Answer:
n = 2 moles (1 sig-fig)
Explanation:
Using the Ideal Gas Law equation (PV = nRT), solve for n (= moles) and substitute data for ...
pressure = P(atm) = 100atm
volume =V(liters) = 50L
gas constant = R = 0.08206L·atm/mol·K
temperature = T(Kelvin) = °C + 273 = (35 + 273)K = 308K
PV = nRT => n = PV/RT = (100atm)(50L)/(0.08206L·atm/mol·K)(308K)
∴ n(moles) = 1.978moles ≅ 2 moles gas (1 sig-fig) per volume data (= 50L) that has only 1 sig-fig. (Rule => for multiplication & division computations round final answer to the measured data having the least number of sig-figs).
Monkey One fell of if the bed.
Answer:
The evidence or effect on a detector of radiation caused by background radiation. In connection with health protection, the background count includes but is not limited to radiations produced by naturally occurring radioactivity and cosmic rays. Dictionary of Military and Associated Terms.Explanation:
Based on this website I would say B. Hope this helps
Answer:
This is a precipitation reaction in which Ni(OH)₂ precipitates.
8.68%
Explanation:
Let's consider the following reaction.
Ni²⁺(aq) + 2 NaOH(aq) ⇄ Ni(OH)₂(s) + 2 Na⁺(aq)
This is a precipitation reaction in which Ni(OH)₂ precipitates.
We can establish the following relations:
- The molar mass of Ni(OH)₂ is 92.71 g/mol.
- 1 mole of Ni(OH)₂ is produced per 1 mole of Ni²⁺.
- The molar mass of Ni²⁺ is 58.69 g/mol.
When 343 mg (0.343 g) of Ni(OH)₂ are collected, the mass of Ni²⁺ that reacted is:
The mass percent of nickel in the 25.0g-sample is: