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Karo-lina-s [1.5K]
3 years ago
7

What is the mass in grams of 0.450 moles of magnesium sulfate, MgSO4?

Chemistry
1 answer:
sergiy2304 [10]3 years ago
7 0

Answer:54.17

Explanation:

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Answer:

Inner

Explanation:

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Part A The first step to engineering is to define the problem. Write down the problem the students have to solve, and describe t
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2 years ago
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20 ML of a gas at 200 K is heated until the new volume is 55ML what is the final temperature of the gas
MrRa [10]

Answer:

T2 = 550K

Explanation:

From Charles law;

V1/T1 = V2/T2

Where;

V1 is initial volume

V2 is final volume

T1 is initial temperature

T2 is final temperature

We are given;

V1 = 20 mL

V2 = 55 mL

T1 = 200 K

Thus from V1/T1 = V2/T2, making T2 the subject;

T2 = (V2 × T1)/V1

T2 = (55 × 200)/20

T2 = 550K

8 0
2 years ago
Which will most likely result if there is increased upwelling in a coastal area
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3 years ago
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In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2
valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

\dfrac{80}{17}=4.706

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

\tt \dfrac{120}{32}=3.75

Mol ratio of reactants(to find limiting reatants) :

\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

\tt \dfrac{6}{5}\times 3.75=4.5

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%

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3 years ago
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