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2 years ago

As electrons gain more energy they can move ____ away from the nucleus

Chemistry

1 answer:

Snowcat [4.5K]2 years ago

4 0

The answer is A. Further away. I hope I helped. Have a great day/night!!!!!! ❤️

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In which of the titrations described below will the first (or only) equivalence point be reached upon the addition of 25.0 mL of

Mnenie [13.5K]

Answer:

1, 3, and 4

Step-by-step explanation:

We must calculate the volume of NaOH needed for each titration.

HCl + NaOH ⟶ NaCl + H₂O

n(HCl) = 25.0 mL × (0.100 mmol/1mL) = 2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

H₂C₂O₄ + NaOH ⟶ NaHC₂O₄

n(H₂C₂O₄) = 50.0 mL × (0.100 mmol/1mL) = 5.00 mmol

n(NaOH) = 5.00 mmol H₂C₂O₄ × (1 mmol NaOH/1 mmol H₂C₂O₄)

= 5.00 mmol NaOH

V(NaOH) = 5.00 mmol × (1 mL/0.100 mmol) = 50.0 mL

HC₂H₃O₂ + NaOH ⟶ NaC₂H₃O₂ + H₂O

n(HC₂H₃O₂) = 25.0 mL × (0.100 mmol/1mL) = 2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl)

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

HBr + NaOH ⟶ NaBr + H₂O

n(HBr) = 25.0 mL × (0.100 mmol/1mL) =2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl)

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

Titrations 1, 3, and 4 reach the first or only equivalence point at 25.0 mL NaOH.

8 0

3 years ago

Read 2 more answers

Operations with units”: Determine the units of the answers to the following problems.

sammy [17]

Answer:

1) km 2) cm^ 3) no units (kg cancels) 4) g/cm^3 5) g 6) g°C

Explanation:

Like units will cancel ((m/km)*(km) = m

Non-cancelled units remain (ml)*(g/ml)*(°C) = g°C

5 0

2 years ago

Is a new can of paint open or closed system

Brums [2.3K]

Answer:

I think it might be a closed system

8 0

2 years ago

If 45.0 mL of ethanol (density =0.789g/mol) initially at 6.0°C mix with 45.0 mL of water (density =1.0 g/mol) initially at 28.0°

Likurg_2 [28]

The final temperature of the mixture : 21.1° C

<h3>Further explanation </h3>

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released

Q in(gained) = Q out(lost)

Heat can be calculated using the formula:

Q = mc∆T

Q = heat, J

m = mass, g

c = specific heat, joules / g ° C

∆T = temperature difference, ° C / K

Q ethanol=Q water

mass ethanol=

$\tt mass=\rho\times V\\\\mass=0.789\times 45=35.505~g$

mass water =

$\tt mass=1~g/ml\times 45~ml=45~g$

then the heat transfer :

$\tt 35.505\times 2.42~J/g^oC\times (t-6)=45\times 4.18~J/g^oC\times (28-t)\\\\85.922t-515.533=5266.8-188.1t\\\\274.022t=5782.33\rightarrow t=21.1^oC$

5 0

2 years ago

Please help! I have less than an hour left!

melisa1 [442]

Carbocation
I guess pls tell me if it wrong

8 0

3 years ago