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Alisiya [41]
3 years ago
8

Many times, during a chemical reaction, energy is given off. this type of chemical reaction can be termed

Chemistry
1 answer:
WITCHER [35]3 years ago
3 0
Physical can be seen
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Explain what each quantum number in a quantum number set tells you about the electron. Compare and contrast the locations and pr
klio [65]
This is a bit long right now.

(n l m s) these are the numbers.

n=2, second orbital level.
l is the type of orbit: l=1 elongated, l=0 spherical
m=0, magnetic number: 0
s=spin, both have spin positive

You need still to round it up. Srry!

4 0
2 years ago
Read 2 more answers
Calculate the mass of 3.5 mol C6H6
Lilit [14]
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
4 0
3 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
When working with hazardous substances in the laboratory your partner splashes a small amount of Sodium Hydroxide onto their clo
alexira [117]

Answer:

a) After helping our partner, we should immediately report the incident to the lab manager or any person in charge of the emergencies occurring in the lab.

b) We should have a copy of the Material Safety Data Sheet to give to the responders. This is because the responder can identify what materials were being used by the person ans what other security measures need to be taken.

5 0
3 years ago
Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.
TEA [102]

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

HCl+KOH\rightarrow KCl+H_2O

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

n_{KOH}=0.00768mol-0.00192mol=0.00576mol

And the resulting concentration of KOH and OH ions as this is a strong base:

[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M

And the resulting pH is:

pH=14+log(0.131)\\\\pH=13.1

Regards!

3 0
3 years ago
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