Energy added = mass*specific heat* rise in temperature.
= 5 * (60-50) * 4.18
= 50*4.18
= 209 J (answer).
Answer:
9.85
Explanation:
M1V1 =M2V2
6.20×v1= 0.470×0.130
v1 = ( 0.470 × 0.130 ) ÷ 6.20
v1 = 0.0098 L × 1000
V1 = 9.8 ml
Answer:
266 liters Carbon iv oxide will be formed
Explanation:
The first thing we need to do here is to calculate the number of moles of oxygen reacted.
This can be calculated using the ideal gas equation;
Mathematically;
PV = nRT
From the question;
P = pressure = 10.5 atm
V = volume = 400 liters
n = number of moles = ?
T = temperature = 125 + 273 = 398K
R = molar gas constant = 0.0821 L•atm•K^-1•mol^-1
Substituting these values, we have ;
n = PV/RT
n = (10.25 * 400)/(398 * 0.0821) = 125 moles
From the question;
15 moles oxygen gave 10 moles CO2
125 moles oxygen will give x moles CO2
x = (125 * 10)/15 = 83 moles
Now, we want to know the volume of CO2, present in 83 moles using the given reaction conditions.
Mathematically;
PV = nRT
V = nRT/P
= (83 * 0.0821 * 398)/10.25 = 265.65 which is approximately 266 Liters
