it decreases the density of the object the air bubbles take up space. it increases the volume of the object slightly but the objects weight remains the same, hence the objects density decreases
Answer:
The answer to your question is π = 12.47 atm
Explanation:
Data
mass of Magnesium citrate = 25.5 g
volume of solution = 244 ml
temperature = 37°C
molar mass C₆H₆MgO₇ = (12 x 6) + (1 x 6) + (24 x 1) + (16 x 7) = 214 g
Osmotic pressure = π = ?
Process
1.- Calculate the moles of magnesium citrate
214 grams ----------------- 1 mol
25.5 grams --------------- x
x = (25.5 x 1)/214
x = 0.119
2.- Calculate molarity
Molarity = moles/volume
Molarity = 0.119/0.244
Molarity = 0.49
3.- Calculate osmotic pressure
π = MRT
π = (0.49)(0.0821)(37 + 273)
π = (0.49)(0.0821)(310)
π = 12.47 atm
it's the last one because it's depended on the rock layers
Answer:
D. Up-Down
Explanation:
Hoped this helped you <\3
Answer:
1 = Q = 7315 j
2 =Q = -21937.5 j
Explanation:
Given data:
Mass of water = 50 g
Initial temperature = 20°C
Final temperature = 55°C
Energy required to change the temperature = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water is 4.18 j/g.°C.
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 = T2 - T1
ΔT = 55°C - 20°C
ΔT = 35°C
Q = 50 g× 4.18 j/g.°C×35°C
Q = 7315 j
Q 2:
Given data:
Mass of metal = 100 g
Initial temperature = 1000°C
Final temperature = 25°C
Energy released = ?
Specific heat capacity = 0.225 j/g.°C
Solution:
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 = T2 - T1
ΔT = 25°C - 1000°C
ΔT = -975°C
Now we will put the values in formula.
Q = 100 g × 0.225 j/g.°C × -975°C
Q = -21937.5 j
Negative sign show that energy is released.
