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coldgirl [10]
3 years ago
8

How do you think a redox reaction works?

Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
3 0
Any chemical reaction in which the oxidation number of an atom,molecule,or ion changes by gaining or losing an electron
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Answer:

D-the measure of temperature in a system

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3 years ago
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How many moles of water are produced from 40.0 g of C2H6?
inn [45]

Answer:

4 moles of water

Explanation:

this is a combustion reaction, so the balanced equation is: 2C2H6 + 7O2 → 4CO2 + 6H2O.

the molar mass of C2H6 is 30.07g, so 40.0 g of C2H6 is 1.33 moles of C2H6.

mole ratio of H2O to C2H6 is 6/2, or 3.

1.33 moles C2H6 * 3 moles H2O/1 mole C2H6 = 4 moles H2O

4 0
2 years ago
A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of
Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa 

NaOH + CH3COOH → CH3COONa + H2O 

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH 

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH 

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L 

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M 

CH3COONa = 0.0076 / 0.071 = 0.1070M 

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74. 

pH using Henderson - Hasselbalch equation: 

pH = pKa + log ([salt]/[acid]) 

pH = 4.74 + log ( 0.1070/0.1493) 

pH = 4.74 + log 0.717 

pH = 4.74 + (-0.14) 

pH = 4.60.

7 0
3 years ago
The central core of an atom is called the __?
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Answer: Atomic Nucleus!

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(Yes, it was from google.)

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Explain why ice is able to float on water
agasfer [191]

Ice floats because it is about 9% less dense than liquid water. ... The heavier water displaces the lighter ice, so ice floats to the top. One consequence of this is that lakes and rivers freeze from top to bottom, allowing fish to survive even when the surface of a lake has frozen over

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