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).
0.4 g/ml.............................
Answer: 0.05
Explanation:Divide the length (1.5 cm) by the number of turns (30)
Answer:
9.25
Explanation:
Let first find the moles of
and 
number of moles of
= 0.40 mol/L × 200 × 10⁻³L
= 0.08 mole
number of moles of
= 0.80 mol/L × 50 × 10⁻³L
= 0.04 mole
The equation for the reaction is expressed as:

The ICE Table is shown below as follows:

Initial (M) 0.08 0.04 0
Change (M) - 0.04 -0.04 + 0.04
Equilibrium (M) 0.04 0 0.04







for buffer solutions
since they are in the same solution


Answer:
Explanation if an object is in motion and more force is applied to it, the object will begin moving faster. If two objects have the same mass and a greater force is applied to one of the objects, the object which receives the greater force will change speeds more quickly.: