A solution contains 90 mL of methanol, 18 mL of propanol, and 2 mL of diethyl ether. Which is the solvent in this solution?

Write the complete and net ionic equations for the following reactions. If only the reactants are given, predict the products an

11111nata11111 [884]

Explanation:

u can put them in order crossed are spectator ions. hope this helps:)

3 0

2 years ago

The crystal structure of an ionic compound depends on the

Murljashka [212]

Answer:

The ion's sizes and their radius ratio.

Explanation:

Ionic compounds take the form of three dimentional arrays by alternating anions and cations in order to minimize the potential energy of the system by maximizing the attractive force between opposite charges.

The resultant geometric structures are known as crystal lattices and their arrangements depend on the ion's sizes and, since they are bound though electrostatic attraction, their radius ratio.

I hope you find this information useful and interesting! Good luck!

4 0

3 years ago

553.5 mL ?:cm^3 how do i get my answer? and what is the answer?

dmitriy555 [2]

Answer:

553.5 cm3

Reason:

1 mL = 1 cm3

6 0

3 years ago

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

3 years ago

Write the complete balanced equation for the reaction between aluminum metal (Al) and oxygen gas (O2). You do not need to make t

Soloha48 [4]

For the answer to the question above, asking to write the complete balanced equation for the reaction between aluminum metal (Al) and oxygen gas (O2)and You do not need to make the subscripts smaller.
My answer would be,
4Al(s) + 3O2(g) --->2 Al2O3(s)

I hope this helps.

3 0

3 years ago