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kobusy [5.1K]
3 years ago
12

A chemist prepares a solution of magnesium fluoride (MgF_2), by measuring out 0.031 g of magnesium fluoride into a 250. mL. Volu

metric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist magnesium fluoride solution. Round your answer to 2 significant digits. _____ mol/l.
Chemistry
1 answer:
AleksAgata [21]3 years ago
3 0

Answer:

The concentration of the magnesium fluoride solution is 0.0020 M

Explanation:

Step 1: Data given

Mass of MgF2 = 0.031 grams

volume = 250 mL = 0.250 L

Molar mass of MgF2 = 62.3 g/mol

Step 2: Calculate moles

Moles MgF2 = mass MgF2 / molar mass MgF2

Moles MgF2 = 0.031 grams / 62.3 g/mol

Moles MgF2 = 4.98 *10^-4 moles

Step 3: Calculate concentration of MgF2

Concentration MgF2 = moles MgF2 / volume

Concentration MgF2 = 4.98 *10^-4 / 0.250 L

Concentration = 0.0020 M

The concentration of the magnesium fluoride solution is 0.0020 M

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Answer: The "bleaching" of plant leaves is caused by air pollution.

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The change in enthalpy for the reaction A - P is x kJ/mol. What does the enthalpy change for the reaction P -A? (A) -x kJ/mol (B
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Answer : The correct option is, (A) -x kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

A\rightarrow P \Delta H=x\text{ kJ/mole}

Now we have to determine the value of \Delta H for the following reaction i.e,

P\rightarrow A \Delta H'=?

According to the Hess’s law, if we reverse the reaction then the sign of \Delta H change.

So, the value \Delta H' for the reaction will be:

\Delta H'=-(x\text{ kJ/mole})

\Delta H'=-x\text{ kJ/mole}

Hence, the value of \Delta H for the reaction is -x kJ/mole.

3 0
3 years ago
A container holds 0.490 m3 of oxygen at an absolute pressure of 5.00 atm. A valve is opened, allowing the gas to drive a piston,
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<u>Answer:</u> The new volume will be 2.04m^3

<u>Explanation:</u>

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2            (at constant temperature)

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=5.00atm\\V_1=0.490m^3\\P_2=1.20atm\\V_2=?m^3

Putting values in above equation, we get:

5atm\times 0.490m^3=1.20\times V_2\\\\V_2=2.04m^3

Hence, the new volume will be 2.04m^3

5 0
3 years ago
The acid dissociation constant, Ka, of HNO3 is 4.0 x104. What does the ka
Shtirlitz [24]

Answer:

i thing its b

Explanation:

3 0
3 years ago
An ideal gas in a closed container initially has a volume V and Temperature T the final tempera is 5/4T and the final pressure i
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Answer:

V_2 = \frac{5V}{8}

Explanation:

I am assuming you are saying what is the final volume of the gas

Known :

Initial volume (V1) = V

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Final temperature (T2) = 5/4 T

Initial pressure (P1) = P

Final pressure (P2) = 2P

<u>Wanted: Final volume (V2)</u>

<u />

<u />\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\frac{PV}{T} = \frac{(2P)V_2}{(5/4)T}\\\frac{V}{1} = \frac{(2)V_2}{5/4}\\5/4V = 2V_2\\\\V_2 = \frac{5V}{8}

5 0
2 years ago
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