Answer:
7.08
Explanation:
To solve this problem we'll use the <em>Henderson-Hasselbach equation</em>:
- pH = pka + log
![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Where is the ratio of [sodium formate]/[formic acid] and pka is equal to -log(Ka), meaning that:
- pka = -log (1.8x10⁻⁴) = 3.74
We<u> input the data</u>:
- 4.59 = 3.74 + log
And<u> solve for </u>:
- 0.85 = log
=
- = 7.08
The question is incomplete, the complete question is;
The student collects the H2(g) produced by the reaction and measures its volume over water at 298 K after carefully equalizing the water levels inside and outside the gas-collection tube, as shown in the diagram below. The volume is measured to be 45.6mL . The atmospheric pressure in the lab is measured as 765 torr , and the equilibrium vapor pressure of water at 298 K is 24 torr .(i) The pressure inside the tube due to the H2(g)
Answer:
741 torr
Explanation:
From the question we can see that the atmospheric pressure in the lab is 765 torr.
The vapour pressure of water = 24 torr
From Dalton's law of partial pressure, we know that;
Total pressure = Pressure of the H2 + Partial pressure of water vapour
Therefore;
Pressure of H2 = Total pressure - Partial pressure of water vapour
Pressure of H2 = 765 torr - 24 torr = 741 torr
Metalloids show both metal and non metal properties
Let's begin with the basic values that will be used in the solution.
The formula of propane is C3H8. It is an alkane, a hydrocarbon with the general formula of CnH2n+2. Notice that hydrocarbons have only Carbon and Hydrogen atoms. Its molar mass (M) is 44 g.
Molar Mass Calculation is done as like that
C=12 g/mol, H=1 g/mol. 1 mole propane has 3 moles Carbon atoms and 8 mole Hydrogen atoms. M(C3H8)= 3*12+ 8*1= 44 g
Combustion reaction of hydrocarbons gives carbon dioxide and water by releasing energy. That energy is called as enthalpy of combustion (ΔHc°).
ΔHc° of propane equals -2202.0 kj/mol. Burning of 1 mole C3H8 releases 2202 kj energy. Minus sign only indicates that the energy is given out ( an exothermic reaction ).
Let's write the combustion reaction.
C3H8 + O2 ---> CO2 + H20 (unbalanced) ΔHc° = -2202 kj/mol
Now, we calculate mole of 20 kg propane. Convert kilogram into gram since we use molar mass is defined in grams.
mole=mass/molar mass ; n=m/M ; n= 20000 g /44 (g/mol)=454 mole
1 mole propane releases 2202 kj energy.
454 mole propane release 2202 kj *454= 1000909 kj
The answer is 1000909 kj.
The solution would be like
this for this specific problem:
<span><span>
E</span>=</span><span>mc</span>ΔT<span> <span>
= (</span>15<span> g</span><span>)(</span>1.91<span> <span>J<span><span>g∘</span>C</span></span>)(</span>25<span><span> ∘</span>C</span>−15<span><span> ∘</span>C</span><span>)
</span></span>= 28.65 * 10
= 286.5
<span>
I hope this helps and if you have any further questions, please don’t hesitate
to ask again. </span>
