Steam rises to the top and stays there
Answer:
gold and copper
Explanation:
but I think there is 1 more
Answer:
The new volume of the balloon will be 6046.28 L
Explanation:
Initial pressure (P1) = 99 kpa
initial volume (V1) = 3000 L
Initial temperature = 39 C = 39 + 273 = 312 K
Final pressure (P2) = 45.5 kpa
Final temperature = 16 C = 16 +273 = 289K
Final volume = ????
To calculate the final volume using the general gas equation
P1 V1 / T1 = P2 V2 / T2
make V2 the subject of the formular
V2 = 99000 ×3000× 289 / 45500×312
V2 = 85833000 /14196
V2 = 6046.28 litres
The reaction must be a + b --> c
Then you can predict a reaction rate, r o the type r = k * a^n * b^m
Given that the reaction rate is not affected by the concentration of b you can state that m = 0 and r = k * a^n.
Now given, that there is a proportional relation between the reaction rate and a (double a gives double rate), then n = 1 and r = k*a. You can verify that if you dobule a r also doubles.
Answer: r = k*a
Answer: 32.94 g
Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.
for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm
V = 15.0 L
T = 350.8 + 273 = 623.8 K
For krypton, 
n = 0.146 moles
for chlorine, 
n = 0.439
From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.
Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.
So, the amount of product formed is calculated from moles of krypton.
Molar mass of krypton tetrachloride is 225.61 gram per mol.
There is 1:1 mol ratio between krypton and krypton tetrachloride.

= 32.94 g of 
So, 32.94 g of the product will form.