Two astronauts would exhale about 40 moles of carbon dioxide daily.
Carbon dioxide reacts with lithium hydroxide in a 1 : 2 mole ratio. Set up a proportion:
1 : 2 = 40 : x
Then, find x:
<span>12=40x
</span>
Cross multiply. x = 80 moles of LiOH per day for both astronauts
Its c ok im sorry if i got the last answer wrong
3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O
<h3>Further explanation
</h3>
Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:
- 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
- 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
- 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:
Balancing C atoms, H and the last O atoms
Reaction
Zn + HNO₃⇒ Zn(NO₃)₂ + NO + H₂O
aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O
Zn : left = a, right =1 ⇒a=1
H : left = b, right = 2d⇒ b=2d (eq 1)
N : left = b, right = 2+c⇒b=2+c (eq 2)
O : left = 3b, right = 6+c+d ⇒3b=6+c+d(eq 3)
3(2d)=6+c+d
6d=6+c+d
5d=6+c (eq 4)
3(2+c)=6+c+d
6+3c=6+c+d
2c=d (eq 5)
5(2c)=6+c
10c=6+c
9c=6
c = 2/3
d = 2 x 2/3
d = 4/3
b = 2 x 4/3
b = 8/3
The equation
aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O to
Zn + 8/3HNO₃⇒ Zn(NO₃)₂ + 2/3NO + 4/3H₂O x 3
3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:


where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>

- half-life length
</span>t - total time elapsed.
First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125
Thus:
<span>

</span>




It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>

</span><span>

</span>

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>