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
Answer:
44 grams of CO₂ will be formed.
Explanation:
The balanced reaction is:
C + O₂ → CO₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:
- C: 1 mole
- O₂: 1 mole
- CO₂: 1 mole
Being the molar mass of each compound:
- C: 12 g/mole
- O₂: 32 g/mole
- CO₂: 44 g/mole
By stoichiometry the following mass quantities participate in the reaction:
- C: 1 mole* 12 g/mole= 12 g
- O₂: 1 mole* 32 g/mole= 32 g
- CO₂: 1 mole* 44 g/mole= 44 g
The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
If 12 grams of C react, by stoichiometry 32 grams of O₂ react. But you have 40 grams of O₂. Since more mass of O₂ is available than is necessary to react with 12 grams of C, carbon C is the limiting reagent.
Then by stoichiometry of the reaction, you can see that 12 grams of C form 44 grams of CO₂.
<u><em>44 grams of CO₂ will be formed.</em></u>
Answer:a) 0.1 mole. b) 4g. c) 2% d) 196 mL
Explanation: in 200mL , 0.1mole
mw NaOH = 40g/mol —> 4g in 0.1 mole
4g in 200mL so 2g in 100mL
density NaOH = 1g/mL so if 4g in 200 mL, 4mL , 196 mL water
The ionic compound of FePO4 is the iron III phosphate. Or you can call it ferric phosphate or ferric orthophosphate.
Just for general informations, It's normally used in steel and metal manufacturing processes, in organic farming, and many other uses.
Hope this Helps :)
Given:
Iron, 125 grams
T
1 = 23.5 degrees Celsius, T2 =
78 degrees Celsius.
Required:
Heat produced in kilojoules
Solution:
The molar mass of iron is 55.8
grams per mole. SO we need to change the given mass of iron into moles.
Number of moles of iron = 125 g/(55.8
g/mol) = 2.24 moles
<span>
Q (heat) = nRT = nR(T2 = T1)</span>
Q (heat) = 2.24 moles (8.314
Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
<u>Q (heat) = 1014.97 Joules or
1.015 kilojoules</u>
<span>This is the amount of heat
produced in warming 125 g f iron.</span>
