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Yuliya22 [10]
3 years ago
9

The density of an object never changes when the temperature of that object changes

Chemistry
1 answer:
nekit [7.7K]3 years ago
6 0
That question shoulda be true
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Predict what will be observed in each experiment below. A student sees tiny bubbles clinging to the inside of an unopened plasti
Arisa [49]

Answer:

A.

Explanation:

Squeezing the bottle creates negative pressure inside the bottle because of the plastic's elasticity that will hasten the extraction of the carbon dioxide from the soda.

6 0
3 years ago
Read 2 more answers
Substance P is carbon.
stepan [7]

Explanation:

substance Q could be <em><u>oxygen (O2)</u></em>

substance R could be <em><u>carbon</u></em><em><u> </u></em><em><u>d</u></em><em><u>i</u></em><em><u>o</u></em><em><u>x</u></em><em><u>i</u></em><em><u>d</u></em><em><u>e</u></em><em><u> </u></em><em><u>(</u></em><em><u>C</u></em><em><u>O</u></em><em><u>2</u></em><em><u>)</u></em>

4 0
3 years ago
12.<br> An object with a mass of 3 kg has a momentum of 75 kg•m/s. What is the velocity?
marissa [1.9K]

Answer:

the velocity is 25 m/s

Explanation:

The computation of the velocity is shown below:

As we know that

Magnitude of Momentum = (mass) × (speed)

75 kg. m/s = 3 kg × speed

So, the speed is

= 75 ÷ 3

= 25 m/s

hence, the velocity is 25 m/s

3 0
3 years ago
What is the temperature -71 oC expressed in Kelvin?<br> Group of answer choices
satela [25.4K]

Answer: 202.15

Explanation:

-71°C + 273.15

= 202.15K

5 0
3 years ago
Propanoic acid, ch3ch2cooh, has a pka =4.9. draw the structure of the conjugate base of propanoic acid and give the ph above whi
xenn [34]

Conjugate base of Propanoic acid (CH_{3}CH_{2}COOH is propanoate where -COOH group gets converted to -COO^{1-}. The structure of conjugate base of Propanoic acid is shown in the diagram.

The p^{H} above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.

p^{H}= p^{k_{a}+ log\frac{[Conjugate base]}{[acid]}=4.9+log\frac{90}{10}=5.85

As 90% conjugate base is present, so propanoic acid present 10%.

8 0
3 years ago
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