Answer:
248 mL
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by water (Qw) and the heat released by the coffee (Qc) is zero.
Qw + Qc = 0
Qw = -Qc [1]
We can calculate each heat using the following expression.
Q = c × m × ΔT
where,
- ΔT: change in the temperature
163 mL of coffee with a density of 0.997 g/mL have a mass of:
163 mL × 0.997 g/mL = 163 g
From [1]
Qw = -Qc
cw × mw × ΔTw = -cc × mc × ΔTc
mw × ΔTw = -mc × ΔTc
mw × (54.0°C-25.0°C) = -163 g × (54.0°C-97.9°C)
mw × 29.0°C = 163 g × 43.9°C
mw = 247 g
The volume corresponding to 247 g of water is:
247 g × (1 mL/0.997 g) = 248 mL
Answer:
1.73 atm
Explanation:
Given data:
Initial volume of helium = 5.00 L
Final volume of helium = 12.0 L
Final pressure = 0.720 atm
Initial pressure = ?
Solution:
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
P₁ × 5.00 L = 0.720 atm × 12.0 L
P₁ = 8.64 atm. L/5 L
P₁ = 1.73 atm
I hope this picture helps
Answer:
Volume of liquid = 28.7 mL
Explanation:
Given data;
Density of solid = 3.57 g/ml
Mass of solid = 19.5 g
Volume of water = 23.2 mL
Total volume when solid is dropped into graduated cylinder= ?
Solution:
Density = mass/ volume
v = m/d
v = 19.5 g/ 3.57 g/ml
v = 5.5 mL
Volume of liquid = volume of water + volume of solid
Volume of liquid = 23.2 mL + 5.5 mL
Volume of liquid = 28.7 mL
Answer:
The answer is <u>applied research</u>
Explanation:
Pure research becomes <u>applied research</u> when scientists develop a hypothesis based on the data and try to solve a specific problem.
This is because the pure research try to understand, predict or explain the behavior of different phenomena <em>(the data)</em> while the applied research try to develop new technologies or methods (<em>hypothesis)</em> to take part, intervene and/or create changes on these phenomena and solve a <em>specific problem.</em>
