Moles = mass / molar mass
<span>moles P = 0.422 g / 30.97 g/mol = 0.01363 mol </span>
<span>moles O = (0.967 g - 0.422g) / 16.00 g/mol = 0.03406 moles </span>
<span>So ratio moles P : moles O </span>
<span>= 0.01363 mol : 0.03406 mol </span>
<span>Divide each number in the ratio by the smallest number </span>
<span>(0.01363 / 0.01363) : (0.03406 / 0.01363) </span>
<span>= 1 : 2.5 </span>
<span>The empirical formula needs to be the smallest whole number ratio of atoms in the molecules. Since you have a non-whole number, multiply the ratio by the smallest number needed to make both number whole numbers. In this case x 2 </span>
<span>2 x (1 : 2.5) </span>
<span>= 2 : 5 </span>
The average atomic mass formula is:
Substituting the values in the formula:
Hence, 
is the average atomic mass of element M.
Explanation:
The given data is as follows.
n = 2 mol, P = 1 atm, T = 300 K
Q = +34166 J, W= -1216 J (work done against surrounding)
= 
Relation between internal energy, work and heat is as follows.
Change in internal energy (
) = Q + W
= [34166 + (-1216)] J
= 32950 J
Also, 
= 
32950 J = 
1321.06 K + 300 K = 
= 1621.06 K
Thus, we can conclude that the final temperature of the gas is 1621.06 K.
Answer:
HCl is the limiting reactant. It will completely be consumed (1.37 moles)
Option D is correct
Explanation:
Step 1: Data given
Mass of Zinc (Zn) = 50.0 grams
Mass of Hydrogen chloride (HCl) = 50.0 grams
atomic mass Zn = 65.38 g/mol
Molar mass HCl = 36.46 g/mol
Step 2: The balanced equation
Zn + 2HCl → ZnCl2 + H2
Step 3: Calculate moles
Moles = mass / molar mass
Moles Zn = 50.0 grams / 65.38 g/mol
Moles Zn = 0.764 moles
Moles HCl = 50.0 grams / 36.46 g/mol
Moles HCl = 1.37 moles
Step 4: Calculate limiting reactant
For 1 mol Zn we need 2 moles HCl to produce 1 mol ZnCl2 and 1 mol H2
HCl is the limiting reactant. It will completely be consumed (1.37 moles)
Zn is in excess. There will react 1.37/2 = 0.685 moles
There will remain 0.764 -0.685 = 0.079 moles
In a galvanic cell, the flow of electrons will be from the anode to cathode through the circuit .
Whether a cell is an electrolysis cell (non-spontaneous chemistry driven by forcing electricity from an external energy source) or a galvanic cell (spontaneous chemistry driving electricity), will determine the charge of the anode and the cathode. Depending on where the electrons encounter resistance and find it difficult to pass, a negative charge may emerge. Therefore, you cannot determine the direction of the current just on the charge on the electrode.
Oxidation and reduction always take place at the anode and cathode, respectively.
An element undergoes oxidation when it surrenders one or more electrons to become more positively charged. These electrons leave the chemicals in any type of cell and travel to the anode, where they enter the external circuit.
An element picks up an electron during reduction to become more negatively charged (less positive, lower oxidation state). These electrons are captured from the external circuit at the cathode in both types of cells.
Therefore, no matter what kind of cell you are dealing with, the oxidizing chemicals at the anode transfer the electrons to the external circuit; these electrons then move through the circuit from the anode to the cathode, where they are captured by the reducing chemicals. The electrons always go from the anode to the cathode via the external circuit.
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