The relation between the volume and the temperature of the gas is given by Charles's law. The final temperature of the gas at 0.75 liters is -193.8°C.
<h3>What is Charles's law?</h3>
Charles's law was derived from the ideal gas equation and is used to state the relationship between the temperature and the volume of the gas. With a decrease in volume the temperature decreases.
If the pressure is kept constant then with an increase in temperature the volume of the gas expands. The law is given as,
V₁ ÷ T₁ = V₂ ÷ T₂
Given,
Initial volume (V₁) = 2.80 L
Initial temperature (T₁) = 23 °C = 296.15 K
Final volume (V₂) = 0.75 L
Final temperature = T₂
Substituting the values above as:
T₂ = (V₂ × T₁) ÷ V₁
= 0.75 × 296.15 ÷ 2.80
= 79.325 K
Kelvin is converted as, 79.325K − 273.15 = -193.8°C
Therefore, the final temperature is -193.8°C.
Learn more about Charle's law, here:
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2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
Answer:
All elements are neutral
Explanation:
The plum pudding model of the atom indicated that all elements are neutral. The model was proposed by J.J Thomson after he conducted his experiment on the gas discharge tube.
From the experiment he discovered electrons which he called cathode rays.
Therefore, he suggested the plum pudding model of the atom.
The model describes negatively charge sphere surrounded by positive charges to balance them.
Water would have a much lower boiling point much like its other hydrides and it would loss its ability to dissolve polar substances plus it couldn't form water columns so no more cohesion between water molecules
Answer:
226.8 mg of mupirocin powder are required
Explanation:
Given that;
weight of standard pack = 22 g
mupirocin by weight = 2%
so weight of mupirocin = 2% × 22 = 2/100 × 22 = 0.44 g
so by adding the needed quantity of mupirocin powder to prepare a 3% w/w mupirocin ointment
mg of mupirocin powder are required = ?, lets rep this with x
Total weight of ointment = 22 + x g
Amount of mupirocin = 0.44 + x g
percentage of mupirocin in ointment is 3?
so
3/100 = 0.44 + x g / 22 + x g
3( 22 + x g ) = 100( 0.44 + x g )
66 + 3x g = 44 + 100x g
66 - 44 = 100x g - 3x g
97 x g = 22
x g = 22 / 97
x g = 0.2268 g
we know that; 1 gram = 1000 Milligram
so 0.2268 g = x mg
x mg = 0.2268 × 1000
x mg = 226.8 mg
Therefore, 226.8 mg of mupirocin powder are required
