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Nina [5.8K]
2 years ago
7

For an ideal gas, evaluate the volume occupied by 0.3 mole of gas.

Chemistry
1 answer:
Serhud [2]2 years ago
8 0

Answer:

V=0.3×22.4=6.72 liters hope this helps

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How many atoms are in 12 moles of I. Check your significant figures.
ELEN [110]

Explanation:

7.2×1024 moleas

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5 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
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Combustion is an exothermic reaction where the reactants are combined with ____
Anna [14]

Answer:

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7 0
3 years ago
Deep under water canyons are called
noname [10]
These are called ocean or underwater trenches <span />
5 0
3 years ago
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Which of the following equations follows the law of conservation of mass?
Dovator [93]

Answer:

option A = C₂H₄ + 3O₂   →    2CO₂   +  2H₂O

Explanation:

Law of conservation of mass:

This law stated that mass can not be created or destroyed in chemical reaction. It just changed from one to another form.

For example:

C₂H₄   +   3O₂   →    2CO₂   +  2H₂O

28 g   + 96 g    =      88 g  +  36 g

    124  g           =      124 g

This reaction correctly hold the law of conservation of mass.

Other options:

C  + 4H₂  →   CH₄

12 g  + 8g  = 16 g

  20 g  =   16 g

This reaction do not hold the law of conservation of mass.

3H₂O  →   3H₂  +  3O₂

  54 g  =   6 g +  96 g

   54 g =   102 g

This reaction do not hold the law of conservation of mass.

2Na  +  Cl  →  NaCl

46 g  + 35.5 g  =   58.5 g

  81.5 g = 58.5 g

This reaction do not hold the law of conservation of mass.

6 0
3 years ago
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