Answer:
Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature.
Explanation:
The expression for Charles's Law is shown below:
This states that the volume of the gas is directly proportional to the absolute temperature keeping the pressure conditions and the moles of the gas constant.
<u>Thus, when the volume of the gas is exposed to a temperature above -273.15 K, the volume increases linearly with the temperature. </u>
<u>For example , if the temperature of the gas is reduced to half, the volume also reduced to half. </u>
<u>At -273.15 K, according to Charles's law, it is possible to make the volume of an ideal gas = 0.</u>
Answer:
The energy required is 3225 Joules.
Explanation:
Given,
mass of lead cube = 500 grams
T₁ = 25°C
T₂ = 75°C
specific heat of lead = 0.129 J/g°C
Energy required to heat the lead can be found by using the formula,
Q = (mass) (ΔT) (Cp)
Here, ΔT = T₂ - T₁ = 75 - 25 = 50
Substituting the values,
Q = (500)(50)(0.129)
Q = 3225 Joules.
Therefore, energy required is 3225 J.
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Answer:
0.036 M
Explanation:
To do this, let's mark the dye as D and bleach as B.
We have the concentrations of both, and we already know that they react in a 1:1 mole ratio. The total volume of reaction is 9 + 1 = 10 mL or 0.010 L, and we hava both concentrations.
The problem already states that the dye reacts completely, so this is the limiting reagent, while bleach is the excess.
To know the remaining amount of bleach, we need to do this with the moles. First, let's calculate the initial moles of D and B:
moles D = 3.4x10⁻⁵ * 0.009 = 3.06x10⁻⁷ moles
moles B = 0.36 * 0.001 = 3.6x10⁻⁴ moles
Now that we have the moles, and that we know that all the dye reacts completely, let's see how many moles of bleach are left:
moles of B remaining = 3.6x10⁻⁴ - 3.06x10⁻⁷ = 3.597x10⁻⁴ moles
These are the moles presents of B after the reaction has been made. The concentration of the same will be:
[B] = 3.597x10⁻⁴ / 0.010
[B] = 0.0357
With 2 SF it would be:
[B] = 3.6x10⁻² M
Answer:- Volume of the gas in the flask after the reaction is 156.0 L.
Solution:- The balanced equation for the combustion of ethane is:
From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.
Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:
= 126 L 
126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.
let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:
= 60.0 L 
Similarly, let's calculate the volume of water vapors formed:
= 90.0 L 
Since ethane is present in excess, the remaining volume of it would also be present in the flask.
Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:
= 30.0 L 
Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L
Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L
Hence. the answer is 156.0 L.
