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3 years ago

Circle all that apply for an anion.

Chemistry

1 answer:

Delvig [45]3 years ago

4 0

Answer:

b. d. f. h

Explanation:

Anions are negatively charged. They are usually non metals with few exceptions. They need to gain electrons in order to form and they are always listed after cation in a compound.

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Part A

Roman55 [17]

These are two questions and two answers

Question 1.

Answer:

7.33 × 10 ⁻³ c

Explanation:

1) Data:

a) m = 9.11 × 10⁻³¹ kg

b) λ = 3.31 × 10⁻¹⁰ m

c) c = 3.00 10⁸ m/s

d) s = ?

2) Formula:

The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:

λ = h / (m.s)

h is Planck's constant: h= 6.626×10⁻³⁴J.s

3) Solution:

Solve for s:

s = h / (m.λ)

Substitute:

s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg × 3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s

To express the speed relative to the speed of light, divide by c = 3.00 10⁸ m/s

s = 2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³

Answer: s = 7.33 × 10 ⁻³ c

Question 2.

Answer:

2.06 × 10 ⁻³⁴ m.

Explanation:

1) Data:

a) m = 45.9 g (0.0459 kg)

b) s = 70.0 m/s

b) λ = ?

2) Formula:

Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:

λ = h / (m.s)

h is Planck's constant: h= 6.626×10⁻³⁴J.s

3) Solution:

λ = h / (m.s)

Substitute:

λ = 6.626×10⁻³⁴J.s / ( 0.0459 kg × 70.0 m/s) = 2.06 × 10 ⁻³⁴ m

As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.

Answer: 2.06 × 10 ⁻³⁴ m.

5 0

3 years ago

The tundra is characterized by _____.

muminat

Extremely cold temperatures

5 0

3 years ago

You are provided with 300.0 mL of a buffer solution consisting of 0.200 M H3BO3 and 0.250 M NaH2BO3.

My name is Ann [436]

Answer:

a. 9.34

b. 9.06

c. 6 mL

Explanation:

Part a.

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

pH = pKa + log [A⁻] / [HA]

where pKa is the negative log of Ka for the weak acid H₃BO₃ and can be obtained from reference tables, [A⁻] and [HA] are the concentrations of the weak conjugate base H₂BO₃⁻ and and the weak acid H₃BO₃ respectively.

Proceeding with the calculations, we have

Ka H₃BO₃ = 5.80 x 10⁻¹⁰

pKa = - log (5.80 x 10⁻¹⁰) = 9.24

[H₂BO₃⁻ ] = 0.250 M

[H₃BO₃] = 0.200 M

pH = 9.24 + log (0.250/0.200) = 9.34

part b.

When 1.0 mL of 6.0 M HCl is added to the buffer , we know that it will react with the conjugate base in the buffer doing what buffers do: keeping the pH within a small range according to the capacity of the buffer:

H₂BO₃⁻ + H⁺ ⇒ H₃BO₃

So lets calculate the new concentrations of acid and conjugate base after reaction and apply the Henderson equation again:

Initial # of moles:

H₃BO₃ = 0.300 L x 0.200 mol/L = 0.06 mol

H₂BO₃⁻ = 0.200 L x 0.250 mol/L = 0.05 mol

mol HCl = 0.001 L x 6.0 mol/L = 0.006 mol

After reaction

H₃BO₃ = 0.06 mol + 0.006 mol = 0.066 mol

H₂BO₃⁻ = 0.05 mol - 0.006 mol = 0.044 mol

New pH

pH = 9.24 + log ( 0.044 / 0.66 ) = 9.06

Note: There is no need to calculate the new concentrations since we have a quotient in the expression where the volumes cancel each other.

Part c.

We will be using the Henderson-Hasselbach equationm again but now to calculate ratio [H₂BO₃⁻] / [HBO₃] that will give us a pH of 10.00. Thenwe will make use of the stoichiometry of the reaction to calculate the volume of NaOH required.

pH = pKa + log[H₂BO₃⁻]-[H₃BO₃]

10.00 = 9.24 + log [H₂BO₃⁻]-[H₃BO₃]

⇒[H₂BO₃⁻] / [H₃BO₃] = antilog (0.76) = 5.75

Initiall # moles:

mol H₃BO₃ = 0.06 mol

mol H₂BO₃ = 0.05 mol

after consumption of H₃BO₃ from the reaction with NaOH:

H₃BO₃ + NaOH ⇒ Na⁺ + H₂BO₃⁻ + H₂O

mol H₃BO₃ = 0.06 - x

mol H₂BO₃⁻ = 0.05+ x mol

Therefore we have the algebraic expression:

[H₂BO₃⁻] / [H₃BO₃] = mol H₂BO₃⁻ / mol HBO₃ = 5.75

( again volumes cancel each other)

0.05 + x / 0.06 - x = 5.75 ⇒ x = 0.044

SO 0.037 mol NaOH were required, and since we know Molarity = mol / V we can calculate the volume of 6.0 M NaOH added:

V = 0.044 mol / 6.0 mol/L = 0.0073 L

V = 7.3 mL

6 0

4 years ago

All atomic nuclei, except those of ordinary hydrogen, contain neutrons. a. True b. False

ozzi

This would be false

4 0

3 years ago

Read 2 more answers

Newton's Third Law states that for every ________ there is an equal and opposite _____________.

slava [35]

Answer:

D. Action, Reaction

Explanation:

Newton's third law states that for every action, there is an equal and opposite reaction.

Point to be noted:

The action and reaction pair act on different bodies and not on the same body.

Example:

When you jump up from the ground, your legs apply an action force on the ground and the ground applies equal and opposite reaction force on you and hence you propel up from the ground.

5 0

3 years ago