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avanturin [10]
3 years ago
9

What are some similarities of solids and gases?

Chemistry
2 answers:
Llana [10]3 years ago
5 0
1) Both have molecules or atoms 
2) Both occupy space and have mass 
3) Both can be ionized 
4) Solids conduct electricity similarly gas can also conduct electricity in discharge tube under low pressure 
5) Both have densities
Nookie1986 [14]3 years ago
3 0
They are both part of the 3 trio (sorry I forgot the name of it)
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How many grams of copper (II) fluoride, CuF2, are needed to make 5.1 liters of a 1.4M solution?
Alexus [3.1K]

Answer: 724.71 grams

Explanation:

Volume of solution (v) = 5.1 liters

Concentration of solution (c) = 1.4M

Amount of CuF2 needed (n) = ?

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

make n the subject formula

n = c x v

n = 1.4M x 5.1 Liters

n = 7.14 moles

Since, 7.14 moles of CuF2 (n) is needed, use the molar mass of CuF2 to get the mass in grams.

The atomic masses of Copper = 63.5g;

and Fluorine = 19g

So, I CuF2 = 63.5g + (19g x 2)

= 63.5g + 38g

= 101.5g/mol

Then, apply the formula

Number of moles = mass in grams / molar mass

7.14 moles = m / 101.5 g/mol

m = 7.14 moles x 101.5 g/mol

m = 724.71g

Thus, 724.71 grams of copper (II) fluoride, CuF2, are needed to make 5.1 liters of a 1.4M solution

7 0
4 years ago
What is true about models in science?
Firlakuza [10]

Answer:

hi

Explanation:

3 0
4 years ago
Read 2 more answers
Which of the following is a chemical change?
avanturin [10]
C is the answer because that is an actual chemical change where the sulfur and iron are not separate from each other anymore
8 0
3 years ago
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2. How many orbitals are in the following sublevels?
stich3 [128]
This answer would be a
6 0
3 years ago
At 517 mm Hg and 24 °C, a sample of gas occuples a volume of 95 ml. The gas is transferred to a 225-ml flask and the temperature
vodomira [7]

Answer:

P_2=194.78mmHg

Explanation:

Hello,

In this case, we employ the combined ideal gas law in order to understand the volume-gas-pressure behavior as shown below:

\frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}

Hence, solving for the final pressure P2, we obtain (do not forget temperature must be absolute):

P_2=\frac{P_1V_1T_2}{V_2T_1}=\frac{517mmHg*95mL*(-8.0+273.15)K}{(24+273.15)K*225mL}\\ \\P_2=194.78mmHg

Best regards.

7 0
4 years ago
Read 2 more answers
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