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choli [55]
3 years ago
5

A voltaic cell differs from an electrolytic cell in that a chemical cell uses

Chemistry
1 answer:
oksian1 [2.3K]3 years ago
7 0

Answer:

3) an applied electric current

Explanation:

An electric source is used in an electrolytic cell.

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Given the equation: HCI + Na2SO4 ---> NaCl + H2SO4 Hint: you should balance the equation If you start with 2.0 g of HCl (hydr
Rasek [7]

Answer:

approx 2.45g

Explanation:

2HCl + Na2SO4 → 2NaCl + H2SO4

 2    :       1           :      2       :      1

0.05                                     0.025 (moles)

⇒ mH2SO4 = 0.025 × 98 = 2.45 g

3 0
3 years ago
In a coffe cup calorimeter, 50.0mL of 0.100M of AgNO3 and 50mL of 0.100M HCl are mixed to yield the following reaction:
Jet001 [13]

Answer:

The enthalpy change of the reaction is -66.88 kJ/mol.

Explanation:

Mass of the solution = m = 100 g

Heat capacity of the solution = c = 4.18 J/g°C

Initial temperature of the solutions before mixing = T_1=22.60^oC

Final temperature of the solution after mixing = T_2=23.40^oC

Heat gained by the solution due to heat released by reaction between HCl and silver nitrate = Q

Q=m\times c\times (T_2-T_1)

Q=100 g\times 4.18 J/g^oC\times (23.40^oC-22.60^oC)=334.4 J

Heat released due to reaction = Q' =-Q = -334.4 J

Moles of silver nitrate = n

Molarity of silver nitrate solution = 0.100 M

Volume of the silver nitrate solution = 50.0 mL = 0.050 L ( 1 mL = 0.001 L)

Moles =Molarity\times Volume (L)

n=0.100 M\times 0.050 L=0.005 mol

Enthalpy change of the reaction = \Delta H

=\Delta H=\frac{-334.4 J}{0.005 mol}=-66,880 J/mol=-66.88 kJ/mol

1 J = 0.001 kJ

The enthalpy change of the reaction is -66.88 kJ/mol.

7 0
3 years ago
Given these reactions,
Andrej [43]

Answer:

-805.8kJ

Explanation:

8 0
3 years ago
CAN ANYONE ELP ME WITH CHEMISTRY PLEASE?? I will be asking multiple questions but even if you can only answer one-- I will still
vlabodo [156]
3, 9, 16 are correct! 17 should be b because buffered means that it resists change in pH
6 0
3 years ago
What is the pressure exerted by 0.122 mol oxygen gas in a 1.50-L in a container at room temperature
dem82 [27]

Answer:

The answer to your question is P = 1.95 atm

Explanation:

Data

Pressure = P = ?

number of moles = n = 0.122

Volume = V = 1.5 l

Temperature = T = 20°C

Constant of ideal gases = 0.082 atm l/mol°K

Process

1.- Convert temperature to °K

T = 20 + 273

T = 293°K

2.- Write the formula of ideal gases

       PV = nRT

-Solve for P

      P = nRT/V

3.- Substitution

     P = (0.122 x 0.082 x 293) / 1.5

4.- Simplification

     P = 2.93 / 1.5

5.- Result

     P = 1.95 atm

7 0
3 years ago
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