Answer:
6moles of water
Explanation:
Given parameters:
Number of moles of oxygen = 3moles
Reaction equation:
2H₂ + O₂ → 2H₂O
Unknown:
Number of moles of water formed = ?
Solution:
To solve this problem;
compare the number of moles in the reaction.
1 mole of oxygen gas will produce 2 mole of water
3 moles of oxygen gas will produce 3 x 2 = 6moles of water.
Hey there!
<span>In this case, the equation of Clapeyron is used :
R = 0.082
Volume in liters :
100.0 mL / 1000 => 0.1 L
</span>
P * V = n * R * T
15.0 * 0.1 = 0.500 * 0.082 * T
1.5 = 0.041 * T
T = 1.5 / 0.041
T = 36.5 K
Answer C
Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
Answer:
The kilogram is the standard unit of mass.