Answer: Option (A) is the correct answer.
Explanation:
A catalyst is a substance which helps in increasing the rate of a reaction without itself getting consumed in the reaction.
A catalyst lowers the activation energy of a reaction so that reactant molecules with less energy can easily participate into the chemical reaction and thus, product formation becomes faster.
Thus, we can conclude that adding a catalyst to the reaction would reduce the activation energy of a chemical reaction.
Answer:
Airline IATA ICAO
Aerosul Linhas Aéreas 2S ASO
ASTA Linhas Aéreas 0A SUL
Azul Linhas Aéreas AD AZU
Azul Conecta 2F (AD) OWT
Answer:
17.92L or 17.92dm³1
Explanation:
number of moles of a gas = volume of that gas/ 22.4(S.V.P)
0.8 mol = x / 22.4
cross multiplication
x = 22.4 × 0.8
x = 17.92
:. the volume of 0.8 mol of nitrogen gas is 17.92dm³ or 17.92L
Answer:
3.568 g of H₂O
Solution:
The Balance Chemical Equation is as follow,
2 H₂ + O₂ → 2 H₂O
Step 1: Calculate the Limiting Reagent,
According to Balance equation,
4.04 g (2 mol) H₂ reacts with = 32 g (1 mol) of O₂
So,
0.40 g of H₂ will react with = X g of O₂
Solving for X,
X = (0.40 g × 32 g) ÷ 4.04 g
X = 3.17 g of O₂
It means 0.4 g of H₂ requires 3.17 g of O₂, while we are provided with 3.2 g of O₂ which is in excess. Therefore, H₂ is the limiting reagent and will control the yield of products.
Step 2: Calculate amount of Water produced,
According to equation,
4.04 g (2 mol) of H₂ produces = 36.04 g (2 mol) of H₂O
So,
0.40 g of H₂ will produce = X moles of H₂O
Solving for X,
X = (0.40 g × 36.04 mol) ÷ 4.04 g
X = 3.568 g of H₂O
Ionization enthalpy, IE, is also called ionization potential is the ability to remove the electron from the neutral gaseous atom. There is a trend observed in the periodic table for the IE value. As we go from left to right in a period, IE vale increases. While moving from top to bottom in a group, IE value decreases.
- The phenomenon of unexpected drop in IE1 values between Groups 2 and 13, in period 2 and period 4 is due to the introduction of d-orbitals in the case of period 4 elements.
- While moving in the period, there is the constant addition of electrons in the nucleus. The shell sie remains constant while electron pull increases from the nucleus, this leads to a reduction in the size of the atom. As the size decreases, it is difficult to remove the electron from the atom, and thus IE value increases in the case of period 2.
- When we study the case of period 4, there is an introduction of d-electrons. As the inner shell electron increases, there is an increase in the shielding effect. This shielding effect tends to decrease the nuclear attraction between the nucleus and outermost electrons. Ultimately this decreases the IE value in the fourth period. Such a phenomenon is absent in the case of group 2 elements.
- If we speak in terms of orbital energy, the IE value decreases while moving from top to bottom in the period. This is due to the fact that, as we go down in the periodic table, the number of shells increases, and the outermost electron is too far from the nuclear attraction, therefore it can be ejected out easily. This marks a decrease in IE value.
To learn more about ionization refer the link:
brainly.com/question/1558319
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