Answer:
P₂ = 1.05 atm
Explanation:
Given data:
Initial temperature = 24.0 °C (24+273 = 297 K)
Initial pressure = 0.900 atm
Final pressure = ?
Final temperature = 75 °C (75 + 273 =348 K)
Volume = constant
Solution:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
0.900 atm / 297 K = P₂/348 K
P₂ = 0.900 atm × 348 K / 297 K
P₂ = 313.2 atm. K /297 K
P₂ = 1.05 atm
The flow rate of water differs from honey due to the texture and thickness of the product. Honey’s thickness doesn’t allow it to move as quickly as the water, therefore affecting the flow rate.
These are three questions and three complete answers
Answer:
a) Cr²⁺: [Ar] 4s² 3d²
b) Cu²⁺: [Ar] 4s² 3d⁷
c) Co³⁺: [Ar] 4s² 3d⁴
Explanation:
<u>a) Cr²⁺</u>
- Number of elecrons of the neutral atom: 24
- Number of electrons of the ion: 24 - charge = 24 - 2 = 22.
Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....
Hence, for 22 electrons you get:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²
- Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:
[Ar] 4s² 3d²
<u>b) Cu²⁺</u>
- Number of elecrons of the neutral atom: 29
- Number of electrons of the ion: 29 - charge = 29 - 2 = 27.
Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....
Hence, for 27 electrons you get:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁷
- Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:
[Ar] 4s² 3d⁷
<u>c) Co³⁺</u>
- Number of elecrons of the neutral atom: 27
- Number of electrons of the ion: 27 - charge = 27 - 3 = 24.
Fill the orbitals in increasing order of energy. Using Aufbau's rules the order is: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ .....
Hence, for 24 electrons you get:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴
- Abbreviated notation: since the last complete level is the number 3s² 3p⁶, you use the noble gas of the period 3, which is Ar, and the configuration is:
[Ar] 4s² 3d⁴
No I can’t
EXAMPLE took the test k12
Answer: The ΔH of the reaction if 51.3 g of 
reacts with excess 
to yield 1387.6 kJ is 432.27kJ
Explanation:
To calculate the moles, we use the equation:
moles of
As is present in excess, is the limiting reagent as it limits the formation of product.
If 3.21 moles of methane releases heat = 1387.6 kJ
Thus 1 mole of methane release=
