To convert boiling water to steam, that would involve heat of vaporization. The heat of vaporization for water at atmospheric conditions is: ΔHvap = <span>2260 J/g.
Molar mass of water = 18 g/mol
Q = m</span>ΔHvap = (1.50 mol water)(18 g/mol)(<span>2260 J/g) = 61,020 J
Time = Q/Rate = (61,020 J)(1 s/20 J) = 3051 seconds
In order to express the answer in three significant units, let's convert that to minutes.
Time = 3051 s * 1min/30 s = <em>102 min</em></span>
Answer:
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
Explanation:
Step 1: Data given
Nitric acid = HNO3
Molar mass of H = 1.01 g/mol
Molar mass of N = 14.0 g/mol
Molar mass O = 16.0 g/mol
Number of moles nitric acid (HNO3) = 0.25 moles
Molairty = 0.10 M
Step 2: Calculate molar mass of nitric acid
Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)
Molar mass HNO3 = 1.01 + 14.0 + 3*16.0
Molar mass HNO3 = 63.01 g/mol
Step 3: Calculate mass of solute use
Mass HNO3 = moles HNO3 * molar mass HNO3
Mass HNO3 = 0.25 moles * 63.01 g/mol
Mass HNO3 = 15.75 grams
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
We can use two equations for this problem.<span>
t1/2 = ln
2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is
decay constant.
20 days = 0.693 / λ
λ = 0.693 / 20 days
(1)
Nt = Nο eΛ(-λt) (2)
Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time
taken.
t = 40 days</span>
<span>No = 200 g
From (1) and (2),
Nt = 200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>
</span>Hence, 50.01 grams of isotope will remain after 40 days.
<span>
</span>
total pressure = ( 705 torr + 231.525torr) = 936.525torr
Answer:
B 1.23 g/cc
Explanation:
For something to float on seawater, the density must be less than 1.03 g/mL. If the object sinks, the density is greater than 1.03 g/mL.
Let’s examine the answer choices. Keep in mind, the ice berg is mostly below the water level.
A. 0.88 g/cc
This is less than 1.03 g/cc, which would result in floating.
B. 1.23 g/cc
This is the best answer choice. The iceberg is mostly beneath the water, but some of it is exposed. The density is greater than 1.03 g/mL, but not so much greater that it would immediately sink.
C. 0.23 g/cc
This is less than 1.03 g/cc, which would produce floating.
D. 4.14 g/cc
This is much greater than 1.03 g/cc and the result would be sinking.
