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melisa1 [442]
3 years ago
5

In a synthesis reation you stat with 1.7L of hydrogen how many liters of water will be produced?

Chemistry
1 answer:
Vesna [10]3 years ago
6 0

check the answer in Google.

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A solution of sulfuric acid has a concentration of 0.0980 g/L. If the density of the acid is 1.84 g/mL, what is the concentratio
mariarad [96]
The answer is 98ppm.

The ppm (Parts per million) is also a concentration unit. 1 ppm is equivalent to 1mg/L
Now, concentration of the solution of sulfuric acid is 0.0980 g/L.
We rewrite, 0.0980 g = 0.0980*1000 mg = 98mg
Therefore, the concentration of the sulfuric acid solution is 98 mg/L = 98 ppm.
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4 years ago
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How many total atoms are in 3Na2SO4?
tigry1 [53]

Answer:

A. 21

Explanation:

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3 years ago
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Which element in group 2 (Alkaline Earth Metals) is the most reactive?
never [62]

Answer:

the reactivity of alkaline earth metals increases from top to the bottom of the group, that's because the atoms get bigger from the top to the bottom so the valence electrons are farther from the nucleus.

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2 years ago
A sample of nitrogen gas with a volume of 2.55 L is collected at a pressure of 670.0 mmHg. If the temperature is held constant,
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U will equate and make V2 the subject

5 0
3 years ago
Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
enyata [817]

Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

                     = 6.35 + log(\frac{0.045}{0.45})

                     = 6.35 + (-1)

                     = 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

                          = 0.045 M

So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

5 0
3 years ago
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