All categories

2 years ago

(7th grade science question)

Chemistry

1 answer:

vesna_86 [32]2 years ago

4 0

Answer:

id say mass

Explanation:

but i d k

You might be interested in

Which of the following is the correct sequence of the planets in order: CLOSEST to the sun --> FURTHEST away from the sun

Strike441 [17]

The answer is A M.V.E.M.J.S.U.N

3 0

3 years ago

Read 2 more answers

Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the

Cloud [144]

Answer : The equilibrium concentration of $H_3O^+$ at $25^oC$ is, $2.154\times 10^{-3}m$ .

Solution : Given,

Equilibrium constant, $K_c=1.8\times 10^{-5}$

Initial concentration of $HC_2H_3O_2$ = 0.260 m

Let, the 'x' mol/L of $H_3O^+$ are formed and at same time 'x' mol/L of $HC_2H_3O_2$ are also formed.

The equilibrium reaction is,

$HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)$

Initially 0.260 m 0 0

At equilibrium (0.260 - x) x x

The expression for equilibrium constant for a given reaction is,

$K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

Now put all the given values in this expression, we get

$1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}$

By rearranging the terms, we get the value of 'x'.

$x=2.154\times 10^{-3}m$

Therefore, the equilibrium concentration of $H_3O^+$ at $25^oC$ is, $2.154\times 10^{-3}m$ .

4 0

2 years ago

Another name for Rows is ...

timurjin [86]

Answer:

periods

Explanation:

Know the periodic table

- Hope that helped! Let me know if you need a further explanation.

3 0

3 years ago

Read 2 more answers

The reaction: 2 SO2(g) + O2(g) --> 2 SO3(g) has an equilibrium constant of K1. What is the K value for the reaction: SO3(g) -

Murljashka [212]

Answer: The value of equilibrium constant for reverse reaction is $(\frac{1}{K_1})^{1/2}$

Explanation:

The given chemical equation follows:

$2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

The equilibrium constant for the above equation is $K_1$

We need to calculate the equilibrium constant for the reverse equation of above chemical equation, which is:

$SO_3(g)\rightarrow SO_2(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of ' $\frac{1}{2}$ ', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(\frac{1}{K_1})^{1/2}$

Hence, the value of equilibrium constant for reverse reaction is $(\frac{1}{K_1})^{1/2}$

8 0

3 years ago

Which is a way that the elements are grouped on the periodic table? (2 points)

Stells [14]

I think the correct answer from the choices listed above is the second option. A way that the elements are grouped on the periodic table would be that elements with similar bonding properties are in the same column. Each column is called a group. The elements in each group have the same number of electrons in the outer orbital.

5 0

3 years ago

Read 2 more answers