Answer:
2445 L
Explanation:
Given:
Pressure = 1.60 atm
Temperature = 298 K
Volume = ?
n = 160 mol
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 08206 L.atm/K.mol
Applying the equation as:
1.60 atm × V = 160 mol × 0.08206 L.atm/K.mol × 298 K
<u>⇒V = 2445.39 L</u>
Answer to four significant digits, Volume = 2445 L
Answer:
0.0025moles
Explanation:
Molarity of a solution (M) = number of moles (n) ÷ volume (V)
According to this question, to make 250 mL of a 0.01 M solution of CaCl, the following number of moles is needed:
Volume = 250mL = 250/1000 = 0.250Litres.
Using; molarity = n/V
0.01 = n/0.250
n = 0.0025
n = 2.5 × 10^-3 moles.
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
Answer:
a)there would be no reaction
Explanation:
The activity series of metals has many functions. The one applicable to this problem is that it can be used to determine whether a reaction will occur or not. Also, based on the positions of metals in the series, we can know how reactive a metal is compared to another.
In a single displacement reaction, a metal replaces another metal based on their position on the activity series. Metals that are higher in the series are generally more reactive than others below them and so will displace them.
Would aluminum replace magnesium to form a new compound or would there be no reaction?
Magnesium is higher than aluminum in the activity series. Therefore it is more reactive than aluminum. No reaction will occur.
Answer: Final temperature = 206∘C
Explanation:
Heat Energy is given as
q= mCΔT
ehere
q= Heat energy = 87.4J
m= mass=1.25g
C=specific heat c= 0.386Jg∘C) ,
ΔT = Change in temperate of which the final temperature= 25.0∘C
q= mCΔT
ΔT = q/mC
ΔT = 87.4/ 1.25 X 0.386=181.14∘C
But,
T final- T initial = ΔT
T final = T initial + ΔT
T final = 25.0∘C +181.14∘C=206.14∘C rounded to 206∘C
