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o-na [289]
3 years ago
8

What is the volume of 0.250 m hydrochloric acid required to react completely with 20.0 ml of 0.250 m ca(oh)2?

Chemistry
1 answer:
Soloha48 [4]3 years ago
5 0
The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol  in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol  / 0.250 mol/L = 40.0 mL 
40.0 mL of 0.250 M HCl is required
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Answer: The molar mass of each gas

Explanation:

Mole fraction is the ratio of moles of that component to the total moles of solution. Moles of solute is the ratio of given mass to the molar mass.

\text{Mole fraction of solute}=\frac{\text{Moles of solute}}{\text{Total Moles}}

Suppose if there are three gases A, B and C.

a) \text{Mole fraction of A}=\frac{\text{Moles of A}}{\text{ Moles of (A+B+C)}}

b) \text{Mole fraction of B}=\frac{\text{Moles of B}}{\text{ Moles of (A+B+C)}}

c) \text{Mole fraction of C}=\frac{\text{Moles of C}}{\text{ Moles of (A+B+C)}}

moles of solute =\frac{\text {given mass}}{\text {Molar mass}}

Thus if mass of each gas is known , we must know the molar mass of each gas to know the moles of each gas.

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Yes it is B,D.

Explanation:

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3 years ago
The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
avanturin [10]

Answer : The actual cell potential of the cell is 0.47 V

Explanation:

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given redox reaction is :

Ni^{2+}(aq)+Zn(s)\rightarrow Ni(s)+Zn^{2+}(aq)

The balanced two-half reactions will be,

Oxidation half reaction : Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction : Ni^{2+}+2e^-\rightarrow Ni

The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

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

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

4 0
3 years ago
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