Because the reaction occurs between a metal and non metal
CL=2,8,7
C
Answer:
Explanation:
Given parameters:
Initial temperature T₁ = 25.2°C = 25.2 + 273 = 298.2K
Initial pressure = P₁ = 0.6atm
Final temperature = 72.4°C = 72.4 + 273 = 345.4K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:
where P and T are temperatures, 1 and 2 are initial and final temperatures.
Input the parameters and solve;
P₂ = 0.7atm
The new volume if the balloon is cooled at constant pressure is 3.98 L.
Charles's law, states that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature if the pressure remains constant.
The new volume is calculated using the Charles law formula
V₁ / T₁ = V₂ / T₂
where,
V₁ = The initial volume of air = 4.24 l
T₁ = 23.00 °C into kelvin = 23 +273 =296 K
T₂ = 5.00 °C into kelvin = 5.00 + 273 = 278 K
V₂ = ?
By making V₂ subject the subject of the formula by multiplying both sides by T₂
V₂ = ( V₁ × T₂ ) / T₁
V₂ = (4.24 L × 278 K) / 296 k
= 3.98 L
Therefore, the new volume, if the balloon is cooled at constant pressure, is 3.98 L.
An air-filled balloon will contract when chilled and expand when heated. This occurs because the gas that makes up the air within the balloon expands when it is warm and contracts when it is cool.
Learn more about Charles law here:
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Answer:
2.6 mole of lithium bromide have 225.81 g.
Explanation:
Given data:
Number of moles of lithium bromide = 2.6 mol
Mass of lithium bromide = ?
Solution:
Formula:
<em>Number of moles = mass/ molar mass</em>
Molar mass of lithium bromide = 86.85 g/mol
Mass = number of moles × molar mass
Mass = 2.6 mol ×86.85 g/mol
Mass = 225.81 g
Thus 2.6 mole of lithium bromide have 225.81 g.
KOH added : 7.5 ml
<h3>Further explanation</h3>
Buffer solution of weak acid HCOOH and strong base KOH
Reaction
initial = 100 ml 0.1 M HCOOH = 10 ml mol HCOOH, and x mlmol of KOH
KOH + HCOOH ⇒ COOHK + H₂O
x 10
x x x x
- 10-x x x
[HCOO - ] = 3[HCOOH]
