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Lera25 [3.4K]
4 years ago
6

NAME:

Chemistry
1 answer:
Fofino [41]4 years ago
3 0
Hah what is this work
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A glass was filled with 10 ounces of water, and 0.01 ounce of the water evaporated each day during a twenty-day period.  What pe
liubo4ka [24]

Answer:

2%

Explanation:

Original mass of water in glass = 10 ounces

amount that evaporates each day = 0.01 ounce

number of days = 20 days

Total amount that evaporated = 0.01 x 20 = 0.2 ounce

percentage of water that evaporated during this period = 0.2/10 x 100

                            = 2%

<em>Hence, the percentage of the original amount of water evaporated during this period is 2%</em>

7 0
3 years ago
Find the force it would take to accelerate an 800-kg car at a rate of 5m/s2.
laila [671]
Force= mass•acceleration

Force=800•5

Force=4000 N
3 0
3 years ago
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What does the law of conservation of energy states.
marin [14]

Answer:

The total energy of an isolated system remains constant; it is said to be conserved over time.

7 0
3 years ago
Indicate the number of significant figures in the measured number 0.0020 <br> A. 4<br> B. 5<br> C. 2
Sergio [31]

<u>Answer:</u>

The correct answer option is C. 2.

<u>Explanation:</u>

We are given the number '0.0020' and we are to indicate the number of significant figures in the given measured number.

According to the rules of significant figures, numbers that are non-zero, zeros between any two significant numbers and the ending zeros in the decimal position are categorized as significant figures.

Since there is one non-zero number and one ending zero in the decimal position, therefore 0.0020 has 2 significant figures.

6 0
3 years ago
A chemistry student weighs out 0.027 kg of an unknown solid compound X and adds it to 550. mL of distilled water at 30.° C. Afte
QveST [7]

Answer:

1. Yes

2.The solubility of X is 34.55g/L

Explanation:

Solubility of  solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm^{3} or 1 Litre of water.

From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved

= 0.019kg*1000 = 19g.

if 19g is required to saturate 550mL at 30°C,

then\frac{19*1000}{550} will saturate 1L

= 34.545g  will saturate 1Litre

The solubility thus is 34.55g/L

 

5 0
3 years ago
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