Answer:
Heat transfer = Q = 62341.6 J
Explanation:
Given data:
Heat transfer = ?
Mass of water = 50.0 g
Initial temperature = 30.0°C
Final temperature = 55.0°C
Specific heat capacity of water = 4.184 J/g.K
Solution:
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 = 55.0°C - 30.0°C
ΔT = 25°C (25+273= 298 K)
Q = 50.0 g × 4.184 J/g.K ×298 K
Q = 62341.6 J
Molar mass SiO2 = 28 + 32 = 60
<span>so moles sand = 3.4 x 10-7 / 60</span>
Answer:
15.98 L
Explanation:
First, you need to find T1, T2, V1 and V2.
T1 = 25 C = 298.15 K (25C + 273.15K)
T2 = 100 C = 373.15 K (100C + 273.15K)
V1 = 20. L
V2 = ? (we are trying to find)
Next, rearrange to fit the formula
V2 = V1 x T1 / T2
Next, fill in with our numbers
V2 = 20. L x 298.15 K / 373.15 K
Do the math and you should get...
15.98 L
- If you need more help or futher explanation please let me know. I would be glad to help!
Answer:
The Law of Definite Proportions ensures that chemical compounds are always created using the same proportions, regardless of the amount of the compound which is being made
Answer:
The purpose of the experiment is to see how water of different temperature and salinity affect the density.
Explanation:
Temperature and salinity directly affect the density of the water. Water of low temperature is more dense than water of high temperature, BUT, (fresh)water with no salt is less dense than (sea)water with more salt, so temperature and salinity change density of water.
