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2 years ago

Chemical reactions idk what to do plz help asap

Chemistry

1 answer:

Rasek [7]2 years ago

7 0

Answer:

it's B

Explanation:

i got it right

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How many total atoms are there in one molecule of C145H293O168?

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Answer:

606 atoms

Explanation:

Add the numbers 145 + 293 + 168

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B. bias

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Explanation:

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2 years ago

1. NaOH mass of a solution of 200g in which its percentage is 25%. What mass of sulfuric acid solution is needed to completely n

Ede4ka [16]

Answer:

$m_{H2SO4$ = 61.25 g

$m_{Na2SO4}$ = 88.75 g

Explanation:

$m_{NaOH}$ = $\frac{200 . 25 }{100}$ = 50 g

⇒ $n_{NaOH}$ = $\frac{50}{40}$ = 1.25 (moles)

2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

2 : 1 : 1 : 2

1.25 (moles)

⇒ $n_{H2SO4}$ = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ $m_{H2SO4}$ = 0.625 × 98 = 61.25 g

$n_{Na2SO4}$ = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ $m_{Na2SO4}$ = 0.625 × 142 = 88.75 g

4 0

3 years ago

A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th

Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

$pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Also, relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log\frac{[conjugate base]}{[acid]}$

= $3.347+ log \frac{0.15}{0.12}$

= 3.44

Therefore, pH of the buffer is 3.44.

(b) No. of moles of HCl added = $Molarity \times volume$

= $11.6 M \times 0.001 L$

= 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

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3 years ago