Answer:
c.boron-11
Explanation:
The atomic mass of boron is 10.81 u.
And 10.81 u is a lot closer to 11u than it is to 10u, so there must be more of boron-11.
To convince you fully, we can also do a simple calculation to find the exact proportion of boron-11 using the following formula:
(10u)(x)+(11u)(1−x)100%=10.81u
Where u is the unit for atomic mass and x is the proportion of boron-10 out of the total boron abundance which is 100%.
Solving for x we get:
11u−ux=10.81u
0.19u=ux
x=0.19
1−x=0.81
And thus the abundance of boron-11 is roughly 81%.
Answer:
1.5e+8 atoms of Bismuth.
Explanation:
We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):
For this, it is necessary to know the values in meters for any of these diameters:
Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.
<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>
1 atom of Bismuth = 320pm in diameter.
<h3>Diameter of a biscuit in meters</h3>
<h3>Resulting Ratio</h3>
How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:
In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.
Answer: (C) Although the average kinetic energy of the colliding substances increases, this has no influence on activation energy.
Explanation:
After increasing the temperature of the reaction , the rate of the chemical reaction increases due to increase in the average kinetic energy of the particles. At increased temperature high proportions of particles can react making the reaction faster.
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.
1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.
2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.
</span>
3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.<span>
</span>
The half life equation is -->P(t) = Pi (0.5) ^ (t/c)
c is equal to the element to reach its half-life (5 seconds)t is equal to the duration of time the element is expose to (20 seconds)Pi is the initial amount (340)0.5 is the base of this exponential function to represent half-life.P(t) is the expression for the function of time
P(20) = 340 (0.5)^20/5P(20) = 340 (0.5) ^4P(20)= 21.25 grams
Fraction = P(t)/Pi = P(20)/Pi =21.25/340 =1/16
Therefore, when given 20 seconds, 340 grams of Fluorine-21 will degrade to 21.25 grams OR 1/16 of its original mass.
Hope this method helps! (This is my answer btw, I think you may have accidentally posted twice?)
