Assume it is 1 litre and weighs 1kg.
2 percent of 1 kg is 20g.
20g divided by molar mass of NaOH.
20g divide by 40 = 0.5 mole
0.5 mole in a litre would be 0.5M
That is the answer: 0.5M
The equilibrium constant is found by [product]/[reactant]
If the equilibrium constant is very small, such as 4.20 * 10^-31, then that means at equilibrium there is very little product and a lot of reactant.
And likewise, if there is a lot of product formed, and very little reactant, then the K value will be very large, which tells us that it is predominantly product.
At equilibrium, for any reaction, there will always be some reactant and some product present. There cannot be zero reactant or zero product. Also keep in mind that the equilibrium constant is dependent on temperature.
At equilibrium, for your reaction, it is predominantly reactants.
For a p type of semiconductor we need a dopant which is from 13th group in periodic table
Al , B, Ga, In Tl
So the correct element will be In : Indium
The other elements belongs to 15th group and hence will give n type semiconductor
Answer:
+4
Explanation:
In PbO2, oxygen exhibits an oxidation number of -2 (since it's not a peroxide or superoxide):
Let the oxidation number of Pb be x. Then, for the compound to be neutral, the oxidation numbers of all atoms should add up to zero.
⇒ x + (−2) + (−2) = 0
x = +4
So the oxidation no. of Pb is +4.
I hope this helps.
To get a result with the best degree of precision, the
number of significant figures should be equal to the smallest number of
significant figures of the given numbers. In this case, the smallest is 3 as
given by the number 9.03 mL.
Therefore density is:
<span>11.50 g / 9.03 mL = 1.27 g/mL</span>
