Correction: The temperature change is from 20 °C to 30 °C.
Answer:
Cp = 1.0032 J.g⁻¹.°C⁻¹
Solution:
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 5016 J
m = mass = 500 g
Cp = Specific Heat Capacity = ??
ΔT = Change in Temperature = 30 °C - 20 °C = 10 °C
Solving eq. 1 for Cp,
Cp = Q / m ΔT
Putting values,
Cp = 5016 J / (500 g × 10 °C)
Cp = 1.0032 J.g⁻¹.°C⁻¹
Answer:
C. increase to 7.
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, the molar relationship is 1 to 1, therefore, the moles are:
Thus, since the entire hydrogen ions are neutralized, the pH C. increase to 7.
Best regards.
In the combustion process using excess oxygen, each mole of methane results to 1 mole of co2 while ethane produces 2 moles of Co2. Under same conditions, these can be translated to volume. Hence the total volume absorbed is 10 cm3 + 20 cm3 = 30 cm3.
Answer:
0.595 M
Explanation:
The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.
Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water
0.0194 = x/x + 55.6
0.0194(x + 55.6) = x
0.0194x + 1.08 = x
x - 0.0194x = 1.08
0.9806x= 1.08
x= 1.08/0.9806
x= 1.1 moles of KNO3
Mole fraction of water= 55.6/1.1 + 55.6 = 0.981
If
xA= mole fraction of solvent
xB= mole fraction of solute
nA= number of moles of solvent
nB = number of moles of solute
MA= molar mass of solvent
MB = molar mass of solute
d= density of solution
Molarity = xBd × 1000/xAMA ×xBMB
Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101
Molarity= 20.6/34.6
Molarity of KNO3= 0.595 M
<span>C.) Protons, neutrons. Hope it helps :)</span>
