Explanation:
Bao + .....Co2... --> BaCO3 + ......
Answer:
If an item is less dense, it floats on the water.
Explanation:
We know this since the more dense something it usually is heavier than if it was less dense. Which ways it down resulting to it sinking.
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).
Answer:
pOH of resulting solution is 0.086
Explanation:
KOH and CsOH are monoacidic strong base
Number of moles of
in 375 mL of 0.88 M of KOH =
= 0.33 moles
Number of moles of
in 496 mL of 0.76 M of CsOH =
= 0.38 moles
Total volume of mixture = (375 + 496) mL = 871 mL
Total number of moles of
in mixture = (0.33 + 0.38) moles = 0.71 moles
So, concentration of
in mixture,
= 
Hence, ![pOH=-log[OH^{-}]=-log(0.82)=0.086](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E%7B-%7D%5D%3D-log%280.82%29%3D0.086)
Answer:
b)15.0°C
Explanation:
Specific Heat of Water=4.2 J/g°C
This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.
∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.
80×4.2 J=336 J
Total Energy Provided=1680 J
The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}
Temperature increase=
=5°C
Initial Temperature =10°C
Final Temperature=Initial + Increase in Temperature
=10+5=15°C