Answer:
2,5
Explanation:
When move from left to right in periodic table atomic number increases. So element to the right of carbon has one electron more than carbon therefore it's electronic configuration would be 2,5
Answer:
The answer to your question is P = 1.64 atm
Explanation:
Data
Volume = 2.5 x 10⁷ L
Temperature = 22°C
Pressure = ?
Moles = 1.7 x 10⁶
R = 0.082 atm L/ mol°K
Process
1.- Convert temperature to °K
Temperature = 22 + 273
= 295°K
2.- Use the Ideal gas law to solve this problem
PV = nRT
- Solve for P
P = nRT / V
- Substitution
P = (1.7 x 10⁶)(0.082)(295) / 2.5 x 10⁷
- Simplification
P = 41123000 / 2.5 x 10⁷
- Result
P = 1.64 atm
Answer:
Newton's First Law of Motion
Explanation:
Newton’s first law of motion is when an object can’t move or change their speed without some type of force. Inertia is included since is caused by unbalanced force.
Answer:
A.
Explanation:
Physical properties are characteristics of a substance that can be observed or measured without changing the composition of the substance. Some examples of physical properties are its volume and its density. There are two types of physical properties, intensive physical properties and extensive physical properties.
Answer:
(a) The system does work on the surroundings.
(b) The surroundings do work on the system.
(c) The system does work on the surroundings.
(d) No work is done.
Explanation:
The work (W) done in a chemical reaction can be calculated using the following expression:
W = -R.T.Δn(g)
where,
R is the ideal gas constant
T is the absolute temperature
Δn(g) is the difference between the gaseous moles of products and the gaseous moles of reactants
R and T are always positive.
- If Δn(g) > 0, W < 0, which means that the system does work on the surroundings.
- If Δn(g) < 0, W > 0, which means that the surroundings do work on the system.
- If Δn(g) = 0, W = 0, which means that no work is done.
<em>(a) Hg(l) ⇒ Hg(g)</em>
Δn(g) = 1 - 0 = 1. W < 0. The system does work on the surroundings.
<em>(b) 3 O₂(g) ⇒ 2 O₃(g)
</em>
Δn(g) = 2 - 3 = -1. W > 0. The surroundings do work on the system.
<em>(c) CuSO₄.5H₂O(s) ⇒ CuSO₄(s) + 5H₅O(g)
</em>
Δn(g) = 5 - 0 = 5. W < 0. The system does work on the surroundings.
<em>(d) H₂(g) + F₂(g) ⇒ 2 HF(g)</em>
Δn(g) = 2 - 2 = 0. W = 0. No work is done.
