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3 years ago

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Which of the following best represents potential energy being converted to kinetic energy

1 answer:

Setler [38]3 years ago

3 0

Where are the options

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The properties of two elements are listed below. Which prediction is supported by the information in the table?

ANEK [815]

K will give up an electron more easily than Br.

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Please answer in details. what class of carbon is coal

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Coal falls under Graphite

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2 years ago

How much water will be required to dissolve 0.3g of benzoic acid at 95oC given that the solubility of BA at 95oC is 68.0g/L?

Ainat [17]

Write as a proportion, showing the relationship of both given information:
68.0g 0.3g
---------- = -----------
1L x ( your answer)

Cross multiply: 68.0g× X = 0.3g × 1L
68.0g (X)= 0.3g/L
Solve for X by dividing both sides by 68.0 g
68.0g (X) = 0.3g/L
------------- ------------------
68.0g 68.0g

Then enter into calculator 0.3/68 and that will be your solution. Make sure you round up.

6 0

2 years ago

Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe

leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

$c = \frac{n}{V}$

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

$M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}$

$M_{KOH} = 56.106 \frac{g}{mol}$

<u>Calculate the moles of potassium hydroxide.</u>

$n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}$

$n_{KOH} = 0.25941(9)mol$

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

$V = \frac{500.0mL}{1}*\frac{1L}{1000mL}$

$V = 0.5000L$

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

$c = \frac{n}{V}$

$c = \frac{0.25941(9)mol}{0.5000 L}$

$c = 0.5188(3) \frac{mol}{L}$ <= Keep an insignificant figure for rounding

$c = 0.5188 \frac{mol}{L}$ <= Rounded up

$c = 0.5188M$ <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

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2 years ago

Use the table to answer the question.

Alex17521 [72]

Answer:

High levels of moisture are contained within an air mass.

Tell me if I'm correct, plz!

Explanation:

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