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Leokris [45]
4 years ago
8

Which of the following observations is usually not evidence of a chemical change ?

Chemistry
1 answer:
Nookie1986 [14]4 years ago
5 0

Answer:

D

Explanation:

Have a nice day

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Is a mountain a deconstructive process
blagie [28]

Answer:

yes

Explanation:

i looked it up.........

6 0
3 years ago
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105
rosijanka [135]
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
6 0
3 years ago
Read 2 more answers
What else supports the idea that an exothermic reaction has occurred?
kodGreya [7K]
The main pieces of evidence that an exothermic reaction has occurred is an increase in temperature due to the release of energy, a release of energy in the form of light, or a release of gas.
4 0
3 years ago
Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri
lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

           CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)

Calculate the amount of CO_{2} dissolved as follows.

             CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}

It is given that K_{CO_{2}} = 0.032 M/atm and P_{CO_{2}} = 1.9 \times 10^{-4} atm.

Hence, [CO_{2}] will be calculated as follows.

           [CO_{2}] = K_{CO_{2}} \times P_{CO_{2}}          

                           = 0.032 M/atm \times 1.9 \times 10^{-4}atm

                           = 0.0608 \times 10^{-4}

or,                        = 0.608 \times 10^{-5}

It is given that K_{a} = 4.46 \times 10^{-7}

As,      K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}

          4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}  

               [H^{+}]^{2} = 2.71 \times 10^{-12}

                      [H^{+}] = 1.64 \times 10^{-6}

Since, we know that pH = -log [H^{+}]

So,                      pH = -log (1.64 \times 10^{-6})

                                 = 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0
3 years ago
How many significant figures are in 5.40
kotykmax [81]
There are 3 significant figures in this value, all values before and after the decimal point are significant. As there is a decimal point, the zeros trailing are also significant.
4 0
3 years ago
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