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3 years ago

A student must make a 3 m acid solution using a 5 m acid solution. which of these is the safest way to make the solution?

1 answer:

8 0

The motivation to abstain from adding water to concentrated acids is that, with a few acids, amid weakening, a considerable measure of warmth is discharged, by adding the corrosive to the water, the generally extensive measure of water will retain the warmth. On the off chance that you added water to concentrated corrosive when you initially beginning pouring the water, it could get sufficiently hot for the little measure of water that was filled all of a sudden bubble and splatter corrosive on you. Concentrated sulfuric corrosive is most famous for doing this, not all acids get that hot on weakening, but rather in the event that you make a propensity for continually adding the corrosive to water for every one of them, you can't turn out badly.

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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.

6 0

3 years ago

Please help with the question attached<br>I need it in an hour

lawyer [7]

Answer:

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4 0

2 years ago

The partial pressure of N2 in the air is 593 mm Hg at 1 atm. What is the partial pressure of N2 in a bubble of air a scuba diver

Evgen [1.6K]

Answer: Partial pressure of $N_{2}$ at a depth of 132 ft below sea level is 2964 mm Hg.

Explanation:

It is known that 1 atm = 760 mm Hg.

Also, $P_{N_{2}} = x_{N_{2}}P$

where, $P_{N_{2}}$ = partial pressure of $N_{2}$

P = atmospheric pressure

$x_{N_{2}}$ = mole fraction of $N_{2}$

Putting the given values into the above formula as follows.

$P_{N_{2}} = x_{N_{2}}P$

$593 mm Hg = x_{N_{2}} \times 760 mm Hg$

$x_{N_{2}}$ = 0.780

Now, at a depth of 132 ft below the surface of the water where pressure is 5.0 atm. So, partial pressure of $N_{2}$ is as follows.

$P_{N_{2}} = x_{N_{2}}P$

= $0.78 \times 5 atm \times \frac{760 mm Hg}{1 atm}$

= 2964 mm Hg

Therefore, we can conclude that partial pressure of $N_{2}$ at a depth of 132 ft below sea level is 2964 mm Hg.

8 0

3 years ago

2. Nitric oxide contains 46.66% nitrogen and 53.34% oxygen. Water contains 11.21% hydrogen and 88.79% oxygen. Ammonia contains 1

Mrrafil [7]

Answer:

The law of reciprocal proportions states that if two elements react individually with a given weight of a third element, the ratio of the masses with which they combine with the third element are either the same or a simple multiple of the ratio of the masses with which they combine with each other

The compounds formed includes;

1) Nitric oxide, NO

Nitrogen = 46.66% × 30.01 = 14

Oxygen = 53.34% × 30.01 = 16

2) Water, H₂O

Hydrogen = 11.21% × 18.01528 = 2

Oxygen = 88.79% × 18.01528 ≈ 16

3) Ammonia, NH₃

Hydrogen = 17.78% × 17.031 ≈ 3

Nitrogen = 82.22% × 17.031 ≈ 14

The ratio of nitrogen to oxygen in nitric oxide = 14:16 = 7:8

The ratio of nitrogen to hydrogen in ammonia = 14:3

The ratio in which hydrogen and oxygen combine with nitrogen = 3/16

The ratio of hydrogen and oxygen combine with each other in water = 2/16

Therefore, the ratio with which hydrogen and oxygen combine with nitrogen, is (2/3) times the ratio with which they combine with each other, which verifies the law of reciprocal proportions

Explanation:

4 0

3 years ago

From the graph of Density vs. Concentration, created in Graph 1, what was the relationship between the concentration of the suga

USPshnik [31]

The graph is not given in the question, so, the required graph is attached below:

Answer:

According to the graph, the relationship between the density of the sugar solution and the concentration of the sugar solution is directly proportional to each other as they both are increasing exponentially.

The graph shows that, the density of sugar solution will increase with the increase in concentration of sugar in the solution.

8 0

3 years ago