Hope u do well :) ik this maybe hard but you’ll get through it , the answer is D !
Evaporation happens<span> when atoms or </span>molecules<span> escape from the liquid and turn into a vapor. Not all of the </span>molecules in a liquid have the same energy. <span>Sometimes a </span>liquid<span> can be sitting in one place (maybe a puddle) and its molecules will become a </span>gas<span>. That's the process called </span>evaporation<span>. It can happen when liquids are cold or when they are warm. It happens more often with warmer liquids. You probably remember that when matter has a higher temperature, the molecules have a higher </span>energy<span>. When the energy in specific molecules reaches a certain level, they can have a </span>phase change<span>. Evaporation is all about the energy in individual molecules, not about the average energy of a system. The average energy can be low and the evaporation still continues. </span>
Answer:
7,94 minutes
Explanation:
If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>
For the zero-order reactions, concentration-time equation can be written as follows:
[A] = - Kt + [Ao]
where:
- [A]: concentration of the reactant A at the <em>t </em>time,
- [A]o: initial concentration of the reactant A,
- K: rate constant,
- t: elapsed time of the reaction
<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>
Data:
K = 4.2 ×10−3atm/s,
[A]o=[HBr]o= 2 atm,
[A]=[HBr]=0 atm (all HBr(g) is gone)
<em>We clear the incognita :</em>
[A] = - Kt + [Ao]............. Kt = [Ao] - [A]
t = ([Ao] - [A])/K
<em>We replace the numerical values:</em>
t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes
So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).
The dense aerosol cloud created caused decreases in the amount of solar radiation that reached the Earths surface.
<u>Answer:</u> The mass of iron in the ore is 10.9 g
<u>Explanation:</u>
We are given:
Mass of iron (III) oxide = 15.6 g
We know that:
Molar mass of Iron (III) oxide = 159.69 g/mol
Molar mass of iron atom = 55.85 g/mol
As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.
To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:
In 159.69 g of iron (III) oxide, mass of iron present is
So, in 15.6 g of iron (III) oxide, mass of iron present will be =
Hence, the mass of iron in the ore is 10.9 g