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8_murik_8 [283]

2 years ago

13

What is an element? Arrow

1 answer:

Butoxors [25]2 years ago

7 0

Answer:

One arrow is positioned in each box according to Hund's Rule which tells us to maximise the number of unpaired electrons in orbitals of the same subshell, and, to give those electrons the same "spin" (parallel spin).

Explanation:

You might be interested in

Under what conditions does the hydrolysis of an amide bond occur?

Mademuasel [1]

Hi.

I did some digging and I think I found what you're looking for.

I found this on Q(uizlet)

basic or acidic conditions and the reactants must be heated.

~

5 0

3 years ago

A molecule of alkene has the chemical formula C3H4. How many carbon-carbon double bonds are present in the molecule?Choices:123

alexdok [17]

2 Carbon double bonds

4 0

3 years ago

Rasheed calculates that his chemical reaction should produce 4 moles of product, but when he does the experiment, he gets only 3

Brut [27]

Answer:

The answer to your question is 75%

Explanation:

Data

Theoretical production = 4 moles

Experimental production = 3 moles

Percent yield = ?

Formula

$Percent yield = \frac{Experimental production}{Theoretical production} x 100$

Substitution

$Percent yield = \frac{3}{4} x 100$

Result

Percent yield = 75 %

8 0

4 years ago

How does the law of conservation of matter apply to chemical equations?

Tasya [4]

Answer: option D.

The total number of atoms of each element on both sides of the

equation must be the same.

Explanation:

3 0

4 years ago

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

8 0

4 years ago