Power measures work and time
power=work/time
Answer:
0.05 mg/mL ( B )
Explanation:
Given data:
20 mg/ml starch
2% solution = 2g of solute is in 100g of solvent
<u>Determine the new concentration in mg/ml </u>
Dilution equation = C1V1 = C2V2
new concentration ; applying the dilution factor
dilution factor = 1 : 400 ; ( 2 /400 )g = 0.005 g of solute is present in every 100 mL
∴ new concentration = 0.00005 g / 1 mL * ( 1000 mg / 1g ) = 0.05 mg/mL
Answer:
stabilizina selection
Explanation:
As given in the question, the galls may be smaller or larger. However the eggs laid in smaller gallsare prey to wasps, hence they are destroyed
Possible contaminate will be Proteus or Salmonella. To differentiate the two genera urease test can be accomplished. However, urease test, Proteus will produce a positive test result.
This test can be done to determine the ability of an organism to split urea. The principle of urease test is that urea is the product of decarboxylation of amino acid. Splitting of urea via hydrolysis will produce ammnonia and CO2. The CO2 produce allows the formation of bubbles.
Answer:
A tissue
Explanation:
Cells are grouped together to carry out specific functions. A group of cells that work together form a tissue.
I hope this helps :")
