Answer:
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
Explanation:
<u>Step 1</u>: Data given
Mass of the metal = 21 grams
Volume of water = 100 mL
⇒ mass of water = density * volume = 1g/mL * 100 mL = 100 grams
Initial temperature of metal = 122.5 °C
Initial temperature of water = 17°C
Final temperature of water and the metal = 19 °C
Heat capacity of water = 4.184 J/g°C
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<u>Step 2: </u>Calculate the specific heat capacity
Heat lost by the metal = heat won by water
Qmetal = -Qwater
Q = m*c*ΔT
m(metal) * c(metal) * ΔT(metal) = - m(water) * c(water) * ΔT(water)
21 grams * c(metal) *(19-122.5) = -100 * 4.184 * (19-17)
-2173.5 *c(metal) = -836.8
c(metal) = 0.385 J/g°C
The metal has a heat capacity of 0.385 J/g°C
This metal is copper.
Answer: Hello i am confused are you asking a question?
Explanation:
Psolution = X · PH_20
= 0.966 · 31.8 torr
= 30.7 torr
Answer:
Explanation:
In this problem, the temperature stays constant. The volume and pressure change, so we use Boyle's Law. This states that the pressure of a gas is inversely proportional to the volume. The formula is:
Now we can substitute any known values into the formula.
Originally, the gas has a volume of 25.0 liters and a pressure of 2.05 atmospheres.
The volume is decreased to 14.5 liters, but the pressure is unknown.
Since we are solving for the new pressure, or P₂, we must isolate the variable. It is being multiplied by 14.5 liters and the inverse of multiplication is division. Divide both sides by 14.5 L .
The units of liters cancel.
The original values of volume and pressure have 3 significant figures, so our answer must have the same.
For the number we found, that is the hundredth place.
The 4 in the thousandth place (in bold above) tells us to leave the 3 in the hundredth place.
The new pressure is approximately <u>3.53 atmospheres.</u>