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kirill115 [55]
3 years ago
6

Label each state of matter occurring in the photographs, and label the arrows to indicate the addition or removal of thermal ene

rgy

Chemistry
2 answers:
mote1985 [20]3 years ago
7 0

Answer: The first one would be a frozen thing like ice since the molecules are so packed, the next would be like this inside of a basketball or car tires, they move around freely(solid), and the last would be a gas since they aren't so many left(gas)

Explanation:

taurus [48]3 years ago
3 0

Answer:

he said it right >:| God damit

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The correct answer is A. :)
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Atoms of the element beryllium would most likely behave similar to the way _______ behaves?
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A student has two solutions of a substance;Solution-1:25M , 400ml and Solution-2:30M , 300M. What is the molarity of the final s
Vlad [161]

Answer:

The molarity of the final solution  is 1.7 M

Explanation:

The parameters given are;

First solution = 400 ml of 1.25 M

Second solution = 300 ml of 2.30 M

Therefore, we have;

First solution contains 400/1000 * 1.25 moles = 0.5 moles of the substance

Second solution contains 300/1000 * 2.30 moles = 0.69 moles of the

Hence the sum of the two solutions contains 0.5 + 0.69 = 1.19 moles of the substance

The volume of the sum of the two solutions = 400 ml + 300 ml = 700 ml

Hence we have the concentration of the final solution presented as follows;

700 ml contains 1.19 moles of the substance

Therefore;

1000 ml will contain 1000/700 * 1.19 = 1.7 moles

The molarity of the final solution = The number of moles per 1000 ml = 1.7 M.

3 0
3 years ago
One gram of Sn was electroplated from a SnCl, solution in 30 minutes. What was the current used?
Andrews [41]

Answer:

0.903 A

Explanation:

We have given mass of Sn = 1 gram

Time t=30 minutes=30×60=1800 sec

According to Faraday law m=zit

Where z is electron chemical equivalent

z is calculated by z=\frac{atomic\ mass}{nF} where n is number of element and F is Faraday constant

So z=\frac{atomic\ mass}{nF}=\frac{118.71}{2\times 96500}=0.000615

So 1=0.000615\times i\times 1800

i =0.903 A  

4 0
4 years ago
An experiment requires 0.52 M NH3(aq). The stockroom manager estimates that 15 L of the base is needed. What volume of 15 M NH3(
Bingel [31]

Answer: 0.52 L of 15 M NH_3(aq) will be used to prepare this amount of 0.52 M base.

Explanation:

But on diluting the number of moles remain same and thus we can use molarity equation.

C_1V_1(stock)=C_2V_2 (to be prepared)

where,

C_1 =concentration of stock solution = 15 M

V_1 = volume of stock solution = ?

C_2 = concentration of solution to be prepared = 0.52 M

V_2 = volume of solution to be prepared = 15 L

15\times V_1=0.52\times 15

V_1=0.52L

Thus 0.52 L of 15 M NH_3(aq) will be used to prepare this amount of 0.52 M base

8 0
3 years ago
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