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Svetach [21]
3 years ago
9

The carbon-magnesium bond in a Grignard reagent is covalent and highly-polarized such that the carbon is negatively charged. Whi

ch of the following statements can be used to describe the Grignard carbanion? Select all that apply.
1. Grignard reagents are weak bases
2. Grignard reagents are weak nucleophiles
3. Grignard reagents are strong bases
4. Grignard reagents are strong nucleophiles
Chemistry
1 answer:
goldenfox [79]3 years ago
8 0

Answer:

Option 3 - Grignard reagents are strong bases and option 4 - Grignard reagents are strong nucleophiles

Explanation:

The carbon in grignard reagent has a very strong carbanionic character which gives grignard reaction it's unique characteristic of a strong base and strong nucleophile.

RCH_{2} - x + M_{g}  ⇒   RCH_{2}^{o- o+} Mgx

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If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M<br> TRIS base, what will be the resulting pH?
Katyanochek1 [597]

<u>Answer:</u> The pH of resulting solution is 8.7

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

  • <u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol

  • <u>For TRIS base:</u>

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol

Volume of solution = 50 + 60 = 110 mL = 0.11 L    (Conversion factor:  1 L = 1000 mL)

  • To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

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

pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})

We are given:

pK_a = negative logarithm of acid dissociation constant of TRIS acid = 8.3

[\text{TRIS acid}]=\frac{0.005}{0.11}

[\text{TRIS base}]=\frac{0.012}{0.11}

pH = ?

Putting values in above equation, we get:

pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7

Hence, the pH of resulting solution is 8.7

6 0
3 years ago
WHAT IT IZ BRAINLY anyways I got a question for y’all do you think someone can forget who they rlly loved ? Can you ever forget
alexdok [17]

Answer:

Its a 50% chance of it happeing

Explanation:

5 0
3 years ago
Yousef measured the height of each seedling on day 1 and day 7. These are his results.
kati45 [8]
That’s a lot of numbers
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3 years ago
If one liquid has a higher volatility than another, and both liquids are at the same
Romashka [77]

Answer:

<u>Option B is correct</u>

Explanation:

Step 1: Define volatility

In chemistry, the term volatility, is a way to describe how readily a substance  transitions from a liquid phase to a gas phase, also called evaporating.

At a given temperature and pressure, a substance with high volatility is more likely to evaporate more quickly , while a substance with a lower volatility is more likely to be a liquid or solid, so not to evaporate or slower.

The higher the volatility, the higher the (vapor) pressure of a compound. Increasing temperature means the vapor pressure will also increase,

Step 2: In this case:

⇒ O<u>ption A is false</u> because the pressure will be higher when volatility is higher.

<u>⇒ Option B is correct</u> because higher volatility means evaporating more quickly

<u>⇒ Option C is false</u> because higher volatility means higher pressure. When pressure increases, the surface tension decreases.

<u>⇒ Option D is false</u> because when the volatility is higher, the liquid/gas escape the container, easier, so there will be less resistance.

3 0
3 years ago
Given that [OH-] = 5.46x10^-9M
grigory [225]
  • pOH=-log[OH-]
  • pOH=-log[5.46×10^{-9}]
  • pOH=-log5.46-log10^-9
  • pOH=-0.74+9
  • pOH=8.26

pH:-

  • 14-pOH
  • 14-8.26
  • 5.74

It's acidic as pH<7

Now

  • -log[H+]=5.74
  • log[H+]=-5.74
  • H+=10^{-5.74}
  • H+=1.81×10^-6M
8 0
2 years ago
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