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3 years ago

How many kL does a 9.51 ´ 109 cL sample contain?

Chemistry

1 answer:

elena-14-01-66 [18.8K]3 years ago

4 0

Answer:

9.51 × 10⁴ kL

Explanation:

Step 1: Given data

Volume of the sample (V): 9.51 × 10⁹ cL

Step 2: Convert "V" to liters

We will use the conversion factor 1 L = 100 cL.

9.51 × 10⁹ cL × (1 L / 100 cL) = 9.51 × 10⁷ L

Step 3: Convert "V" to kL

We will use the conversion factor 1 kL = 1000 L.

9.51 × 10⁷ L × (1 kL / 1000 L) = 9.51 × 10⁴ kL

9.51 × 10⁹ cL is equal to 9.51 × 10⁴ kL.

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Write the condensed structural formula of all isomers of hexane and name them

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Answer:

Isomers

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Explanation:

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3 years ago

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3 years ago

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hram777 [196]

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4 years ago

In an aqueous chloride solution cobalt(II) exists in equilibrium with the complex ion CoCl42-. Co2 (aq) is pink and CoCl42-(aq)

kolezko [41]

Answer:

1. This reaction is (A) Exothermic .

2. When the temperature is decreased the equilibrium constant, K: (A) Increases

3.When the temperature is decreased the equilibrium concentration of Co2: (A) Increases

Explanation:

$CoCl4^2-(aq) Co_2+(aq) + 4Cl^-(aq)$

1. The pink color predominates at low temperatures, indicating that the commodity is preferred.

This is a reaction that is exothermic.

2. As the decrease in the temperature , the equilibrium constant , K ;

equilibrium constant = $\frac{product}{reactant}$ = $\frac{[CO^2^+][Cl^-^4]}{CoCl^2^-_4}$

As the temperature drops, the concentration of $Cl^-$ and $CO^2^+$ rises, and K rises as well , thus it increases .

3. The equilibrium concentration of $CO^2^+$ decreases as the temperature decreases:

When the temperature is lowered, the equilibrium shifts to the right , that is it increases.

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3 years ago

How many L of 3.0 M H2SO4 solution can be prepared by using 100.0 mL OF 18 M H2SO4?

max2010maxim [7]

Answer: A volume of 600 mL of 3.0 M $H_{2}SO_{4}$ solution can be prepared by using 100.0 mL OF 18 M $H_{2}SO_{4}$ .

Explanation:

Given: $V_{1}$ = ?, $M_{1} = 3.0 M\\$

$V_{2} = 100.0 mL$ , $M_{2} = 18 M$

Formula used to calculate the volume is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\3.0 M \times V_{1} = 18 M \times 100.0 mL\\V_{1} = \frac{18 M \times 100.0 mL}{3.0M}\\= 600 mL$

Thus, we can conclude that a volume of 600 mL of 3.0 M $H_{2}SO_{4}$ solution can be prepared by using 100.0 mL OF 18 M $H_{2}SO_{4}$ .

6 0

3 years ago