Answer:
The mass of water = 219.1 grams
Explanation:
Step 1: Data given
Mass of aluminium = 32.5 grams
specific heat capacity aluminium = 0.921 J/g°C
Temperature = 82.4 °C
Temperature of water = 22.3 °C
The final temperature = 24.2 °C
Step 2: Calculate the mass of water
Heat lost = heat gained
Qlost = -Qgained
Qaluminium = -Qwater
Q = m*c*ΔT
m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)
⇒with m(aluminium) = the mass of aluminium = 32.5 grams
⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C
⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C
⇒with m(water) = the mass of water = TO BE DETERMINED
⇒with c(water) = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C
32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9
-1742.1 = -7.95m
m = 219.1 grams
The mass of water = 219.1 grams
Ans: The final volume of the balloon is 4.5 L
<u>Given:</u>
Volume of balloon inflated with 3 breaths = 1.7 L
<u>To determine:</u>
Volume of balloon after a total of 3+5 = 8 breaths
<u>Explanation:</u>
Volume of the balloon per breath = 1.7 L * 1 breath/3 breaths = 0.567 L
Final volume of balloon after 8 breaths = 0.567 L * 8 breath/1 breath
= 4.536 L
Answer:
P1 = 2.5ATM
Explanation:
V1 = 28L
T1 = 45°C = (45 + 273.15)K = 318.15K
V2 = 34L
T2 = 35°C = (35 + 273.15)K = 308.15K
P1 = ?
P2 = 2ATM
applying combined gas equation,
P1V1 / T1 = P2V2 / T2
P1*V1*T2 = P2*V2*T1
Solving for P1
P1 = P2*V2*T1 / V1*T2
P1 = (2.0 * 34 * 318.15) / (28 * 308.15)
P1 = 21634.2 / 8628.2
P1 = 2.5ATM
The initial pressure was 2.5ATM
We first calculate for the number of moles of NaOH by dividing the given mass by the molar mass of NaOH which is equal to 40 g/mol. Solving,
moles of NaOH = (68.4 g/ 40 g/mol) = 1.71 moles NaOH
Then, we divide the calculate number of moles by the volume in liters.
molarity = (1.71 moles NaOH / 0.875 L solution)
molarity = 1.95 M
The H3O+ in a 0.050M solution of Ba(OH)2 is calculated as below
write the equation for the dissociation of Ba(OH)2
Ba(OH)2 = Ba^2+ +2OH^-
calculate the OH- concentration
by use of mole ratio between Ba(OH)2 to OH^- which is 1:2 the concentration of OH = 0.050 x2 = 0.1 M
by use of the formula ( H3O+)(OH-) = 1 x10 ^-14
by making H3O+ the subject of the formula
H3O+ = 1 x10^-14/ OH-
substitute for OH-
H3O+ = (1 x10^-14 )/0.1
= 1 x10^-3 M
