Which substance is a strong electrolyte?

Rank the following atoms in order of increasing electronegativity.a. Se, O, S b. P, Na, Cl c. Cl, S, F d. O, P, N

Papessa [141]

Answer:

a)- Se, S, O

b)- Na, P, Cl

c)- S, Cl, F

d)- P, N, O

Explanation:

We can solve this problem by looking at the Periodic Table. In a group, electronegativity increases from bottom to top; whereas in periods it increases from left to right.

Thus, we order the elements from lower to higher electronegativity as follows:

a. Se, O, S ⇒ order: Se, S, O

Because they are all in the same group. Se is near the bottom, followed by S and O is at the top.

b. P, Na, Cl ⇒ order: Na, P, Cl

They are in the same period. Na is at the left, followed by P and Cl is nearest the right.

c. Cl, S, F ⇒ order: S, Cl, F

P and Cl are in the same period, and P is at the left, so it has the lowest electronegativity. F is in the same group of Cl, but at the top. F has the highest electronegativity.

d. O, P, N ⇒ order: P, N, O

N and P are in the same group, but P is at the bottom so it has the lower electronegativity. N and O are in the same period, but O is at the right, so it is the most electronegative.

8 0

3 years ago

What is the correct formula that would result from the combination of the two ionic species?

Soloha48 [4]

Answer:

CsCl

Explanation:

The ionic species are:

Cs⁺ and Cl⁻

6 0

2 years ago

8. The time period of artificial satellite in a circular orbit of radius R is T. The radius of the orbit in which time period is

Elena-2011 [213]

Explanation:

It is given that,

The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :

$T^2\propto R^3$

The radius of the orbit in which time period is 8T is R'. So, the relation is given by :

$(\dfrac{T}{T'})^2=(\dfrac{R}{R'})^3$

$(\dfrac{T}{8T})^2=(\dfrac{R}{R'})^3$

$\dfrac{1}{64}=(\dfrac{R}{R'})^3$

$R'=4\times R$

So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.

4 0

3 years ago

How much time would it take for 336 mg of copper to be plated at a current of 5.6 A ? Express your answer using two significant

schepotkina [342]

Answer:

1.8 × 10² s

Explanation:

Let's consider the reduction that occurs upon the electroplating of copper.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We will establish the following relationships:

1 g = 1,000 mg
The molar mass of Cu is 63.55 g/mol
When 1 mole of Cu is deposited, 2 moles of electrons circulate.
The charge of 1 mole of electrons is 96,486 C (Faraday's constant).
1 A = 1 C/s

The time that it would take for 336 mg of copper to be plated at a current of 5.6 A is:

$336mgCu \times \frac{1gCu}{1,000mgCu} \times \frac{1molCu}{63.55gCu} \times \frac{2mole^{-} }{1molCu} \times \frac{94,486C}{1mole^{-}} \times \frac{1s}{5.6C} = 1.8 \times 10^{2} s$

3 0

3 years ago

I created a typical vinegar and baking soda experiment. About 100 ml of white vinegar, and 100ml of baking soda. Once the reacti

olganol [36]

It is 20 because then you will put 30 inside of 20milimeeter and put 70 years into the other pint and put it learnt into the cup

7 0

2 years ago