Answer:
The procedure with the simple calorimeter to compare the energy released by different fuels is described below in detail.
Explanation:
If the response delivers heat (rxn < 0), then heat is consumed by the calorimeter (calorimeter > 0), and its temperature improves. Conversely, if the response consumes heat (rxn > 0), suddenly heat is carried from the calorimeter to the arrangement (calorimeter < 0), and the temperature of the calorimeter drops.
Answer: Belongs to the group 2A
Explanation:
As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.
Not only loses easily the first electron, but the second too
To remove the third electron you requiered a huge amount of energy
Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.
The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.
Now the question: group I or group II ?
The elements of group I have low ionization energies for the first electron but high energies for the second ones.
Being all that said, the unknown element belongs to the Group 2A
Answer:
The pressure increases to 3.5 atm.
Solution:
According to Gay-Lussac's Law, " At constant volume and mass the pressure of gas is directly proportional to the applied temperature".
For initial and final states of a gas the equation is,
P₁ / T₁ = P₂ / T₂
Solving for P₂,
P₂ = P₁ T₂ / T₁ ----- (1)
Data Given;
P₁ = 3 atm
T₁ = 27 °C + 273 = 300 K
T₂ = 77 °C + 273 = 350 K
Putting values in eq. 1,
P₂ = (3 atm × 350 K) ÷ 300 K
P₂ = 3.5 atm
Aye i can’t see the full question dawg, you gotta send a full picture of it
