5)

Please help solve this I’ll award brainliesttt

Elza [17]

Answer:

types of elements nacl MGS

behavior shared between atoms tranferred from one to another

properties high melting poin low melting point

example SI-o metal + non metal

example H-H only non metal elements

Explanation:

3 0

1 year ago

Which of the following you do NOT need to pay attention while using a centrifuge?

Phoenix [80]

Answer:

The radius of the centrifuge.

Explanation:

Hello,

Since the radius of the centrifuge is just a design parameter, it wouldn't be a cause of failure because it is used to know how many tubes could be fitted in into the centrifuge. On the other hand, keeping you attention away from other factors could turn into a failure as long as the sample could be poured down or just turn out inadequate for the expected results.

Best regards.

8 0

3 years ago

An empty graduated cylinder weighs 26.5 grams and when it is filled with an unknown liquid to the 45.8 ml mark the cylinder and

Softa [21]

Answer:

24.2 is the density of the liquid

7 0

3 years ago

g, Assuming the precipitate is totally insoluble in water, which aqueous ions will be present in the solution (collected in the

Allushta [10]

Answer:

Cl⁻, Na⁺, OH⁻

Explanation:

The titration is:

CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)

In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.

Thus, ions present after the equivalence point are: Cl⁻, Na⁺ (Don't react, spectator ions), and OH⁻.

3 0

2 years ago

Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid

Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

$35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol$

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

$\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M$

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb) where Cb is the initial concentration of the base

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

7 0

2 years ago