Answer:
1.37 x 
CFU/mL
Explanation:
First, the dilution factor needs to be calculated.
Since four 9 ml dilution blanks were prepared, the dilution factor that yielded 137 colonies is of 
.
Next is to divide the colony forming unit from the dilution by the dilution factor:
137/ = 137 x 
In order to get the CFU/ml, divide the CFU from the dilution by the plated volume (1 mL) from the final dilution tube.
137 x /1 = 1.37 x
Hence, the CFU/ml present in the original <em>E. coli </em> sample is 1.37 x .
cfu/ml = (no. of colonies x dilution factor) / volume of culture plate
1) You need to use the atomic mass of copper.
You can find it in a periodic table. It is 63.546 amu.
2) The atomic mass is the weigthed mass of the different isotopes.
This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:
=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 * atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.
3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x
=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278
=> 63.546 = 62.9296x + 64.9278 - 64.9278x
=> 64.9278x - 62.9296 = 64.9278 - 63.546
=> 1.9982x = 1.3818
=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%
=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%
Answer:
Cu-63 69.15%;
Cu-65 : 30.85%
<span>Assuming that there are 36 strontium and 24 phosphate, there
aren’t any equal cations and anoins because in theory only one ionic bond is
formed by a strontium with each phosphate ion. To the point that a cation will
eventually have an excess.</span>
H20. 2 of hydrogen and oxygen
The value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
Aa we know that, 125mL of 0.06M Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl.
Given, T = 25°C.
<h3>Chemical equation:</h3>
Pb(NO3)2 + NaCl ---- NaNO3 + PbCl2
PbCl2 in aqueous solution split into following ions
PbCl2 ------ Pb(+2) + 2Cl-
Q = [Pb(+2)] [Cl-]^2
The Concentration of Pb(+2) ions and Cl- ions can be calculated as
[Pb(+2)] = 0.06 × 125/200
= 0.0375
[Cl-] = 0.02 × 75/200
= 0.0075
By substituting all the values, we get
[0.0375] [0.0075]^2
= 2.11 × 10^(-6).
Thus, we calculated that the value of Q for 125.0 ml of 0.0500 m Pb(NO3)2 is mixed with 75.0 ml of 0.0200 m NaCl at 25°C is 2.11 × 10^(-6).
learn more about Ions:
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