Answer:
982.5 kg/m³
Explanation:
When the temperature of a fluid increases, it dilates, and because of the variation of the volume, it's density will vary too. The density can be calculated by the expression:
ρ₁ = ρ₀/(1 + β*(t₁ - t₀))
Where ρ₁ is the final density, ρ₀ the initial density, β is the constant coefficient of volume expansion, t₁ the final temperature, and t₀ the initial temperature.
At t₀ = 4°C, the water desity is ρ₀ = 1,000 kg/m³. The value of the constant for water is β = 0.0002 m³/m³ °C, so, for t₁ = 93°C
ρ₁ = 1,000/(1 + 0.0002*(93 - 4))
ρ₁ = 1,000/(1+ 0.0178)
ρ₁ = 982.5 kg/m³
Answer: The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster. Effusion rate is inversely proportional to molar mass.
Explanation:
Answer:
Option D: More than 30 seconds
Explanation:
The enzyme CATALASE is found in almost all living organisms. CATALASE helps in the decomposition of one substance into another substance. CATALASE will breaks down hydrogen peroxide into water and oxygen.
When the potatoes were boiled it will surely produce no bubbles because the heat would have degrade the enzyme - catalase While the potatoes at room temperature potato produced the most bubbles because catalase works best at a room temperature.
If the potato solution was boiled for 10 minutes and cooled for 10 minutes before being tested, the average time for the disks to float to the surface of the hydrogen peroxide solution would be MORE THAN 30 SECONDS
Answer:
14.9075 g, 28.67%, 0.11%
Explanation:
The mean concentration of calcium = summation x / frequency
= ( 14.92 + 1491 + 14.88 + 14.92 ) /4 = 14.9075 g
Standard deviation = √(summation (x - μ)² /n) = √ ( ((14.92 - 14.9075)² +(14.91 - 14.9075)² + (14.88 - 14.9075)² + ( 14.92 - 14.9075)²) / 4) = 0.0164
b) percent error = abs(14.9075 - 20.90) / 20.90 × 100 = 28.67%
c) relative standard deviation = standard deviation / mean × 100 = 0.0164 / 14.9075 × 100 = 0.11%
d) The accuracy of the measure is the measurement compared to the actual which according to the standard set by the instructor (5%error) is not very accurate because the percent error is high (28.67%) while the relative standard deviation is quite low ( 0.11%) which means the measurement precision is very high.
The student will have to redo the experiment because the experiment was not too accurate since the percent error is way higher than the set value (5%) although the precision was high.
