A) James Cook.
B) He put his sailors on a strict diet to see if they would get scurvy.
C) Sauerkraut.
D) He told others of this diet and that none of his sailors died of scurvy.
E) Chemicals can be found almost anywhere and almost anyone can be a scientist.
<span>A 50-gram sample with a half-life of 12 days will have a remaining mass of 25 grams after its 12-day half-life.
Every cycle of a half-life, the sample will lose half of its mass, so if the half-life, itself, is 12 days and the time period passing is 12 days, one half-life has passed and the material will be halved.</span>
2AgNO3+K2CrO4⇒Ag2CrO4(s)+2KNO3
Hence by mixing 0.0024M AgNO3 and 0.004M
K2CrO4, we will have Ag2CrO4 which is precipitated out and leave us with
0.0024M KN03 which is mixed with (0.004-0.0024/2)M, it can be 0.0028M, of K2Cr04
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.
Answer:
50 mg
Explanation:
First, we have to calculate the partial pressure of O₂ (pO₂) using the following expression.
pO₂ = P × X(O₂) = 1.13 atm × 0.21 = 0.24 atm
where,
P: total pressure
X(O₂): mole fraction of oxygen
Then, we can calculate the concentration of O₂ in water (C) using Henry's law.
C = k × pO₂ = (1.3 × 10⁻³ M/atm) × 0.24 atm = 3.1 × 10⁻⁴ M
where,
k: Henry's constant for O₂
The mass of oxygen in a 5.00 L bucket with a concentration of 3.1 × 10⁻⁴ M is: (MM 32.0 g/mol)
