The pressure of the gas is obtained as 48 atm.
<h3>What is the total pressure?</h3>
Now we know that;
Number of moles of CH4 = 48.0 grams /16 g/mol = 3 moles
Number of moles of H2 = 56.0 grams/2 g/mol = 28 moles
Total number of moles present = 3 moles + 28 moles = 31 moles
Using;
PV =nRT
P = total pressure
V = total volume
n = total number of moles
R = gas constant
T = temperature
P = nRT/V
P = 31 * 0.082 * 286/15
P = 48 atm
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I could not break unless you hit it with a sledge hammer
Answer:
Q = 1455.12 Joules.
Explanation:
Given the following data;
Mass = 300 grams
Initial temperature = 22.3
Final temperature = 59.9°C
Specific heat capacity = 0.129 J/gºC.
To find the quantity of energy;
Where,
Q represents the heat capacity.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
dt represents the change in temperature.
dt = T2 - T1
dt = 59.9 - 22.3
dt = 37.6°C
Substituting the values into the equation, we have;
Q = 1455.12 Joules.
Answer:
pH = 6.999
The solution is acidic.
Explanation:
HBr is a strong acid, a very strong one.
In water, this acid is totally dissociated.
HBr + H₂O → H₃O⁺ + Br⁻
We can think pH, as - log 7.75×10⁻¹² but this is 11.1
acid pH can't never be higher than 7.
We apply the charge balance:
[H⁺] = [Br⁻] + [OH⁻]
All the protons come from the bromide and the OH⁻ that come from water.
We can also think [OH⁻] = Kw / [H⁺] so:
[H⁺] = [Br⁻] + Kw / [H⁺]
Now, our unknown is [H⁺]
[H⁺] = 7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]
[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) / [H⁺]
This is quadratic equation: [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴
a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴
(-b +- √(b² - 4ac) / (2a)
[H⁺] = 1.000038751×10⁻⁷
- log [H⁺] = pH → 6.999
A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.
Answer:
70.0°C
Explanation:
We are given;
- Amount of heat generated by propane as 104.6 kJ or 104600 Joules
- Mass of water is 500 g
- Initial temperature as 20.0 ° C
We are required to determine the final temperature of water;
Taking the initial temperature is x°C
We know that the specific heat of water is 4.18 J/g°C
Quantity of heat = Mass × specific heat × change in temperature
In this case;
Change in temp =(x-20)° C
Therefore;
104600 J = 500 g × 4.18 J/g°C × (x-20)
104600 J = 2090x -41800
146400 = 2090 x
x = 70.0479
=70.0 °C
Thus, the final temperature of water is 70.0°C
