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nydimaria [60]
1 year ago
13

You run an experiment in which two substances chemically react in a closed system. you run the same experiment in an open system

. how would the masses of the products in each experiment compare? explain.
Chemistry
1 answer:
m_a_m_a [10]1 year ago
6 0

If we run an experiment in which two substances chemically react in a closed system and then we run the same experiment in an open system, then the masses of the products in each experiment will be different because the open system allows the interchange of matter and energy with the media.

<h3>What is an open system?</h3>

An open system is an interrelated group pf parts that work together to interchange matter and energy with the surrounding environment, while a closed system does not generate an exchange of matter and energy.

Therefore, with this data, we can see that an open system interchange energy and matter with the environment, while closes systems do it.

Learn more about open systems here:

brainly.com/question/28891854

#SPJ1

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2 years ago
Identify the limiting reactant when 32. 0 g hydrogen is allowed to react with 16. 0 g oxygen
Mkey [24]

Answer:

Oxygen is limiting reactant

Explanation:

2 H2  +   O2  ======> 2 H2 O

from this equation (and periodic table) you can see that

  4 gm of H combine with 32 gm O2  

     H / O  =  4/32 = 1/8

       32 /16    =  2/1    shows O is limiter

         for 32 gm H you will need 256 gm O   and you only have 16 gm

3 0
2 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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