Answer:
When a front passes over an area, it means a change in the weather. Many fronts cause weather events such as rain, thunderstorms, gusty winds, and tornadoes. At a cold front, there may be dramatic thunderstorms. At a warm front, there may be low stratus clouds. Usually, the skies clear once the front has passed.
Explanation:
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<u>61.25 grams</u> of CO can be formed from 35 grams of oxygen.
The molecular mass of oxygen is <u>16 gmol⁻¹</u>
The molecular mass of carbon monoxide is<u> 28 gmol⁻¹</u>
Explanation:
The molar mass of carbon monoxide is molar mass of C added to that of O;
12 + 16 = 28
= 28g/mol
The molar mass of oxygen is 16 g/mol while that of oxygen gas (O₂) is 32 g/mol
Since the ration oxygen to carbon monoxide is 1: 2 moles, we begin to find out how many moles of carbon monoxide are formed by 35 g of oxygen;
35/32 * 2
= 70/32 moles
Then multiply by the molar mass of carbon monoxide;
70/32 * 28
= 61.25 g
Answer: D. transverse
Explanation:
Light is a transverse wave, while sound is a longitudinal wave.
Answer:
Pressure for H₂ = 11.9 atm
Option 5.
Explanation:
We determine the complete reaction:
2Al(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
As we do not know anything about the HCl, we assume that the limiting reactant is the Al and the acid is the excess reagent.
Ratio is 2:3.
2 moles of Al, can produce 3 moles of hydrogen
Therefore 4.5 moles of Al must produce (4.5 . 3) / 2 = 6.75 moles
Now we can apply the Ideal Gases law to find the H₂'s pressure
P . V = n . R . T → P = (n . R .T) / V
We replace data: (6.75 mol . 0.082L.atm/mol.K . 300K) / 14L
Pressure for H₂ = 11.9 atm
Answer: 1) 
Equilibrium constant is defined as the ratio of the product of concentration of products to the product of concentration of reactants each term raised to their stochiometric coefficients.
![K_{eq}=\frac{[H_2S]^2}{H_2]^2\times [S_2]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D)
where [] = concentration in Molarity=
Thus ![[H_2S]=\frac{68.5}{1.0}=68.5M](https://tex.z-dn.net/?f=%5BH_2S%5D%3D%5Cfrac%7B68.5%7D%7B1.0%7D%3D68.5M)
![[H_2]=\frac{0.50}{1.0}=0.50M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%5Cfrac%7B0.50%7D%7B1.0%7D%3D0.50M)
![[S_2]=\frac{0.020}{1.0}=0.020M](https://tex.z-dn.net/?f=%5BS_2%5D%3D%5Cfrac%7B0.020%7D%7B1.0%7D%3D0.020M)
![K_{eq}=\frac{[68.5]^2}{0.50]^2\times [0.020]}=938450](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5B68.5%5D%5E2%7D%7B0.50%5D%5E2%5Ctimes%20%5B0.020%5D%7D%3D938450)
As the value of K is greater than 1, the reaction is product favored.
2) 
![K_{eq}=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![K_{eq}=\frac{[0.500]^2}{[0.0250]}=10](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5B0.500%5D%5E2%7D%7B%5B0.0250%5D%7D%3D10)
3) 
![K_{eq}=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ctimes%20%5BH_2%5D%5E3%7D)
4) Reactions which do not continue to completion are called equilibrium reactions as the rate of forward reaction is equal to the rate of backward direction.