1) Balanced equation
C3H8 + 5O2 -> 3 CO2 + 4 H2O
2) 0.700 L C3H8
Given the pressure and temperature do not change, the molar ratio is equivalent to volume ratio
1molC3H8 / 5 mol O2 => 1 L C3H8 / 5 L O2
0.700 L C3H8 / x L O2 = 1 L C3H8 / 5 L O2 => x = 0.700 L C3H8 * 5 L O2 / 1 L C3H8
x = 3.500 L O2
3) CO2 produced
1 L C3H8 / 3 L CO2 = 0.700 L C3H8 / x L CO2 =>
x = 0.700 L C3H8 * 3 L CO2 / 1 L C3H8 = 2.100 L CO2
4) Water vapor produced
1) 1 L C3H8 / 4 L H2O = 0.700 LC3H8 / x L H2O =>
x = 0.700 L C3H8 * 4 L H20 / 1 L C3H8 = 2.800 L H2O
The hydronium ions increase and the ph goes down.
i think the answer is COS2 tell me if im right or wrong
Answer:
3.99 g
Explanation:
The following data were obtained from the question:
Half life (t½) = 2 years.
Original amount (N₀) = 128 g
Time (t) = 10 years
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 years.
Decay constant (K) =.?
K= 0.693/t½
K = 0.693/2
K = 0.3465 year¯¹.
Finally, we shall determine the amount remaining after 10 years i.e the amount remaining when they arrive on Earth. This can be obtained as follow:
Original amount (N₀) = 128 g
Time (t) = 10 years
Decay constant (K) = 0.3465 year¯¹.
Amount remaining (N) =..?
Log (N₀/N) = kt/2.3
Log (128/N) = (0.3465 × 10)/2.3
Log (128/N) = 1.5065
Take the antilog of 1.5065
128/N = Antilog (1.5065)
128/N = 32.1
Cross multiply
128 = 32.1 × N
Divide both side by 32.1
N = 128/32.1
N = 3.99 g
Therefore, the amount remaining is 3.99 g
