Answer:
ΔT = 98.84 °C
Explanation:
Given data:
Heat absorbed = 255 J
Mass of gold = 20.0 g
Specific heat capacity of gold = 0.129 J/g.°C
Temperature change = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
by putting values,
255 J = 20.0 g × 0.129 J/g.°C × ΔT
255 J = 2.58 J / °C × ΔT
ΔT = 255 J / 2.58 J / °C
ΔT = 98.84 °C
Answer:
D
Explanation:
When a dilution is made, a volume of the stock solution is collected and then is mixed to the solvent. The total amount of the solute (number of moles or mass), must be equal in the volume of the sample and at the final volume, because of the Lavoiser's law (the matter can't be created nor destructed).
The mass or the number of moles is the concentration (C) multiplied by the volume (V), so, if 1 is the sample of the stock solution, and 2 the diluted solution:
C1*V1 = C2*V2
The final volume of the solution is 10 mL. So, let's identify the volume needed for each stock solution.
Acetaminophen
C1 = 500 ug/L
C2 = 50 ug/L
500*V1 = 50*10
V1 = 1 mL
Dextromethorphan
C1 = 1 mM = 1000 uM
C2 = 20 uM
1000*V1 = 20*10
V1 = 0.2 mL
So, the volume of water needed is the total less the volume of the stocks solutions less the volume of the embryo water:
V = 10 - 1 - 1 - 0.2 = 7.8 mL
Thus, to to the solution, it's necessary to add at 1 mL of the embryo water 1 ml of 500 ug/I acetaminophen and 0.2 ml of 1 mM dextromethorphan, and 7.8 ml of water.
Is it multiple choice? If not, the main reason would be the ability to escape from prey, or move to a different climate based on the individual needs.
Answer:
mass H2O2 = 0.31g H2O2
mass unreacted = 0.0745g BaO2
Explanation:
Mw BaO2 = 169.33 g / mol
Mw HCl = 36.46 g/mol
Mw H2O2 = 34.0147 g/mol
LR:
⇒ mol BaO2 = 1.50g BaO2 * mol / 169.33 g = 8.86E-3 mol BaO2 / 1 = 8.86E-3 mol BaO2
⇒ mol HCl = 25.0 mL * 0.0272g / mL * mol / 36.46g = 0.0186 mol HCl / 2 = 9.3E-3 mol HCl.....L.R
mol H2O2:
⇒ 0.0186mol HCl * (mol H2O2 / 2mol HCl) = 9.325E-3 mol H2O2
⇒ g H2O2 = 9.325E-3mol H2O2 * 34.0147g H2O2 / mol H2O2 = 0.31g H2O2
mass of wich reagent is left unreacted:
mol that react BaO2 = 0.0186mol HCl * (mol BaO2 / 2mol HCl) = 9.3E-3mol BaO2
mol that unreacted BaO2 = 9.3E-3 - 8.86E-3 = 4.4E-4 mol BaO2
g unreacted BaO2 = 4.4E-4mol BaO2 * (169.33g BaO2 / mol BaO2) = 0.0745g BaO2
