Ask question

Ask question

All categories

2 years ago

13

How does a lithium cation compare to a lithium atom?

2 answers:

ExtremeBDS [4]2 years ago

6 0

The correct answer is gonna be C) A lithium cation ion is smaller

A lithium cation has lost its valence electrons, which causes the remaining electrons to be pulled in stronger by the positive charge in the nucleus. As they get closer to the nucleus, the overall size of the atom is decreased.

Assoli18 [71]2 years ago

3 0

The correct answer is c

You might be interested in

A diver swims 5. 875 feet up to the surface of the water. Which number represents the diver’s position in relation to the surfac

Tanzania [10]

Answer

Negative 5 and StartFraction 7 over 8 EndFractionfeet

7 0

2 years ago

About how long is 1 meter?

lyudmila [28]

It should be 3 feet... I did something like this one day in school but I’m trying to remember

3 0

3 years ago

Read 2 more answers

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

2 years ago

Review this reaction:H2SO4 + NaOH ?.What are the product(s)?Na2SO4H2ONa2SO4 and H2ONaH and SO4OHH3SO5

muminat

i think its MIDDLE FINGERS UP IN THE SKY AND AT THESE AHOLE MODERATORS

6 0

3 years ago

Read 2 more answers

The reducing agent in the reaction described in Fe + 2HCl → FeCl2 + H2 is A. H2. B. Fe. C. HCl. D. FeCl2.

Fed [463]

The reducing agent is b). Fe

7 0

3 years ago

Read 2 more answers