Climate is the weather that occur over a long period in a particular place.
Answer:
d. Temperature and number of molecules of gas
Step-by-step explanation:
Boyle's Law states, "The volume of a fixed mass of a gas is inversely proportional to the pressure if the temperature remains constant."
Let's examine the words.
"… volume…is inversely proportional to the pressure …" This means that volume and pressure are the <em>variables</em>.
"… fixed mass of a gas …" means that the number of molecules is constant.
"… temperature remains constant" speaks for itself.
a, c, and e are <em>wrong</em>, because pressure is a variable.
b is <em>wrong</em>, because volume is a variable.
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K = 
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M
Answer:
I think the layers of the atmosphere does temperature increase with increasing height. It is affected by convection because it heats the lower atmosphere. It is affected by conduction because the is the transfer of thermal energy. I guess
Hope this help!
Answer: The mass percentage of
is 5.86%
Explanation:
To calculate the mass percentage of
in the sample it is necessary to know the mass of the solute (
in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).
To calculate the mass of the solute, we must take the mass of the
precipitate. We can establish a relation between the mass of
and
using the stoichiometry of the compounds:

Since for every mole of Tl in
there are two moles of Tl in
, we have:

Using the molar mass of
we have:

Finally, we can use the mass percentage formula:
