Answer:
When the graduation line denotes the volume contained in the calibrated vessel, the ware is marked “TC”. When the graduation line indicates the volume delivered from the vessel, the ware is marked “TD”.
Explanation:
2H2(g) + O2(g) → 2H2O(l ) Chemical equations give the following information about chemical reactions. Chemical equations show the formulas for the substances that take part in the reaction.
The final temperature : T₂ = 680.625 K
<h3>Further explanation</h3>
Given
V₁=240 ml
T₁ = 90 + 273 = 363 K
V₂ = 450 ml
Required
The final temperature
Solution
Charles's Law
When the gas pressure is kept constant, the gas volume is proportional to the temperature

Input the value :
T₂ = V₂T₁/V₁
T₂ = 450 x 363 / 240
T₂ = 680.625 K
7.2 / 2.4 = 3 half-lives
(1/2)3 = 0.125 (the amount remaining after 3 half-lives)
100.0 g x 0.125 = 12.5 g remaining
For the following reaction, 76.0 grams of barium chloride are allowed to react with 67.0 grams of potassium sulfate.
The reaction consumes _____ moles of barium chloride. The reaction produces _____ moles of barium sulfate and _____ moles of potassium chloride.
Answer: a) The reaction consumes 0.365 moles of barium chloride.
b) The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.
Explanation:
To calculate the moles :


According to stoichiometry :
1 mole of
require 1 mole of 
Thus 0.365 moles of
will require=
of 
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 1 moles of
give = 1 moles of 
Thus 0.365 moles of
give =
of 
As 1 moles of
give = 2 moles of 
Thus 0.365 moles of
give =
of 
Thus the reaction consumes 0.365 moles of barium chloride. The reaction produces 0.365 moles of barium sulfate and 0.730 moles of potassium chloride.