Answer:
Explanation:
We are asked to find the specific heat capacity of a sample of lead. The formula for calculating the specific heat capacity is:
The heat absorbed (Q) is 237 Joules. The mass of the lead sample (m) is 22.7 grams. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature. The temperature increases <em>from</em> 29.8 °C <em>to </em>95.6 °C.
- ΔT = final temperature -inital temperature
- ΔT= 95.6 °C - 29.8 °C = 65.8 °C
Now we know all three variables and can substitute them into the formula.
- Q= 237 J
- m= 22.7 g
- ΔT = 65.8 °C
Solve the denominator.
- 22.7 g * 65.8 °C = 1493.66 g °C
Divide.
The original values of heat, temperature, and mass all have 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place tells us to round the 8 up to a 9.
The specific heat capacity of lead is approximately <u>0.159 Joules per gram degree Celsius.</u>
Answer:
The cuvette was blank with the solution so that the spectrometer will only read the solute absorbance. This also ensures that the spectrometer will ignore other absorbance fluctuations that normally occur due to the chemical make-up of water. The spectrometer only considered the absorbance of 
as indicated on the spectrum. The reaction between the 
and the 
are both clear liquids that form the orange liquid product which creates the absorbance spectrum. Because the color of the solution is orange, it reflects this and similar colors while absorbing blueish hues. We can find the absorption of only the by pre-rinsing the cuvette with each solution we intend to measure before placing it in the spectrometer. Also, wipe each cuvette with a kimwipe to remove all fingerprints that could effect the data collection.
Explanation:
The cuvette was blank with the solution so that the spectrometer will only read the solute absorbance. This also ensures that the spectrometer will ignore other absorbance fluctuations that normally occur due to the chemical make-up of water. The spectrometer only considered the absorbance of as indicated on the spectrum.
Answer:
pHe = 3.2 × 10⁻³ atm
pNe = 2.5 × 10⁻³ atm
P = 5.7 × 10⁻³ atm
Explanation:
Given data
Volume = 1.00 L
Temperature = 25°C + 273 = 298 K
mHe = 0.52 mg = 0.52 × 10⁻³ g
mNe = 2.05 mg = 2.05 × 10⁻³ g
The molar mass of He is 4.00 g/mol. The moles of He are:
0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol
We can find the partial pressure of He using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 3.2 × 10⁻³ atm
The molar mass of Ne is 20.18 g/mol. The moles of Ne are:
2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol
We can find the partial pressure of Ne using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 2.5 × 10⁻³ atm
The total pressure is the sum of the partial pressures.
P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm
Answer: Increases.
Explanation: As the temperature of a liquid or solid increases its vapor pressure also increases. Conversely, vapor pressure decreases as the temperature decreases.
