Answer:
The differences in atmospheric pressure generate winds.
Explanation:
At the Equator, the sun warms the water and land more than it does the rest of the world. Warm equatorial air rises higher into the atmospher and migrates towards the poles. The complex relationships between fronts cause different types of winds and weather patterns.
I hope this helps :)
<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>
Answer:
<h2>2.44 L</h2>
Explanation:
The volume can be used by using the formula for Boyle's law which is
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we're finding the final volume
We have
We have the final answer as
<h3>2.44 L</h3>
Hope this helps you
The solution before dilution and after dilution contains same number of moles, and water is added for dilution.
Option B
<h3><u>Explanation:</u></h3>
Suppose before dilution, the solution contains x moles of KCl in Y liter of water. Now as the concentration got halved, then the solution contains x moles of KCl in 2Y kiters of solution. So the number of moles of KCl in the solution remained constant.
Again, as the solution is diluted to half of the concentration, water must have been added with the solution to make it dilute.
Density
=mass÷volume
=19.67÷5.90
=3.33 g/ml
