I feel like it is important, because it is always nice to learn about
new things and keep your mind open, to expand your knowledge.
I hope this helps ^-^
PV = nRT
⇒P1/T1 = P2/T2
⇒500/300 = 100/T2
⇒ T2 = 100×300/500 = 60K
Answer is D
Answer:
3'700,000 cfu
Explanation:
One way to count the amount of bacteria in a medium is by doing a dilution of the sample and count how many colonies growth. Each colony is a cfu (Colony forming units).
In the problem, you count 37 colonies. The dilution was 1:100,000. That means the bacteria present in the soap is:
37 colonies × (100,000 / 1) = <em>3'700,000 cfu</em>
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I hope it helps!
Answer:
0.077 M
Explanation:
Data Given :
The concentration of half normal (NaCl) saline = 0.45g / 100 g
So,
Volume of Solution = 100 g = 100 mL
Volume of Solution in Liter = 100 mL / 1000
Volume of Solution = 0.1 L
molar mass of NaCl = 58.44 g/mol
Molarity:
Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M
Formula used for Molarity
M = moles of solute / Liter of solution . . . . . . . . . . (1)
Now to find number of moles of Nacl
no. of moles of NaCl = mass of NaCl / molar mass
no. of moles of NaCl = 0.45g / 58.44 g/mol
no. of moles of NaCl = 0.0077 g
Put values in the eq (1)
M = moles of solute / Liter of solution . . . . . . . . . . (1)
M = 0.0077 g / 0.1 L
M = 0.077 M
So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M
