From the reduction standard potentials;
The emf of Zinc = -0.76 V
and the emf of Aluminium = -1.66 V
In a galvanic cell the component with lower standard reduction potential gets oxidized and that it is added to the anode compartment.
Therefore. the voltage of a galvanic cell made with zinc and aluminium will be;
Voltage =Ered- Eoxd
= -0.76 - (-1.66)
= 0.9 V
DE = dH - PdV
<span>2 H2O(g) → 2 H2(g) + O2(g) </span>
<span>You can see that there are 2 moles of gas in the reactants and 3 moles of gas in the products. </span>
<span>1 moles of ideal gas occupies the same volume as 1 mole of any other ideal gas under the same conditions of temp and pressure. </span>
<span>Since it is done under constant temp and pressure that means the volume change will be equal to the volume of 1 mole of gas </span>
<span>2 moles reacts to form 3 moles </span>
<span>The gas equation is </span>
<span>PV = nRT </span>
<span>P = pressure </span>
<span>V = volume (unknown) </span>
<span>n = moles (1) </span>
<span>R = gas constant = 8.314 J K^-1 mol^-1 </span>
<span>- the gas constant is different for different units of temp and pressure (see wikki link) in this case temp and pressure are constant, and we want to put the result in an equation that has Joules in it, so we select 8.314 JK^-1mol^-1) </span>
<span>T = temp in Kelvin (kelvin = deg C + 273.15 </span>
<span>So T = 403.15 K </span>
<span>Now, you can see that PV is on one side of the equation, and we are looking to put PdV in our dE equation. So we can say </span>
<span>dE = dH -dnRT (because PV = nRT) </span>
<span>Also, since the gas constant is in the unit of Joules, we need to convert dH to Joules </span>
<span>dH = 483.6 kJ/mol = 483600 Joules/mol </span>
<span>dE = 483600 J/mol - (1.0 mol x 8.314 J mol^-1K-1 x 403.15 K) </span>
<span>dE = 483600 J/mol - 3351.77 J </span>
<span>dE = 480248.23 J/mol </span>
<span>dE = 480.2 kJ/mol </span>
Explanation:
The molarity of a solution is defined like the number of moles of solute per liters of solution.
molarity = moles of solute/(volume of solution in L)
We know the volume of solution in L.
volume of solution = 0.65 L
To go from the mass of our solute in grams to moles we have to use its molar mass.
mass of NaCl = 63 g
molar mass of NaCl = 58.44 g/mol
moles of NaCl = 63 g * 1 mol/(58.44 g)
moles of NaCl = 1.078 moles
Finally we can find the molarity of the solution
molarity = moles of NaCl/(volume of solution)
molarity = 1.078 moles/(0.65 L)
molarity = 1.66 M
Answer: the molarity of the solution is 1.66 M.
Answer:
KOH contains only one K, so a mole of KOH contains one mole of K. How many atoms of K are in 1 mole of KOH? There are 6.022 × 1023 atoms of potassium in every mole of potassium. Since one mole of KOH contains one mole of K, the answer is 6.022×1023 atoms of K.
Explanation:
Answer:
4.3043x10^24
Explanation:
Atoms(H) = (3.575 mol H2S/1)(2 mol H/1mol H2S)(6.02x10^23/1molH)
Simple stoicheometry problem. Sorry for taking so long, took Chem sophomore year, its been a while
