Answer:
Extensive properties, such as mass and volume, depend on the amount of matter being measured. Intensive properties, such as density and color, do not depend on the amount of the substance present.
Answer:
a) 6.02 × 10^23
b)6.02×10^23
c)3.01×10^24
d) 3.612×10^24
In every mole of a substance there are 6.02×10^23 entities.
in ( c) there are 5 carbon atoms . 5 is multiplied by the number of moles of the molecule and then multiplied by 6.02×10^23. ( using the formula N=nL )
Answer:
The answer is 139.1 g of sodium azide.
Explanation:
To solve this question we need to find first the stoichometry of the reaction to find the moles involved.
1. The balanced formula:
2NaN3 ⇒ 3N2 + 2Na
(important: the nitrogen is a gas, therefore, is always diatomic N2)
2. The moles:
We need to convert the 90 g of N2 into moles to know the relations of amounts between the other compounds in the chemical equation
90 g of N2 x (1 mol of N2/ 28 g of N2) = 3.21 mol of N2
3. Stoichometry:
The relations of the amounts of moles of N2 and NaN3 using the balanced formula
3.21 mol of N2 x (2 mol of NaN3/3 mol of N2) = 2.14 mol of NaN3
4. Moles into grams
then to answer the question we convert the moles value of NaN3 into grams of NaN3
2.14 mol of NaN3 x ( 65 g of NaN3/ 1 mol of NaN3) = 139.1 g of NaN3.
and this is the answer of the question.
Answer:
Explanation:
Mᵣ: 58.44 278.11
Pb(NO₃)₂ + 2NaCl ⟶ PbCl₂ + 2NaNO₃
m/g: 26.3
1. Moles of NaCl
(b) Moles of PbCl₂
(c) Theoretical yield of PbCl₂
(d) Percent yield
Answer:
![\frac{[F^{-}]}{[HF]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BF%5E%7B-%7D%5D%7D%7B%5BHF%5D%7D)
is larger
Explanation:
, where 
is the acid dissociation constant.
For a monoprotic acid e.g. HA, ![K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
and ![\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
So, clearly, higher the value , lower will the the 
In this mixture, at equilibrium, ![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
will be constant.
of HF is grater than of HCN
Hence, ![(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})](https://tex.z-dn.net/?f=%28%5Cfrac%7BF%5E%7B-%7D%7D%7B%5BHF%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HF%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29%3E%28%5Cfrac%7BCN%5E%7B-%7D%7D%7B%5BHCN%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HCN%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29)
So, is larger
