Answer: c
Explanation:
If their is a question with the anwser “both ___” it’s that
I believe it would be 7.731. x 10^5? That is if they want you to evaluate.
Answer:
Heat transfer = Q = 62341.6 J
Explanation:
Given data:
Heat transfer = ?
Mass of water = 50.0 g
Initial temperature = 30.0°C
Final temperature = 55.0°C
Specific heat capacity of water = 4.184 J/g.K
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 55.0°C - 30.0°C
ΔT = 25°C (25+273= 298 K)
Q = 50.0 g × 4.184 J/g.K ×298 K
Q = 62341.6 J
Answer:
Limiting reactant: O2
grams NO2 produced = 230.276 g NO2
grams of NO unused = 26.67 gNO
Explanation:
2NO + O2 --> 2NO2
Step 1: Determine the molar ratio NO:O2
molar ratio NO:O2 = 5.895: 2.503 = 2.35
stoichiometric molar ratio NO:O2 = 2:1
So, O2 is the limiting reactant.
Step2: Determine the grams of NO2:
?g NO2 = moles O2 x (2moles NO2/1 mol O2) x (MM NO2/ 1 mol NO2) = 2.503 x 2 x 46 = 230.276 g NO2
Step 3: Determine the amount of excess reagent unreacted
moles excess NO reacted = moles O2 x (2 moles NO/1 mol O2) = 2.503 x 2 = 5.006 moles NO reacted
moles NO unreacted = total moles NO - moles NO reacted = 5.895-5.006 =0.889 moles NO unreacted
mass NO unreacted = moles NO unreacted x MM NO = 0.889 x 30 =26.67 g NO unreacted
