The balanced equation for the reaction between KOH and HBr is as follows;
KOH + HBr --> KBr + H₂O
stoichiometry of KOH to HBr is 1:1
number of KOH moles reacted - 0.25 mol/L x 0.015 L = 0.00375 mol
according to molar ration
number of KOH moles reacted = number of HBr moles reacted
number of HBr moles reacted - 0.00375 mol
if 12 mL of HBr contains - 0.00375 mol
then 1000 mL of HBr contains - 0.00375 mol / 12 mL x 1000 mL = 0.313 mol
therefore molarity of HBr is 0.313 M
Answer:
Halogen / salt-former
Explanation:
Bromine is classified as an element in the 'Halogens' section which can be located in group 7 of the Periodic Table. The term "halogen" means "salt-former" and compounds containing halogens are called "salts".
Answer:
Yes
Explanation:
By definition, the equilibrium constanct, Kc, for the reaction A ⇒ 2B is
= [A]^1 / [B]^2
Substitute [A] = 4 and [B] = 2 in the equation,
[A]^1 / [B]^2
= 4^1 / 2^2
= 1
= Kc
So yes the reaction is at equilibrium.
Answer:
b
Explanation:
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Answer:
3 × 10⁴ kJ
Explanation:
Step 1: Write the balanced thermochemical equation
C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(g) ΔH = -2220 kJ
Step 2: Calculate the moles corresponding to 865.9 g of H₂O
The molar mass of H₂O is 18.02 g/mol.
865.9 g × 1 mol/18.02 g = 48.05 mol
Step 3: Calculate the heat produced when 48.05 moles of H₂O are produced
According to the thermochemical equation, 2220 kJ of heat are evolved when 4 moles of H₂O are produced.
48.05 mol × 2220 kJ/4 mol = 2.667 × 10⁴ kJ ≈ 3 × 10⁴ kJ
