Ask question

Ask question

All categories

3 years ago

6

Drag each label to the correct location. An element’s position in the periodic table provides information about its atomic struc

ture. Use what you know about atomic structure to complete the table. Refer to the periodic table as needed.

2 answers:

kykrilka [37]3 years ago

5 0

Answer:

m

Explanation:

m

Mekhanik [1.2K]3 years ago

4 0

Answer:

Wa hatjejr jdbdj aldvdid so ini

You might be interested in

Hey! If someone could help me out I would really appreciate it even if it was just one answer. I'm really stuck on 3. Its second

Sphinxa [80]

Answer:

I would go for number 5 . It sounds more logical to me.

7 0

2 years ago

How many valence electrons does group 18 have

Rus_ich [418]

Answer:

group 18 has 8 valence electrons, that's why they are stable.

8 0

3 years ago

Read 2 more answers

A buffer was prepared by mixing 1.00 mole of ammonia and 1.00 mole of ammonium chloride to form an aqueous solution with a total

lidiya [134]

Answer:

Idek sorry

Explanation:

So sorry i think its 1.00

3 0

2 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

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

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

3 years ago

How many atoms of gold are in a pure gold coin weighing 12.0g?

grigory [225]

12.0g x 1 mol / 63.546g = 0.188839581mol

<span>So, for every 1 mole, we have 6.022 x 10^23 of whatever we're measuring. This gives us a conversion factor of (1 mole / 6.022 x 10^23 atoms) or (6.022 x 10^23 atoms / 1 mole).
</span>
0.188839581 mol x (6.022 x 10^23 atoms) / 1 mol = 1.137191955 x 10^23

<span>Remember from before that we are limited to 3 significant figures. Since our calculations are complete, we can now round down to: 1.14 x 10^23 </span>

<span>That should be your answer!

Hope it helps!
xo</span>

3 0

3 years ago