12.3g 02times1 mol 02/32.0 g 02=mol 02
.38mol/x=20.95/100
x=1.8mol air...
PV=nRT
101.3kPa times V =1.8 mol times 8.31[kPa *L]/{k*mol}*273k
V=40.3Lair
<h2>Selective & Differential Medium</h2>
Explanation:
- Selective media allow specific types of organisms to develop, and inhibit the development of different living beings. The selectivity is cultivated in a few ways.For model, living beings that can use a given sugar are handily screened by making that sugar the main carbon source in the medium. On the other hand,selective hindrance of certain sorts of microorganisms can be accomplished by adding dyes, anti-infection agents, salts or explicit inhibitors which influence the digestion or enzyme systems of the living beings
- Differential media are utilized to separate firmly related life forms or groups of living beings. owing to the pre of specific colors or synthetic compounds in the media, the creatures will deliver trademark changes or development designs that are utilized for ID or separation. An assortment of particular and differential media are utilized in clinical, demonstrative and water contamination research facilities, and in food and dairy laboratories
- Selective media because elevated NaCI level is designed to help grow selective bacteria.differential media because the fermented sugar gives off a yellow halo which allows for differentiate between bacteria
100°C because all the molecules are moving the fastest past each other
Answer:
The mass of water = 219.1 grams
Explanation:
Step 1: Data given
Mass of aluminium = 32.5 grams
specific heat capacity aluminium = 0.921 J/g°C
Temperature = 82.4 °C
Temperature of water = 22.3 °C
The final temperature = 24.2 °C
Step 2: Calculate the mass of water
Heat lost = heat gained
Qlost = -Qgained
Qaluminium = -Qwater
Q = m*c*ΔT
m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)
⇒with m(aluminium) = the mass of aluminium = 32.5 grams
⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C
⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C
⇒with m(water) = the mass of water = TO BE DETERMINED
⇒with c(water) = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C
32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9
-1742.1 = -7.95m
m = 219.1 grams
The mass of water = 219.1 grams
Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)
v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol
n(Na₂S)=n{Cd(NO₃)₂}=cv
m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv
m(Na₂S)=78.046*0.0100*25.00≈19.5 mg