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lorasvet [3.4K]
3 years ago
7

What would be the definition of concentration?​

Chemistry
1 answer:
MariettaO [177]3 years ago
8 0

Answer:

the action or power of focusing one's attention or mental effort.

Explanation:

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Which element has atoms that can form halide ions
Nostrana [21]
Iodine..............................
8 0
3 years ago
Read 2 more answers
Calculate the second volumes. <br> 7.03 Liters at 31 C and 111 Torr to STP
DIA [1.3K]

The second volume :    V₂= 0.922 L

<h3> Further explanation </h3><h3>Given </h3>

7.03 Liters at 31 C and 111 Torr

Required

The second volume

Solution

T₁ = 31 + 273 = 304 K

P₁ = 111 torr = 0,146 atm

V₁ = 7.03 L

At STP :  

P₂ = 1 atm

T₂ = 273 K

Use combine gas law :

P₁V₁/T₁ = P₂V₂/T₂

Input the value :

0.146 x 7.03 / 304 = 1 x V₂/273

V₂= 0.922 L

4 0
3 years ago
the chemical compound C2F4 is used to make PTEE (Teflon). How manyC2F4 molecules are in 485 kilograms of this material?
monitta

Answer:

molecules=2.92x10^{27}moleculesC_2F_4

Explanation:

Hello,

In this case, we use the Avogadro's number to compute the molecules of C2F4 whose molar mass is 100 g/mol contained in a 485-kg sample as shown below:

molecules=485kgC_2F_4*\frac{1000gC_2F_4}{1kgC_2F_4} *\frac{1molC_2F_4}{100gC_2F_4}*\frac{6.022x10^{23}molecules C_2F_4}{1molC_2F_4}  \\\\molecules=2.92x10^{27}moleculesC_2F_4

Best regards,

5 0
3 years ago
How many grams of KNO3 are required to prepare 0.250 L of 0.70 M solution?​
drek231 [11]

Answer:

about 3

Explanation:

just did this for the points just being honiest

4 0
2 years ago
A sample containing 2.30 mol of Ne gas has an initial volume of 8.00 L. What is the final volume, in liters, when the following
marta [7]

Answer:

a. 4,00L

b. 16,00L

c. 12,31L

Explanation:

Avogadro's law says:

\frac{V_1}{n_1} =\frac{V_2}{n_2}

a. If initial conditions are 2,30mol and 8,00L and you lose one-half of atoms, that means you have 1,15mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{1,15mol}

<em>V₂ = 4,00L</em>

b. If initial conditions are 2,30mol and 8,00L and you add 2,30mol, that means you have 4,60mol:

\frac{8,00L}{2,30mol} =\frac{V_2}{4,60mol}

<em>V₂ = 16,00L</em>

c. 25,0g of Ne are:

25,0g × (1mol / 20,1797g) = 1,24 moles of Ne. That means you have 2,30mol - 1,24mol = 3,54mol of Ne

\frac{8,00L}{2,30mol} =\frac{V_2}{3,54mol}

<em>V₂ = 12,31L</em>

I hope it helps!

6 0
3 years ago
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