conjugate acid. pKa = 14.0, so it doesn't happen.

c. NaC≡N will dissociate, and CN will gain a proton and forms HCN as conjugate acid. pKa = 9.40, so it doesn't happen.

d. NaCH2(CO)N(CH3)2 will dissociate and forms CH3(CO)N(CH3)2 as conjugate acid. pKa = -0.19, so it doesn't happen.

e. H2O must gain one proton and forms H3O+. pKa = -1.7, so it doesn't happen.

f. CH3CH2Li will dissociate, and the acid will be CH3CH3. pKa = 50, so it happens.

6 0
3 years ago
How many joules of heat are required to heat 100.0g of room temperature water to the boiling point
Yuri [45]

Answer:

To convert 100.0 g of water at 20.0 °C to steam at 100.0 °C requires 259.5 kJ of energy. Let me know if this helped?

4 0
3 years ago
Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K
Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-

And the undergoing chemical reaction:

MgCl_2+2NaF\rightarrow MgF_2+2NaCl

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2

Next, the moles of magnesium chloride consumed by the sodium fluoride:

n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M

[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M

Thereby, the reaction quotient is:

Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

6 0
3 years ago
2.
Arturiano [62]

Answer:

they are producers and they have means to keep themselves warm

Explanation:

I took test and saw answer.

6 0
3 years ago
Convert 3828 ml to L
crimeas [40]

Answer:

divide the volume value by 1000

So 3828/1000=3.828

3 0
3 years ago
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