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Alchen [17]
3 years ago
8

What term describes a substance that does the dissolving

Chemistry
1 answer:
NARA [144]3 years ago
5 0
 A <span>substance that does the dissolving is called a </span>solute. 
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Write a complete set of quantum numbers for the fifth electron added to a hydrogen ion (i.e., the fifth electron in any electron
ankoles [38]
N=
l=
m(l)=
m(s)=

start with H^+ (no electrons) , then adding 5 electrons will be  1s2 2s2 2p1
so for the 5th electron 
n = 2
l = 1
ml = -1
ms = 1/2

5 0
3 years ago
Read 2 more answers
What is the predicted change in the boiling point of water when 2.10 g of
daser333 [38]

Answer:

A: 0.0028°C

Explanation:

Hopefully this helps!

6 0
2 years ago
In one experiment, the reaction of 1.00 mercury and an excess of sulfur yielded 1.16g of a sulfide of mercury
Nuetrik [128]

<u>Answer and Explanation:</u>

Mercury combines with sulfur as follows -

Hg + S = HgS

Hg = 200,59

S = 32,066 Therefore 1.58 g of Hg will react with -

1.58 multiply with 32,066 divide by 200,96 of sulfur.

= 0.25211 g S

This will form 1.58 + 0.25211 g HgS  = 1.83211 g HgS

The amount of S remaining = 1.10 - 0.25211  = 0.84789 g

5 0
3 years ago
Based on the law of conservation of mass, which of the following statements is correct? Matter is neither created nor destroyed.
Dmitrij [34]

The first statement (Matter is neither created nor destroyed) is correct.

The second statement would violate the law of conservation of mass (I will refer to this as LCM), as it would mean matter can "flow" into the universe, but not out, meaning the total matter will never be less than it was before.

The third statement violates LCM because it means matter is created during a reaction, which is not true.

The last statement violates LCM because it means matter is lost during a reaction, which is not true.

7 0
3 years ago
Consider the reaction: N2(g) + O2(g) ⇄ 2NO(g) Kc = 0.10 at 2000oC Starting with initial concentrations of 0.040 mol/L of N2 and
IrinaVladis [17]

Answer:

0.011 mol/L

Explanation:

This can be solved with something called an ICE table.

I = initial

C = change

E = equilibrium

Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.

x amount of N₂ reacts.  Since the stoichiometry is 1:1, x amount of O₂ also reacts.  This produces 2x of NO.

After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.

Here it is in table form:

\left[\begin{array}{cccc}&N2&O2&NO\\I&0.04&0.04&0\\C&-x&-x&+2x\\E&0.04-x&0.04-x&2x\end{array}\right]

Now we can use the equilibrium constant:

Kc = [NO]² / ( [N₂] [O₂] )

Substituting:

0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )

Solving:

0.10 = (2x)² / (0.04 - x)²

√0.10 = 2x / (0.04 - x)

(√0.10) (0.04 - x) = 2x

(√0.10)(0.04) - (√0.10)x = 2x

(√0.10)(0.04) = 2x + (√0.10)x

(√0.10)(0.04) = (2 + √0.10)x

x = (√0.10)(0.04) / (2 + √0.10)

x = 0.0055

At equilibrium, the concentration of NO is 2x.  So the answer is:

[NO] = 2x

[NO] = 0.011

The equilibrium concentration of NO is 0.011 mol/L.

3 0
3 years ago
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