Answer:
Limiting - Na Excess - Cl2
Explanation:
Answer:
Explanation:
We want the energy required for the transition:
CO 2
(
s
)
+
Δ
→
C
O
2
(
g
)
Explanation:
We assume that the temperature of the gas and the solid are EQUAL.
And thus we simply have to work out the product:
2
×
10^
3
⋅
g
×
196.3
⋅
J
⋅
g
−
1 to get an answer in Joules as required.
What would be the energy change for the reverse transition:
C
O
2
(
g
)
+
→
C
O
2
(
s
)
?
The mass of plutonium that will remain after 1000 years if the initial amount is 5 g when the half life of plutonium-239 (239pu, pu-239) is 24,100 years is 2.5 g
The equation is Mr=Mi(1/2)^n
where n is the number of half-lives
Mr is the mass remaining after n half lives
Mi is the initial mass of the sample
To find n, the number of half-lives, divide the total time 1000 by the time of the half-life(24,100)
n=1000/24100=0.0414
So Mr=5x(1/2)^1=2.5 g
The mass remaining is 2.5 g
- The half life is the time in which the concentration of a substance decreases to half of the initial value.
Learn more about half life at:
brainly.com/question/24710827
#SPJ4
According to the law of conservation of mass, the quantity of the elements, involved in chemical reactions does not change. For example,
H2O2 - > H2O + O2
is wrong, because there are two O atoms on the first side of the equation, and three on the other. To correct it, coefficients must be added, until the amount of both H and O atoms is equal on both sides.
2H2O2 - > 2H2O + O2
Explanation:
Since methane gas is at 1 atm and 273 K, it is at standard temperature and pressure(STP).
One mole of every gas occupies 22.4 dm^3 at STP, and vice versa. So,
22.4 dm^3 at STP of CH4=1 mol=12+4(1)=16 g
0.462 L(0.462 dm^3) at STP of CH4
=(16 g×0.462 dm^3)/22.4 dm^3
=0.33 g
