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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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The compound is used in medicine as a source of magnesium ions, which are essential for many cellular activities. Magnesium chloride has also been used as a cathartic and in alloys. To low
6 0
4 years ago
A 16.0 mL sample of a 1.04 M potassium sulfate solution is mixed with 14.3 mL of a 0.880 M barium nitrate solution and this prec
stiks02 [169]
Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol

Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol

Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.

The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%

Answer: 
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%

I apologize for the mistake previous to this update.

5 0
3 years ago
What kind of a bond is oc2
nexus9112 [7]

Answer:

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8 0
3 years ago
A positive test with 2,4 dinitrophenylhydrazine could indicate the presence of which of the following functional groups ?
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8 0
3 years ago
The recommended daily allowance (RDA) of the trace metal magnesium is 410 mg/daymg/day for males. Express this quantity in μg/da
Ira Lisetskai [31]

Answer:

410mg/day *\frac{1000 ug}{1 mg}

Recommended daily Amount (RDA) of magnesium is  410,000 μg/day.

Explanation:

There are 1000μg in 1 mg and 1000 mg in 1g

1 mg=1000μg

The Recommended daily Amount (RDA) is 410 mg/day of magnesium. Converting 410 mg/day into μg/day

410mg/day *\frac{1000 ug}{1 mg}

It will become 410,000 μg/day

So Recommended daily Amount (RDA) of magnesium is  410,000 μg/day.

4 0
3 years ago
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