Answer:
See explanation
Explanation:
1 mole of a gas occupies 22.4 L
x moles occupies 16.8 L
x = 1 mole * 16.8 L/22.4 L
x = 0.75 moles
number of moles = mass/molar mass
mass = number of moles * molar mass
mass = 0.75 moles * 30.01 g/mol = 22.5075 g = 2.25 * 10^1 g
the coefficient of the scientific notation answer = 2.25
the exponent of the scientific notation answer = 1
significant figures are there in the answer = 6
the right most significant figure in the answer = 3
2.
number of moles = 12.5g/38g/mol = 0.3289 moles
1 mole occupies 22.4 L
0.3289 moles occupies 0.3289 moles * 22.4 L/1 mole
= 7.36736 L = 7.36736 * 10^0 L= 7.37 * 10^0 L
the coefficient of the scientific notation answer =7.37
the exponent of the scientific notation answer = 0
significant figures are there in the answer = 6
the right most significant figure in the answer= 3
Decomposed organic matter ,mud,silt ,and sand
Empirical formula is calculated as follows
calculate the moles of each element, that is % composition/ molar mass
molar masses ( Si= 28.09g/mol , Cl= 35.5 g/mol, I=126.9 g/mol)
moles of silicon = 7.962/28.09g/mol= 0.283 moles
moles of chlorine = 20.10 / 35.5g/mol = 0.566 moles
moles of iodine= 71.94 / 126.9 g/mol= 0.567 moles
divide each mole with smallest mole (0.283)
that is silicon = 0.283/0.283= 1 mole
chlorine = 0.566/0.283= 2 mole
Iodine= o.567/0.283= 2 moles
empirical formula is therefore= SiCl2I2
Since the half-reaction is occurring in a basic solution, add 32OH− to each side of the equation to eliminate the H+ ions.
P₄ +16H₂O + 32OH⁻ ⟶ 4PO₃⁻⁴ + 32H⁺ +32OH⁻
Final reaction :
P₄ + 32OH⁻ ⟶ 4PO₃⁻⁴ + 16H₂O + 20e⁻
A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.
The concept of half-reactions is used to describe what occurs in an electrochemical cell, such as a Galvanic cell battery. Half-reactions can be written to describe both the metal undergoing oxidation (known as the anode) and the metal undergoing reduction (known as the cathode).
Half-reactions are often used as a method of balancing redox reactions. For oxidation-reduction reactions in acidic conditions, after balancing the atoms and oxidation numbers, one will need to add H+ ions to balance the hydrogen ions in the half reaction.
For oxidation-reduction reactions in basic conditions, after balancing the atoms and oxidation numbers, first treat it as an acidic solution and then add OH- ions to balance the H+ ions in the half reactions (which would give H2O).
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