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Ymorist [56]
3 years ago
6

A lab technician needs to create 570.0 milliliters of a 2.00 M solution of magnesium chloride (MgCl2). To make this solution, ho

w many grams of magnesium chloride does the technician need?
Refer to the periodic table for help. Express your answer to three significant figures.
Chemistry
1 answer:
love history [14]3 years ago
7 0

Answer:

108.3g

Explanation:

Given parameters:

Volume of solution = 570mm = 0.57L (1000mm = 1L)

Molarity of solution = 2M

Unknown

Mass of MgCl₂

Solution

We first find the  number of moles in the given concentration of magnessium chloride using the expression below:

       Number of moles of MgCl₂= Molarity x volume = 2 x 0.57 = 1.14moles

Using this known moles, we can the unknown mass of MgCl₂ the technician would require:

       Mass of MgCl₂ required = number of moles of MgCl₂ x molar mass

Molar mass of MgCl₂:

Atomic mass of Cl = 35.5g

Atomic mass of Mg = 24g

MgCl₂ = 24 + (35.5x2) = 95gmol⁻¹

Mass of MgCl₂ required = 1.14mole x 95gmole⁻¹ = 108.3g

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Astroturf is a durable artificial surface used to cover athletic fields. A soccer field 0.06214- mile-long by 253 ft wide is cov
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Answer:

The weight of the Astroturf is 179,684.31 Newtons.

Explanation:

Length of a soccer field = 0.06214 mile = 328.0992 feet

(1 mile = 5280 feet)

Breadth of a soccer field  = 253 feet

Length of a Astroturf which soccer field is to be covered, l = 328.0992 feet

Breadth of a Astroturf which soccer field is to be covered ,b = 253 feet

Thickness of a Astroturf with which soccer field is to be covered = h

h = ½ inch = 0.5 inch = 0.041665 feet

(1 inches = 0.08333 feet)

Volume of the Astroturf ,V= l × b × h

V=328.0992 ft\times 253 ft\times 0.041665 ft=3,458.574 ft^3

Mass of the Astroturf = m

Density of the Astroturf = d = 187 oz/ft^3

d=\frac{m}{V}

m=d\times V= 187 oz/ft^3\times 3,458.574 ft^3=646,753.35 oz

1 oz = 0.0283495 kg

m=646,743.35 oz=646,743.35\times 0.0283495 kg=18,335.13 kg

Weight of the Astroturf = W

W = mg

=W=18,335.13 kg\times 9.8 m/s^2=179,684.31 N

The weight of the Astroturf is 179,684.31 Newtons.

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olchik [2.2K]

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R = 272pm = (272 × (10^-12))m = (2.72 × (10^-10))m

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a = (6.28157 × (10^-10))m = 628pm

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Answer:

Answer in explanation

Explanation:

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C. The products are on the right side of the yields arrow and not at the left side

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