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NeX [460]
3 years ago
11

A chemist adds of a magnesium fluoride solution to a reaction flask. Calculate the mass in micrograms of magnesium fluoride the

chemist has added to the flask. Round your answer to significant digits.
Chemistry
1 answer:
Yuri [45]3 years ago
6 0

The given question is incomplete, the complete question is:

A chemist adds 35.0mL of a 6.19 * 10^−4/mmol magnesium fluorideMgF2 solution to a reaction flask. Calculate the mass in micrograms of magnesium fluoride the chemist has added to the flask. Round your answer to

3 significant digits.

Answer:

The correct answer is 1.35 microgram.

Explanation:

Based on the given information,

The volume of magnesium fluoride given is 35 ml, and the concentration of magnesium fluoride is 6.19 × 10⁻⁴ mmol/L.

Now the moles of MgF₂ can be determined by using the formula,

Moles = Concentration × Volume

Moles of MgF₂ = Concentration of MgF₂ × Volume of MgF₂

= 6.19 × 10⁻⁴ mmol/L × 35 ml × L/1000 ml

= 217 × 10⁻⁷ mmol

The molecular mass of magnesium fluoride is 62.3 gram per mole

Thus, the mass of MgF₂ is,

= 217 × 10⁻⁷ mmol × 62.3 g/mol

= 13500 × 10⁻⁷ mg

= 1.35 microgram

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which of the following describes something an engineer would probably do first when developing new technology meant to reduce oc
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Option B. is the right answer.

Explanation:

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3 years ago
Insoluble compounds in a chemical reaction are called?
Rzqust [24]
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What volume will be occupied by .756 mole of gas at 109 kPa and 30.5 degrees C
olga2289 [7]

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3 years ago
A sample of nitrogen gas had a volume of 500. mL, a pressure in its closed container of 740 torr, and a temperature of 25 °C. Wh
iren2701 [21]

Answer:  Thus the new volume of the gas is 530  ml

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas =  740 torr

P_2 = final pressure of gas = 760 torr

V_1 = initial volume of gas = 500 ml

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 25^oC=273+25=298K

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Now put all the given values in the above equation, we get:

\frac{740\times 500}{298}=\frac{760\times V_2}{323}

V_2=530ml

Thus the new volume of the gas is 530 ml

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3 years ago
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