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goldenfox [79]
2 years ago
15

If 13.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

Chemistry
1 answer:
Paul [167]2 years ago
4 0

6.349 g mass of anhydrous magnesium sulfate will remain.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Molar mass MgSO₄.7 H₂O = 246.52 g/mol

Moles =\frac{mass}{molar \;mass}

Moles =\frac{13.0 g}{246.52}

0.0527 moles

Molar mass MgSO₄ = 120.4 g/mol

Mass of anhydrous magnesium sulfate :

( 0.0527 x 120.4 ) => 6.349 g

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

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vekshin1

Answer 1) In the given reaction of sulfuric acid


2NO_{(g)} + O_{2}_{(g)} ---->  2NO_{2}_{(g)}


2NO_{2}_{(g)} + 2SO_{2}_{(g)} ---->  2NO}_{(g)} + 2SO_{3}_{(g)}


On addition of nitrogen monoxide gas the reaction rate increases and more amount of product is formed.


So, it is clear that NO is the catalyst in this reaction.


Answer 2) This can be proven that NO is catalyst because it increases the rate of the reaction, but it is not consumed during the reaction, and it also gets regenerated at the end of reaction.


Hence, nitrogen mono oxide is considered as the catalyst in the given reaction.


Answer 3) It increases the rate of reaction by decreasing the activation energy of the reaction. Also it can be clearly seen in this reaction the NO is reacting with oxygen to lower the energy of activation. So, it is providing an alternative pathway for proceeding the reaction. This all confirms the assumptions of NO being the catalyst.


2NO_{(g)} + O_{2}_{(g)} ---->  2NO_{2}_{(g)}

4 0
3 years ago
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How many grams are present in a sample of aluminum?
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4 years ago
Which of the following methods would be the best method for separating oil from water?
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Teniendo en cuenta los siguientes fenómenos: ebullición del agua- movimiento de un cuerpo- disolución de sal en agua- combustión
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Answer:

Las siguientes son reacciones químicas;

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Explanation:

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3 0
2 years ago
A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
Read 2 more answers
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