Answer:
variable: not consistent or having a fixed pattern; liable to change.
Explanation:
The correct answer is 221.06 °C hot.
If P₁ is the pressure at T₁ and P₂ is the pressure at T₂ then,
P₁/T₁ = P₂/T₂
It is given that P₁ = 2.38 atm
T₁ = 15.2 degree C = 273 + 15.2 = 288.2 K
P₂ = 4.08 atm
T₂ = x
Thus, 2.38 / 288.2 = 4.08 / x
x = (4.08 × 288.2) / 2.38
x = 494.06 K
x = 494.06 - 273 °C = 221.06 °C
Therefore, the tire would get 221.06 °C hot.
Answer:
3S₈ + 28Br₂ => 8S₃Br₇
Explanation:
Start with either sulfur (S) or bromine (Br) and balance ...
3S₈ + Br₂ => 8S₃Br₇ or S₈ + 7/2Br₂ => S₃Br₇
Balance the remaining reactant ...
3S₈ + 56/2Br₂ => 8S₃Br₇
Remove fractions by multiplying by the fraction's denominator
2(3S₈ + 56/2Br₂ => 8S₃Br₇) => 6S₈ + 56Br₂ => 16S₃Br₇
Reduce to smallest whole number ratio => standard equation at STP ...
3S₈ + 28Br₂ => 8S₃Br₇
Answer:
1120 gm
Explanation:
6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)
(a) Balance the equation.
(b) How many grams of oxygen are required to react with 10 moles of ethane for a complete
combustion reaction?
FIRST, CORRECT THE EQUATION THEN BALANCE
2C2H6(G) + 7O2------------> 4CO2 + 6H2O
so for 10 moles of ethane, we need
7 X 5 = 35 MOLES O2
=35 MOLES O2
O2 HAS A MOLAR MASS OF 2X16 = 32 gm
35 MOLES OF O2 HAS A MASS OF 35 X 32 =1120 gm
I believe the answer is true
