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matrenka [14]
3 years ago
12

What would be the charge on an ion whose neutral atom has an electron configuration of 2.8.7?

Chemistry
1 answer:
larisa86 [58]3 years ago
8 0

Answer:

A

Explanation:

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Which gas dissolves more easily in water? Oxygen or carbon dioxide?
Alborosie
Carbon dioxide dissolves faster 
6 0
4 years ago
Read 2 more answers
How can equilibrium of a closed system chemical reaction be disturbed? Select all that apply.
charle [14.2K]

The correct answers are :

Changing the volume of the system.

Changing the temperature of the system.

Equilibrium will remain unaffected if the concentration of products and reactants are kept the same, and the temperature of the system is kept constant.

As the system is closed, we cannot add or remove products or reactants.

Change in temperature will shift the chemical equilibrium towards the reactant or product depending on whether the reaction is exothermic or endothermic.

Also change in volume  will shift the chemical equilibrium of a chemical reaction if the reactants or products or both are gases.

8 0
4 years ago
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If the molecule C6H12 does not contain a double bond, and there are no branches in it, what will its structure look like?
NARA [144]
I have attached a photo of the structure. 
You can get better at solving problems like this by practicing a lot! 




5 0
4 years ago
If an atom with a mass of 47 amu has 22 neutrons, what atom is it?
LenaWriter [7]

Answer:

Manganese

Explanation:

At Mass - No neutrons = Atomic Number = #protons in nucleus

47 - 22 = 25 => At. No. 25 is Manganese (Mn)

7 0
3 years ago
An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.
Tju [1.3M]

Answer:

a) \Delta S

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

\delta S=\frac{\delta Q}{T}

\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

W=n*R*T*ln(V2/V1)

Replacing:

\Delta S=\frac{n*R*T*ln(V2/V1)}{T}

\Delta S=n*R*ln(V2/V1)}

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

\Delta S

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0
4 years ago
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