<u>Answer:</u> The mass percent of water in the original air sample is 4.16 %
<u>Explanation:</u>
We are given:
Volume of moist air = 12.0 L
Volume of dry air = 11.50 L
Volume of water lost = (12.0 - 11.50) L = 0.50 L
Here, the percent by mass ratio will be equal to the percent by volume ratio as the number of moles is directly related to the volume.
To calculate the volume percentage of water in the air, we use the equation:
Putting values in above equation, we get:
Hence, the mass percent of water in the original air sample is 4.16 %
27) mechanical weathering
28) overgrowth of algae that produce toxins
29) Weathering breaks down and loosens the surface minerals of rock so they can be transported away by agents of erosion such as water, wind and ice. There are two types of weathering:mechanical and chemical. Mechanical weathering is the disintegration of rock into smaller and smaller fragments.
30) I'M NOT SURE
hope this helps
IN exchange can u mark me as brainlist
They tend to pick at it until it is in small enough pieces to swallow. Then their gizzard (rough muscle) in their stomach acts as internal teeth.
Answer:
5.00 g of solute will remain undissolved at the bottom of the container
Explanation:
From the question, the solubility of the solute in the given solvent is 45.0 grams of solute per 500 grams of solvent.
Now if i pour 50.0 grams of solute into 800 grams of solvent, it means that only 45 g will dissolve in 500 g of solvent leaving the additional 5 g undissolved.
Hence, 5 g of solute will remain undissolved at the bottom of the container.
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.
