The correct answer that would best complete the given statement above would be option 2. <span>The relationship between molecular velocities and temperature is a direct relationship. In other words, their relationship is directly proportional. Hope that this is the answer that you are looking for. </span>
<span>C2Br2
First, we need to determine how many moles of the gas we have. For that, we'll use the Ideal Gas Law which is
PV = nRT
where
P = pressure (1.10 atm = 111458 Pa)
V = volume (10.0 ml = 0.0000100 m^3)
n = number of moles
R = Ideal gas constant (8.3144598 (m^3 Pa)/(K mol) )
T = Absolute temperature
Solving for n, we get
PV/(RT) = n
Now substituting our known values into the formula.
(111458 Pa * 0.0000100 m^3) / (288.5 K * 8.3144598 (m^3 Pa)/(K mol))
= (1.11458/2398.721652) mol
= 0.000464656 mol
Now let's calculate the empirical formula for this compound.
Atomic weight carbon = 12.0107
Atomic weight bromine = 79.904
Relative moles carbon = 13.068 / 12.0107 = 1.08802984
Relative moles bromine = 86.932 / 79.904 = 1.087955547
So the relative number of atoms of the two elements is
1.08802984 : 1.087955547
After dividing all numbers by the smallest, the ratio becomes
1.000068287 : 1
Which is close enough to 1:1 for me to consider the empirical formula to be CBr
Now calculate the molar mass of CBr
12.0107 + 79.904 = 91.9147
Finally, let's determine if the compound is actually CBr, or something like C2Br2, or some other multiple. Using the molar mass of CBr, multiply by the number of moles and see if the result matches the mass of the gas. So
91.9147 g/mol * 0.000464656 mol = 0.042708701 g
0.0427087 g is a lot smaller than 0.08541 g. So the compound isn't exactly CBr. Let's divide them to see what the factor is.
0.08541 / 0.0427087 = 1.99982673
1.99982673 is close enough to 2 to within the number of significant digits we have for me to claim that the formula for the unknown gas isn't CBr, but instead is C2Br2.</span>
Carbohydrates, proteins, lipids, and Nucleic Acids.
Answer: Yes
Explanation: Silicon is never found as a free element naturally, meaning it always occurs with other elements.
Answer:
No photoelectric effect is observed for Mercury.
Explanation:
From E= hf
h= Plank's constant
f= frequency of incident light
Threshold Frequency of mercury= 435×10^3/ 6.6×10^-34 × 6.02×10^23
f= 11×10^14 Hz
The highest frequency of visible light is 7.5×10^14. This is clearly less than the threshold frequency of mercury hence no electron is emitted from the mercury surface
