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Sati [7]
3 years ago
8

5.0 liters of a 0.1 M Ca(OH), solution

Chemistry
1 answer:
LiRa [457]3 years ago
3 0

Answer:

0.5mol

Explanation:

mole = concentration × volume

mole = 0.1 × 5.0 = 0.5mol

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Which of the following reactions could be used to extract the lead from the lead nitrate?
alexgriva [62]

Answer:

Option b. Decomposition

Followed by a reduction process using charcoal

Explanation:

Lead can be obtained from lead nitrate by thermal decomposition of lead nitrate as shown below:

2Pb(NO3)2 —> 2PbO + 4NO2 + O2

The PbO obtained is reduced by charcoal(C) to obtain the metallic Pb as shown below:

2PbO + C —> Pb + CO2

4 0
3 years ago
(4.9 x 10-2) (9.80 x 102) =
Viktor [21]
The answer is 46981.2
5 0
3 years ago
What is the formula for vanillic acid​
liberstina [14]

the formula for tht is C8H8O4

3 0
3 years ago
A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a
Semenov [28]

<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol

  • The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

  • Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

  • To calculate the molar mass of metal for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0
3 years ago
What is the voltage for the nonspontaneous reaction between silver (Ag) and copper (Cu) and their ions?
gayaneshka [121]
For an non spontaneous reaction between silver (Ag) and copper (Cu) and their ions, Cu is the oxidizing agent while Ag+ is the reducing agent,
The following reactions will take place;
Anode Cu = Cu+2 + 2e-   E= +0.34 volts
Cathode; Ag+ + e = Ag    E = +0.80 volts
The net reaction will be Cu + 2Ag+ = Cu+2 + 2Ag
Thus, the voltage will be
  = +0.80 - (+0.34)
5 0
3 years ago
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