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faltersainse [42]
3 years ago
5

When 2-phenylacetaldehyde is added dropwise to benzaldehyde in the presence of sodium hydroxide they undergo a crossed aldol rea

ction to form 3-hydroxy-2,3-diphenylpropanal?

Chemistry
1 answer:
egoroff_w [7]3 years ago
7 0
Carbonyl Compounds containing alpha proton readily donates alpha acidic proton when treated with strong Base. In given statement <span>2-phenylacetaldehyde is converted into Enolate and Enolates acts as a nucleophile and attacks carbonyl group of Benzaldehyde. The mechanism of given reaction is as follow, the movement of electrons is shown by RED arrows,</span>

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How many hydrogen atoms are in 35.0 grams of hydrogen gas? How many hydrogen atoms are in 35.0 grams of hydrogen gas? 4.25 × 102
Ede4ka [16]

Answer: 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{35.0g}{2g/mol}=17.5moles

1 mole of hydrogen (H_2) = 2\times 6.023\times 10^{23}=12.05\times 10^{23} atoms

17.5 mole of hydrogen (H_2) = \frac{12.05\times 10^{23}}{1}\times 17.5=2.12\times 10^{25} atoms

There are 2.12\times 10^{25} atoms of hydrogen are there in

35.0 grams of hydrogen gas.

8 0
3 years ago
How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2
Korolek [52]

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
4 0
3 years ago
50 POINTS!! PLEASE HELP!
inn [45]
Okay so the answer will be.
8 0
3 years ago
NO + H2-&gt; N2 + H2O<br> Balance equation?
Oksanka [162]

Answer:

2NO + 4H-> N2 + 2 H2O

Explanation:

Both sides must be equal. :)

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3 years ago
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Answer:

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