<span>Answer: has only one identity
Compounds like water (H2O) has 2 kinds of atoms which are hydrogen and oxygen. They also have a chemical bond and their bond is not able to be separated by physical means. It different with a mixture of water with sugar. The water and sugar are two different molecules but they don't bind each other. If you evaporate the water, the sugar will remains so it </span>can be separated by physical means.<span>.</span>
initial volume of the argon sample = 5.93L according to Boyle's law
What is Boyle's law ?
Boyle's law, also known as Mariotte's law, is a relationship describing how a gas will compress and expand at a constant temperature. The pressure (p) of a given quantity of gas changes inversely with its volume (v) at constant temperature, according to this empirical connection, which was established by the physicist Robert Boyle in 1662. In equation form, this means that pv = k, a constant.
According to Boyle's law
P1/V1 = P2/V2
P1 = initial pressure
P2 = final pressure
V1 =initial volume
V2= final volume
V1 = P1*V2/P2
V1 = 2.32*18.3/7.16 = 5.93L
initial volume of the argon sample = 5.93L according to Boyle's law
To know about Boyle's law from the link
brainly.com/question/26040104
#SPJ4
Answer:
63.9 grams. Yes, the Nacl was converted. Maximum possible ppm is 540ppm.
Explanation:
I this is college level chemistry not regular high school chem.
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
Answer:
81 °C
Explanation:
This is a calorimetry question so a few things you will need for this. The calorimetry equation q=mcΔT & the specific heat of water (4.2J/g•°C). Other definitions are:
q = heat added/released by a sample
m = mass of sample
c=specific heat of sample
ΔT = change in temperature
from here we can rearrange the equation to state:
q/(mc) = ΔT
1200J/((20.0g)(4.2J/g•°C)) = ΔT
14°C = ΔT
If the starting temperature was 95.0°C and we know that the temperature was cooled by 14°C then the final temperature of the water would be 81.
