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geniusboy [140]
3 years ago
15

C. If 62.9 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were

Chemistry
1 answer:
melomori [17]3 years ago
8 0

Answer:

Mass = 76.176 g

Explanation:

Given data:

Mass of lead(II) chloride produced = 62.9 g

Mass of lead(II) nitrate used = ?

Solution:

Chemical equation:

Pb(NO₃)₂  +  2HCl     →     PbCl₂ + 2HNO₃

Number of moles of lead(II) chloride:

Number of moles = mass/molar mass

Number  of moles = 62.9 g/ 278.1 g/mol

Number of moles = 0.23 mol

Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:

                            PbCl₂        :          Pb(NO₃)₂

                               1             :             1

                             0.23         :            0.23

Mass of Pb(NO₃)₂:

Mass = number of moles ×  molar mass

Mass = 0.23 mol × 331.2 g/mol

Mass = 76.176 g

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Answer:

a.all four carbon atoms.

Explanation:

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5 0
2 years ago
Please help!
Grace [21]
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I cannot find a definitive source that connects all this together, but the conduct of the fruit in refrigerators confirms what I am saying.

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3 0
3 years ago
What kind of liquid promote corrosion
MaRussiya [10]
Is this the whole answer?
7 0
3 years ago
The freezing point of ethanol, CH3CH2OH, is -117.300 °C at 1 atmosphere. Kf(ethanol) = 1.99 °C/m
MAXImum [283]

Answer : The molecular weight of this compound is 891.10 g/mol

Explanation :  Given,

Mass of compound = 12.70 g

Mass of ethanol = 216.5 g

Formula used :  

\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=i\times T_f\times\frac{\text{Mass of compound}\times 1000}{\text{Molar mass of compound}\times \text{Mass of ethanol}}

where,

\Delta T_f = change in freezing point

T_f^o = temperature of pure ethanol = -117.300^oC

T_f = temperature of solution = -117.431^oC

K_f = freezing point constant of ethanol = 1.99^oC/m

i = van't hoff factor = 1   (for non-electrolyte)

m = molality

Now put all the given values in this formula, we get

(-117.300)-(-117.431)=1\times 1.99^oC/m\times \frac{12.70g\times 1000}{\text{Molar mass of compound}\times 216.5g}

\text{Molar mass of compound}=891.10g/mol

Therefore, the molecular weight of this compound is 891.10 g/mol

7 0
2 years ago
4NH3 + 3O2 --> 2N2 + 6H2O
larisa86 [58]

The balanced reaction is:

4NH3 + 3O2 --> 2N2 + 6H2O

 <span>We are given the amount of reactants to be used for the reaction. This will be the starting point of our calculation.</span>

83.7g of O2 ( 1 mol / 32 g) = 2.62 mol O2

2.81 moles of NH3

From the balanced reaction, we have a 4:3 ratio of the reactants. The limiting reactant would be oxygen. We will use the amount for oxygen for further calculations.

<span>2.62 mol O2</span><span> (6 mol H2O  / 3 mol O2) (18.02 g H2O / 1 mol H2O) = 94.42 g H2O</span>

8 0
2 years ago
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