Group 18 is the only group in periodic table in which all the elements are nonmetals. This group contains F, Cl, Br, I and At and also the other name of this group is halogen which means salt producer.
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Answer:
The value of Q must be less than that of K.
Explanation:
The difference of K and Q can be understood with the help of an example as follows
A ⇄ B
In this reaction A is converted into B but after some A is converted , forward reaction stops At this point , let equilibrium concentration of B be [B] and let equilibrium concentration of A be [A]
In this case ratio of [B] and [A] that is
K = [B] / [A] which is called equilibrium constant.
But if we measure the concentration of A and B ,before equilibrium is reached , then the ratio of the concentration of A and B will be called Q. As reaction continues concentration of A increases and concentration of B decreases. Hence Q tends to be equal to K.
Q = [B] / [A] . It is clear that Q < K before equilibrium.
If Q < K , reaction will proceed towards equilibrium or forward reaction will
proceed .
Answer:
At 0.58 L of 0.540 M NaOH solution contain 12.5 g NaOH.
Explanation:
Given data:
At volume = ?
Mass of NaOH = 12.5 g
Molarity of solution = 0.540 M
Solution:
First of all we will calculate the number of moles of sodium hydroxide.
Number of moles = mass/molar mass
Number of moles = 12.5 g / 40 g/mol
Number of moles = 0.3125 mol
Volume of NaOH:
Molarity = number of moles / volume in L
Now we will put the values.
0.540 M = 0.3125 mol / volume in L
volume in L = 0.3125 mol / 0.540 mol/L
volume in L = 0.58 L
Answer:
21.2 gm
Explanation:
calculate the mass of butane needed to produce 64.1 g of carbon dioxide to three significant figures and appropriate units
butane is the hydrocarbon C4H10
in combustion, we react hydrocarbons with O2 to form CO2 and H2O
so
C4H10 + O2----------------> CO2 + H2O
BALANCE
2C4H10 + 1302--------> 8CO2 + 10 H2O
the molar mass of CO2 is 12 + 16X2 = 44
64.1 gm of CO2 is
64.1/44 = 1.46 MOLES OF CO2,
FOR EVERY 8 MOLES OF CO2 WE NEED 2 MOLES OF BUTANE IT IS A
8:2 OR 4:1 RATIO. THE MOLES OF C4H10 ARE 1/4 THE MOLES OF CO2
SO
THE MOLES OF C4H10 H10 ARE 1.46/4 =0.365 MOLES
THE MOLAR MASS OF BUTANE IS 58.12
0.365 MOLES OF C4H10 HAS A MASS OF 0.365 X 58.12 = 21.2 gm