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patriot [66]
4 years ago
6

a scientist mixes to clear substances together and notices the following about the final solution it is cloudy some bubbles are

produced the temperature of the final solution is the same as the initial temperature of both substances is this a physical or chemical change​
Chemistry
1 answer:
AlexFokin [52]4 years ago
8 0

Answer:

chemical

Explanation:

it has bubbles

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3 years ago
When you use a crowbar to lift a large rock, you are working against the force called
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Gravity. The rock is being pushed down and you are counteracting that force with a crowbar pushing it upward.
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3 years ago
Please help! Best answer will receive brainliest. 15 points.
9966 [12]

Answer:

Melting of ice is a physical change and during this change latent heat of fusion, that is, heat required to convert 1 g of a solid substance into liquid state will keep the temperature constant unless and until all the solid changes into liquid. On the other hand, a change in which a new product is formed due to change in chemical composition of the reacting species is known as a chemical change. During a chemical change, there will occur change in temperature. So, when a substance is set on fire then there will occur change in temperature of the substance. Hence, it means it is a chemical change.

Explanation:

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3 0
4 years ago
Which of the following is NOT<br> necessary to define a coordinate<br> system?
gulaghasi [49]

Answer:

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8 0
2 years ago
The rate constants of some reactions double with every 10 degree rise in temperature. Assume that a reaction takes place at 271
AfilCa [17]

Answer : The activation energy for the reaction is, 119.7 J

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 271 K

K_2 = rate constant at 281 K = 2K_1

Ea = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 271 K

T_2 = final temperature = 281 K

Now put all the given values in this formula, we get:

\log (\frac{2K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{271K}-\frac{1}{281K}]

Ea=119.7J

Therefore, the activation energy for the reaction is, 119.7 J

3 0
4 years ago
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