Answer:
44.8 L
Explanation:
Using the ideal gas law equation:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
At Standard temperature and pressure (STP);
P = 1 atm
T = 273K
Hence, when n = 2moles, the volume of the gas is:
Using PV = nRT
1 × V = 2 × 0.0821 × 273
V = 44.83
V = 44.8 L
The smallest unit that make up the matter is atoms.
PLEASE MAKE IT BRAINLEST I NEED ONLY 1 LAST BRAINLIST FOR MY PROMOTION
Answer:
V2 = 2.88L
Explanation:
P1= 78atm, V1= 2L, T1= 900K, P2= 45atm, V2=? T2= 750K
Applying the general gas equation
P1V1/T1 = P2V2/T2
Substitute the above
(78*2)/900= (45*V2)/750
V2= (78*2×750)/(900*45)
V2= 2.88L
Answer:
The answers are in the explanation.
Explanation:
The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:
Increasing temperature of ice from -10°C - 0°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g
Q = 2.06J/g°C*10°C*10g
Q = 206J
Change from solid to liquid:
The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:
Q = 333.55J/g*10g
Q = 3335.5J
Increasing temperature of liquid water from 0°C - 100°C:
Q = S*ΔT*m
Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g
Q = 4.18J/g°C*100°C*10g
Q = 4180J
Change from liquid to gas:
The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:
Q = 2260J/g*10g
Q = 22600J
Increasing temperature of gas water from 100°C - 120°C:
Q = S*ΔT*m
Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g
Q = 1.87J/g°C*20°C*10g
Q = 374J
Total Energy:
206J + 3335.5 J + 4180J + 22600J + 374J =
30695.5J =
30.7kJ
<u>Answer:</u> The correct answer is 
<u>Explanation:</u>
We are given:
The substance having highest positive 
potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.
Chromium will undergo oxidation reaction and will get oxidized.
The half reactions for the above cell is:
Oxidation half reaction: 
Reduction half reaction: 
( × 3)
Net equation: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the 
of the reaction, we use the equation:
Putting values in above equation, we get:
Hence, the correct answer is
