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3 years ago

Which statements describe band theory? Check

Chemistry

1 answer:

torisob [31]3 years ago

8 0

Answer:

B and C is corrent i got it right on edge

Explanation:

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50 Points!! 25 Each Points Please Help ASAP!! View Attached Image!! Will Mark Brainliest If All Are Answered!!

Tema [17]

Answer:

Match the words to the definitions.

Explanation:

1. F

2. A

3. C

4. E

5. B

6. D

7. B

8. F

9. A

10. E

11. D

12. C

13. B

4 0

3 years ago

Which equation represents a neutralization reaction?

sergij07 [2.7K]

A neutralization reaction is in essence a reaction between an acid an a base which usually results in pH neutral compounds being formed (a salt and water). Therefor the answer would be OPTION A (base + acid → salt + water)

6 0

4 years ago

What happens to the composition of an atom to cause it to become positively or negatively charged?

cupoosta [38]

The addition and subtraction of negatively charged electrons can easily change an atom’s charge, because they perpetually spin in valence shells outside the nucleus. It is easier for a neighboring atom to share or steal an electron rather than a positively charged proton, which is found in the nucleus. It requires a strong energy input to split a proton free from other protons and neutrons. thus, the atoms lose or gain electrons from neighboring ones and become what is known as "ions". Hope it helped!

6 0

4 years ago

A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to

melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the Henderson-Hasselbach (H-H) equation:

pH = pKa + log $\frac{[CH_3COO^-]}{[CH_3COOH]}$

For acetic acid, pKa = 4.75.

We calculate the original number of moles for acetic acid and acetate, using the given concentrations and volume:

CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻

CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to calculate the new pH, by using the new concentrations:

pH = 4.75 + log $\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L}$ = 5.54

6 0

3 years ago

Sample A: 300 mL of 1M sodium chloride

yarga [219]

Answer:

sample B contains the larger density

Explanation:

Given;

volume of sample A, V = 300 mL = 0.3 L

Molarity of sample A, C = 1 M

volume of sample B, V = 145 mL = 0.145 L

Molarity of sample B, C = 1.5 M

molecular mass of sodium chloride, Nacl = 23 + 35.5 = 58.5 g/mol

Molarity is given as;

$C = \frac{moles \ of \ solute, \ mol}{liters \ of \ solvent} \\\\Moles \ of \ solute \ for \ sample \ A = 1 \times 0.3 = 0.3 \ mol\\\\Moles \ of \ solute \ for \ sample \ B = 1.5 \times 0.145 = 0.2175 \ mol$

The reacting mass for sample A = 0.3mol x 58.5 g/mol = 17.55 g

The reacting mass for sample B = 0.2175 mol x 58.5 g/mol = 12.72 g

The density of sample A $= \frac{mass}{volume} = \frac{17.55}{0.3} = 58.5 \ g/L$

The density of sample B $= \frac{mass}{volume} = \frac{12.72}{0.145} = 87.72 \ g/L$

Therefore, sample B contains the larger density

5 0

3 years ago