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Reika [66]
2 years ago
12

After moving down a group in the periodic table, the number of valence electrons

Chemistry
1 answer:
Delvig [45]2 years ago
8 0

the number of Valence electrons remains the same

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Atoms- 1 2 3 4
liq [111]

Answer:

The ones with 8 protons

Explanation:

Since there are two of them with 8 protons, we can assume they are the same element. The first 8 proton element has 10 neutrons while the second has 11. This makes them isotopes of one another

7 0
3 years ago
A solution of 1.50 g of solute dissolved in 25.0 mL of H₂O at 25°C has a boiling point of 100.45°C. (d) Find the van’t Hoff fact
givi [52]

The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

<h3>What is the value of van 't Hoff factor?</h3>

For most non-electrolytes dissolved in water, the Van 't Hoff factor is essentially $ 1 $ . For most ionic compounds dissolved in water, the Van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.

Which has highest Van t Hoff factor?

The Van't Hoff factor will be highest for

  A. Sodium chloride.

  B. Magnesium chloride.

  C. Sodium phosphate.

  D. Urea.

To learn more about van 't Hoff factor off factor here:

brainly.com/question/22047232

#SPJ4

4 0
2 years ago
Which of the following is accurate in describing the placement and classification of iodine?
fenix001 [56]
You can answer this question by only searching the element in the periodic table.

The atomic number of iodine, I, is 53. It is placed in the column 17 (this is the Group) and row 5 (this is the Period).

The conclusion is that the iodine is located in Period 5, Group 17, and is classified as a nonmetal.
4 0
3 years ago
How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut
san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = <em>8 = Concentration in Ethyl acetate / Concentration in water</em>

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

<em>Where X is the amount of material in grams that comes to the organic phase.</em>

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

<em>Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.</em>

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

<em>In the second extraction, you will extract 0.50g of material</em>

Thus, after the two extraction you will lost:

4.44g + 0.50g = <em>4.94g of material</em>

<em></em>

6 0
3 years ago
In the uncatalyzed reaction N2O4 (g) ⇌ 2 NO2 (g) the pressure of the gases at equilibrium are PN2O4 = 0.377 atm and PNO2 = 1.56
larisa86 [58]

Answer:

The pressures will remain at the same value.

Explanation:

A catalyst is a substance that alter the rate of a chemical reaction. It either speeds up the or slows down the rate of a chemical reaction.

While a catalyst affects the rate, it is noteworthy that it has no effect on the equilibrium position of the chemical reaction. A catalyst works by creating an alternative pathway for the reaction to proceed. Most times, it decreases the activation energy needed to kickstart the chemical reaction.

Hence, we know that it has no effect on the equilibrium position. Factors affecting equilibrium position includes, temperature and concentration of reactants and products( pressure in terms of gases).

The reactants and the products here are gaseous, and as such pressure affects the equilibrium position. Now, we have established that the equilibrium position is unaffected. And as such the pressure affecting it does not change.

Thus, we have established that the pressure of the products and reactants are unaffected and as such they remain at their value unaffected.

8 0
4 years ago
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