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katrin2010 [14]

3 years ago

7

The volume of a pond being studied for the effects of acid rain is 35 kiloliters (kL). There are 1,000 liters (L) in 1 kL and 1

× 10^6 microliters (mL) in 1 L.
What is the volume of this pond in microliters?

1 answer:

Lerok [7]3 years ago

4 0

Answer:

35,000,000,000 mL

Explanation:

You first multiply 35 times 1000.

35,000 L

Now you multiply 35,000 times 10^6

35,000,000,000 mL

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Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.

b)

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ = 0.195 M

Concentration of $SbCl_3$ = $6.98\times 10^{-2} M$

Concentration of $Cl_2$ = $6.98\times 10^{-2} M$

On adding more $[SbCl_5$ to 0.370 M at equilibrium :

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.370 M $6.98\times 10^{-2}M$

At equilibrium:

(0.370-x)M $(6.98\times 10^{-2}+x)M$

The equilibrium constant of the reaction = $K_c$

$K_c=2.50\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}$

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

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3 years ago

Problem page in each of the molecules drawn below one chemical bond is colored red. decide whether this bond is likely to be pol

Slav-nsk [51]

The question is incomplete, here is the complete question:

Problem page in each of the molecules drawn below one chemical bond is colored red. Decide whether this bond is likely to be polar or not. if the bond is likely to be polar, write down the chemical symbol for the atom which will have more negative charge.

The image is attached below.

<u>Answer:</u>

<u>For carbon dioxide molecule:</u> The bond is considered as polar and the elecvtronegative atom is oxygen.

<u>For water molecule:</u> The bond is considered as polar and the elecvtronegative atom is oxygen.

<u>Explanation:</u>

There are two types of covalent bonds:

<u>Polar covalent bond:</u> This bond is formed when difference in electronegativity between the atoms is present. When atoms of different elements combine, it results in the formation of polar covalent bond. <u>For Example:</u> $CO_2,NO_2$ etc..
<u>Non-polar covalent bond:</u> This bond is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. <u>For Example:</u> $N_2,O_2$ etc..

<u>In carbon dioxide molecule:</u>

The given bond is present between C and O atom.

Electronegativity value of C = 2.5

Electronegativity value of O = 3.5

Electronegativity difference = (3.5 - 2.5) = 1

As, electronegativity difference is present. So, the bond is considered as polar and the elecvtronegative atom is oxygen.

<u>In water molecule:</u>

The given bond is present between H and O atom.

Electronegativity value of H = 2.1

Electronegativity value of O = 3.5

Electronegativity difference = (3.5 - 2.1) = 1.4

As, electronegativity difference is present. So, the bond is considered as polar and the elecvtronegative atom is oxygen.

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In order to change a solid to a liquid or a liquid to a gas, what is needed?

Neporo4naja [7]

Heat; rather, or change of the molecules to make them move faster

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3 years ago

How many moles of solute are contained in a 250.0-milliliter solution that has a concentration of 0.85 M?

adoni [48]

Answer : The correct option is, 0.21 moles

Solution : Given,

Molarity of the solution = 0.85 M = 0.85 mole/L

Volume of solution = 250 ml = 0.25 L $(1L=1000ml)$

Molarity : It is defined as the number of moles of solute present in one liter of the solution.

Formula used :

$Molarity=\frac{\text{moles of solute}}{\text{Volume of solution in liters}}$

Now put all the given values in this formula, we get the moles of solute of the solution.

$0.85mole/L=\frac{\text{moles of solute}}{0.25L}$

$\text{moles of solute}=0.2125=0.21moles$

Therefore, the moles of solute is, 0.21 moles

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