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rewona [7]
3 years ago
15

1. Which of the following statements describe cations? Select all that apply.

Chemistry
2 answers:
Wittaler [7]3 years ago
5 0

Answer:

<h3><em>N</em><em>E</em><em>G</em><em>A</em><em>T</em><em>I</em><em>V</em><em>E</em><em>L</em><em>Y</em><em> </em><em>C</em><em>H</em><em>A</em><em>R</em><em>G</em><em>E</em><em>D</em><em> </em><em>I</em><em>O</em><em>N</em><em>S</em></h3>

Explanation:

<h2><em>I</em><em>H</em><em>O</em><em>P</em><em> </em><em>N</em><em>A</em><em>K</em><em>A</em><em>T</em><em>U</em><em>L</em><em>O</em><em>N</em><em>G</em><em> </em><em>#</em><em>L</em><em>I</em><em>K</em><em>E</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>F</em><em>O</em><em>L</em><em>L</em><em>O</em><em>W</em></h2>
Crazy boy [7]3 years ago
5 0
Hey :)

Cations are positively charged ions and they have lost electrons

Hope this helps!
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Suppose that 0.1000 mole each of H2and I2are placed in a 1.000-L flask, stoppered, and the mixture is heated to 425oC. At equili
Katen [24]

<u>Answer:</u> The value of equilibrium constant for the given reaction is 56.61

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.100 moles

Initial moles of hydrogen gas = 0.100 moles

Volume of container = 1.00 L

Molarity of the solution is calculated by the equation:

\text{Molarity of solution}=\frac{\text{Number of moles}}{\text{Volume}}

\text{Molarity of iodine gas}=\frac{0.1mol}{1L}=0.1M

\text{Molarity of hydrogen gas}=\frac{0.1mol}{1L}=0.1M

Equilibrium concentration of iodine gas = 0.0210 M

The chemical equation for the reaction of iodine gas and hydrogen gas follows:

                         H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>                0.1    0.1

<u>At eqllm:</u>          0.1-x   0.1-x   2x

Evaluating the value of 'x'

\Rightarrow (0.1-x)=0.0210\\\\\Rightarrow x=0.079M

The expression of K_c for above equation follows:

K_c=\frac{[HI]^2}{[H_2][I_2]}

[HI]_{eq}=2x=(2\times 0.079)=0.158M

[H_2]_{eq}=(0.1-x)=(0.1-0.079)=0.0210M

[I_2]_{eq}=0.0210M

Putting values in above expression, we get:

K_c=\frac{(0.158)^2}{0.0210\times 0.0210}\\\\K_c=56.61

Hence, the value of equilibrium constant for the given reaction is 56.61

6 0
3 years ago
How many salt and water would you use to prepare 10% salt solution?
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Answer:

We can make 10 percent solution by volume or by mass. A 10% of NaCl solution by mass has ten grams of sodium chloride dissolved in 100 ml of solution. Weigh 10g of sodium chloride. Pour it into a graduated cylinder or volumetric flask containing about 80ml of water.

Explanation:

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4 0
3 years ago
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Answer:

★ v = d/t

★ v = 90/30

★ v =3

<u>Acc</u><u> </u><u>to</u><u> </u><u>question</u>

★ momentum = mass * velocity

★ m = 25×3

★ m = 75kgm/s

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