A scientific theory focuses on explaining a phenomenon, while a law will have a generalised statement about said phenomenon. Often times, we use laws to help us develop theories. =)
Answer:
1.7 bar
Explanation:
We can use the <em>Ideal Gas Law</em> to calculate the individual gas pressure.
pV = nRT Divide both sides by V
p = (nRT)/V
Data: n = 1.7 × 10⁶ mol
R = 0.083 14 bar·L·K⁻¹mol⁻¹
T = 22 °C
V = 2.5 × 10⁷ L
Calculations:
(a) <em>Change the temperature to kelvins
</em>
T = (22 + 273.15) K
= 295.15 K
(b) Calculate the pressure
p = (1.7 × 10⁶ × 0.083 14 × 295.15)/(2.5× 10⁷)
= 1.7 bar
Explanation:
your answer is Kelvin because it is the SI unit of temperature
<u>Answer:</u> The correct answer is Option b.
<u>Explanation:</u>
To calculate the amount of heat absorbed or released, we use the following equation:
.....(1)
where, q = amount of heat absorbed or released.
m = mass of the substance
c = heat capacity of water = 4.186 J/g ° C
= Change in temperature
We are given:
![m=30g\\\Delta T=[40-0]^oC=40^oC\\q=?J](https://tex.z-dn.net/?f=m%3D30g%5C%5C%5CDelta%20T%3D%5B40-0%5D%5EoC%3D40%5EoC%5C%5Cq%3D%3FJ)
Putting values in equation 1, we get:
q = 5023.2 J
We are given:
![m=40g\\\Delta T=[40-30]^oC=10^oC\\q=?J](https://tex.z-dn.net/?f=m%3D40g%5C%5C%5CDelta%20T%3D%5B40-30%5D%5EoC%3D10%5EoC%5C%5Cq%3D%3FJ)
Putting values in equation 1, we get:
q = 1674.4 J
Heat gained by Trial 1 than trial 2 = 
Hence, the amount of heat gained in Trial 1 about 3347 J more than the heat released in Trial 2.
Thus, the correct answer is Option b.
Answer:
Mass of chemical = 1.5 mg
Explanation:
Step 1: First calculate the concentration of the stock solution required to make the final solution.
Using C1V1 = C2V2
C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution
C1 = C2V2/V1
C1 = (6 * 25)/ 0.1
C1 = 1500 ng/μL = 1.5 μg/μL
Step 2: Mass of chemical added:
Mass of sample = concentration * volume
Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL
Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg
Therefore, mass of sample = 1.5 mg
