Answer:
Glucose, found in the food animals eat, is broken down during the process of cellular respiration into an energy source called ATP. When excess ATP and glucose are present, the liver converts them into a molecule called glycogen, which is stored for later use.
100ml volume of 0.0150m hcl solution is requires to titrate 150ml of a 0.0100m caoh2 solution.
Dilution is a solution of decreasing the concentration of a solute in the solution by adding more solvent to the solution. We can use the expression for dilute formula,
C1 V1 =C2 V2
where C1 is the initial concentration,C2 is the final concentration,V1 is the initial volume and V2 is the final volume. Here given, volume of 0.0150M(C1) HCL solution is required to titrate 150ml(V2) of a 0.0100M(C1) Caoh2 solution.
While diluting a solution from a high concentration substance to a low concentration substance we always use the formula of dilution.so, putting all value give in the expression we get the volume of the final concentration.
V1= C2 V2/ C1
= 0.0100m . 150ml /0.0150M
= 100ml
The volume of the hcl solution is 100ml.
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Answer:Calculations must be done first because energy depends on both temperature and mass.
Explanation:
Answer:
104.352°C
Explanation:
Data Given:
Boiling point of water = 100.0°C
Kb (boiling point constant = 0.512°C/m
Concentration of the Mg₃(PO₄)₂ = 8.5 m
Solution:
Formula Used to find out boiling point
ΔTb = m.Kb . . . . . . (1)
where
ΔTb = boiling point of solution - boiling point of water
So,
we can write equation 1 as under
ΔTb = Tb (Solution) -Tb (water)
As we have to find out boiling point so rearrange the above equation
Tb (Solution) = m.Kb + Tb (water) . . . . . . . (2)
Put values in Equation 2
Tb (Solution) = (8.5 m x 0.512°C/m ) + 100.0°C
Tb (Solution) = 4.352 + 100.0°C
Tb (Solution) = 104.352°C
so the boiling point of Mg₃(PO₄)₂ 8.5 m solution = 104.352°C
Answer:
it is actually the cell membrane.
Explanation:
the cell wall can only be found in plants
