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ivolga24 [154]
3 years ago
11

The compound mgso4, epsom salts, is used to soothe sore feet and muscles. how many grams will you need to prepare a bath contain

ing 5.40 mol of epsom salts?
Chemistry
2 answers:
VLD [36.1K]3 years ago
8 0

Answer: Mass= n x molar mass= 4.00 moles x 120.4 g/mol = 482 g

Explanation:

number of moles = mass/ molar mass

Gala2k [10]3 years ago
3 0

Answer:

648 grams of epsom salt will be needed.

Explanation:

Mass of magnesium atom = 24  g/mol

Mass of sulfur atom= 32 g/mol

Mass of oxygen atom= 16 g/mol

Molar mass of epsom salt: 24 g/mol+32 g/mol+4\times 16 g/mol=120 g/mol

Moles of epsom salt in water = 5.40 mol

Mass of 5.40 moles of  epsom salt : Molar mass × Moles:

=120 g/mol\times 5.40 mol=648 g

648 grams of epsom salt will be needed.

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Calculate the volume of 2.65 moles of N2O gas at STP.
Reptile [31]

Answer:

63.6 dm³

Explanation:

Volume of gas= no. of moles ×24dm³

Thus, volume of N₂O

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3 0
3 years ago
If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe
pishuonlain [190]

Answer:

50

Explanation:

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ:           30.01     32.00   46.01

               2NO   +   O₂ ⟶ 2NO₂

Mass/g:  80.00     16.00

2. Calculate the moles of each reactant  

\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

\text{Moles of NO}_{2} =  \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}

(b) Mass of NO reacted

\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

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What are the limitations of the van der Waals equation.​
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