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2 years ago

Will give brainliest

Chemistry

1 answer:

Dafna1 [17]2 years ago

3 0

sorry i know this:

4. 2 methyl 2 heptene

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3 years ago

How many moles of HNO3 are present if 4.90×10−2 mol of Ba(OH)2 was needed to neutralize the acid solution?

Amiraneli [1.4K]

Answer:

0.098 moles

Explanation:

Let y represent the number of moles present

1 mole of Ba(OH)₂ contains 2 moles of OH- ions.

Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.

To get the y moles, we then do cross multiplication

1 mole * y mole = 2 moles * 0.049 mole

y mole = 2 * 0.049 / 1

y mole = 0.098 moles of OH- ions.

1 mole of OH- can neutralize 1 mole of H+

Therefore, 0.098 moles of HNO₃ are present.

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3 years ago

How does the freezing point of water from the ocean compare to distilled water one buys at the store?

Lelechka [254]

Due to prescence of any impurity, there will be change in physical properties of any liquid.

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3 years ago

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podryga [215]

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3 years ago

A 1.50 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 0.100 moles

alexgriva [62]

Answer : The pH of the solution is, 3.41

Explanation :

First we have to calculate the moles of $HF$ .

$\text{Moles of HF}=\text{Concentration of HF}\times \text{Volume of solution}$

$\text{Moles of HF}=0.250M\times 1.50L=0.375mol$

Now we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (6.8\times 10^{-4})$

$pK_a=4-\log (6.8)$

$pK_a=3.17$

The reaction will be:

$HF+OH^-\rightleftharpoons F^-+H_2O$

Initial moles 0.375 0.100 0.375

At eqm. (0.375-0.100) 0 (0.375+0.100)

= 0.275 = 0.475

Now we have to calculate the pH of solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[F^-]}{[HF]}$

Now put all the given values in this expression, we get:

$pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]$

$pH=3.41$

Thus, the pH of the solution is, 3.41

8 0

3 years ago