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SVEN [57.7K]
2 years ago
11

Which statement about oxidation and reduction in a voltaic cell is true?

Chemistry
2 answers:
Verizon [17]2 years ago
6 0
<span>Oxidation is the loss of electrons and corresponds to an increase in oxidation state. A reduction is the gain of electrons and corresponds to a decrease in oxidation state. Balancing redox reactions can be more complicated than balancing other types of reactions because both the mass and charge must be balanced. Redox reactions occurring in aqueous solutions can be balanced by using a special procedure called the half-reaction method of balancing. In this procedure, the overall equation is broken down into two half-reactions: one for oxidation and the other for reduction. The half-reactions are balanced individually and then added together so that the number of electrons generated in the oxidation half-reaction is the same as the number of electrons consumed in the reduction half-reaction.</span>
Dafna11 [192]2 years ago
3 0
It's C ..........................
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Density is d= m/vol. If a material has a mass of 65.5 g and a volume of 32.5 ml, it has a density of
ryzh [129]

Using the given formula, the density of the material is 2.015 g/mL

<h3>Calculating Density </h3>

From the question, we are to determine the density of the material

From the given formula

Density = Mass / Volume

And from the given information,

Mass = 65.5 g

and volume = 32.5 mL

Putting the parameters into the equation,

Density = 65.5/32.5

Density = 2.015 g/mL

Hence, the density of the material is 2.015 g/mL.

Learn more on Calculating density here: brainly.com/question/24772401

#SPJ1

8 0
2 years ago
How much energy (kJ) is required to change 0.18 mole of ice (s) at 0 C to water (l) at 0 C?
Dmitriy789 [7]

Answer:25,06 kJ of energy must be added to a 75 g block of ice.

ΔHfusion(H₂O) = 6,01 kJ/mol.

T(H₂O) = 0°C.

m(H₂O) = 75 g.

n(H₂O) = m(H₂O) ÷ M(H₂O).

n(H₂O) = 75 g ÷ 18 g/mol.

n(H₂O) = 4,17 mol.

Q = ΔHfusion(H₂O) · n(H₂O)

Q = 6,01 kJ/mol · 4,17 mol

Q = 25,06 kJ.

Explanation:

6 0
2 years ago
A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d
Karo-lina-s [1.5K]

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

8 0
2 years ago
Which of the following is the smallest volume? (2 points) 2500 mL
Virty [35]
So to put them all in the same units we have 
<span>2500 mL </span>
<span>250 mL </span>
<span>25mL </span>
<span>2,500,000,000mL </span>

<span>So the third one is the smallest</span>
6 0
3 years ago
Read 2 more answers
SO NOT HALAL MODE :(
amid [387]

Answer:

umm

Explanation:

5 0
3 years ago
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