Answer:
1. 7.256g of NaCl
2. 47.33g of Cl2
Explanation:
2 moles of Na reacts to produce 2 moles of NaCl
8 moles of Na will still produce 8 moles of NaCl
Mass of NaCl = molar mass of Nacl/moles of Nacl
=58.5/8
=7.256g of NaCl
From the equation, 2 moles of Na reacts with 1 mole of Cl2
3/2 moles of Cl2 will react with 3 moles of Na
Mass of Cl2 = 71/1.5
=47.33g of Cl2
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
Answer:
✓ scholastics
Explanation:
you d.ont need a expla.nation rig.ht un.less y.ou wan.na re.ad for an h.our
<u>Answer:</u>
The common name for the compound H2O is water.
<u>Explanation:</u>
The systemic name of H2O is Dihydrogen monoxide.
Answer:
32.1 g
Explanation:
Step 1: Write the balanced combustion reaction
C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O
Step 2: Calculate the moles corresponding to 97.4 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
97.4 g × 1 mol/44.01 g = 2.21 mol
Step 3: Calculate the moles of butane that produced 2.21 moles of carbon dioxide
The molar ratio of C₄H₁₀ to CO₂ is 1:4. The moles of C₄H₁₀ required are 1/4 × 2.21 mol = 0.553 mol
Step 4: Calculate the mass corresponding to 0.553 moles of C₄H₁₀
The molar mass of C₄H₁₀ is 58.12 g/mol.
0.553 mol × 58.12 g/mol = 32.1 g
