Answer:
Explanation:
a) 3.00 mol Cu x 2 mol Ag/ 1 mol Cu = 6 mol Ag
b) 3.00 mol Cu x 2 mol Ag/1 mol Cu x 107.87g Ag/ 1 mol Ag = 647.22g Ag
c) 3.00 mol Cu x 1 mol Cu(NO3)2/ 1 mol Cu = 3 mol Cu(NO3)2
d) 3.00 mol Cu x 1 mol Cu(NO3)2/ 1 mol Cu x 187.54g/ 1 mol Cu(NO3)2 =562.644g Cu(NO3)2
Answer:
0.238 M
Explanation:
A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.
C₁.V₁ = C₂.V₂
C₁ × 24.00 mL = 0.220 M × 52.00 mL
C₁ = 0.477 M
The concentration of Pb(ClO₃)₂ is:

Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature
Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules
The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).
The answer to this question would be: resistance
When a patient developing a resistance to a certain drug, the same amount of doses will not exert the same effect as before. The effect will be lower, thus the doctor will need to increase the dose to increase the effect