Answer:
ΔE = 150 J
Explanation:
From first law of thermodynamics, we know that;
ΔE = q + w
Where;
ΔE is change in internal energy
q is total amount of heat energy going in or coming out
w is total amount of work expended or received
From the question, the system receives 575 J of heat. Thus, q = +575 J
Also, we are told that the system delivered 425 J of work. Thus, w = -425 J since work was expended.
Thus;
ΔE = 575 + (-425)
ΔE = 575 - 425
ΔE = 150 J
Answer:
The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. ... The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant.
Explanation:
The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. ... The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant.
1 mole CO2 = 44.0096 grams CO2
<span>2.1 mol CO2 x (44.0096 grams CO2/1 mole CO2) = 92.4 grams CO2</span>
1) Left up: a chemical change. We can see new substance (red-blue) is formed from one blue and one red atom.
In chemical change new substances are formed, the atoms are rearranged and the reaction is followed by an energy change.
2) Left down: a chemical change. We can see new substance (red-blue) is formed from two blue and one red atoms.
3) MIddle: a physical change. There is no new substance. Bonds are not broken.
4) Right up: a chemical change. Bonds are broken.
5) Right down: a physical change. Change of state of matter.
Answer:
The percent isotopic abundance of Ir-193 is 60.85 %
The percent isotopic abundance of Ir-191 is 39.15 %
Explanation:
we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193
First of all we will set the fraction for both isotopes
X for the isotopes having mass 193
1-x for isotopes having mass 191
The average atomic mass is 192.217
we will use the following equation,
193x + 191(1-x) = 192.217
193x + 191 - 191x = 192.217
193x- 191x = 192.217 - 191
2x = 1.217
x= 1.217/2
x= 0.6085
0.6085 × 100 = 60.85 %
60.85% is abundance of Ir-193 because we solve the fraction x.
now we will calculate the abundance of Ir-191.
(1-x)
1-0.6085 =0.3915
0.3915× 100= 39.15 %
