Answer:
<em>The</em><em> </em><em>Typical</em><em> </em><em>Oxidation</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>Oxygen</em><em> </em><em>is</em><em> </em><em>–</em><em>2</em><em>.</em>
<em>And</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>question</em><em> </em><em>said</em><em>.</em><em>.</em><em>.</em><em> </em><em>its</em><em> </em><em>Oxidation</em><em> </em><em>changes</em><em> </em><em>to</em><em> </em><em>–</em><em>1</em><em> </em><em>when</em><em> </em><em>it</em><em> </em><em>is</em><em> </em><em>in</em><em> </em><em>Peroxides</em><em> </em><em>and</em><em> </em><em>–</em><em>½</em><em> </em><em>w</em><em>h</em><em>e</em><em>n</em><em> </em><em>i</em><em>t</em><em>s</em><em> </em><em>i</em><em>n</em><em> </em><em>SuperOxides</em><em>.</em>
<em>Correct</em><em> </em><em>Answer</em><em> </em><em>:</em><em> </em><em>Option</em><em> </em><em>D</em><em>.</em>
Answer:
1. 9.57 × 10^-9 moles.
2. 7.38mol
Explanation:
1.) To find the number of moles there are in the number of particles in an atom, we divide the number of particles (nA) by Avagadro's constant (6.02 × 10^23)
Hence, to find the number of moles (n) of Manganese (Mn), we say:
5.76 x 10^15 atoms ÷ 6.02 × 10^23
5.76/6.02 × 10^(15-23)
= 0.957 × 10^-8
= 9.57 × 10^-9 moles.
2.) Mole = mass/molar mass
Molar mass of sodium chloride (NaCl) = 23 + 35.5
= 58.5g/mol
mole = 431.6 g ÷ 58.5g/mol
mole = 7.38mol
D: The atomic mass number
Answer:
2 a
element pure form cannot broken down
3 b
Chemical change chemical R×N
(R×N = reaction)
4c homogeneous solution
Yield = Yield (reaction yield): A measure of a chemical reaction's efficiency, as a ratio of moles of product to moles of reactant. Usually expressed as a percentage. % Yield = Moles of product.
Explanation:
We can write the balanced equation for the synthesis reaction as
H2(g) + Cl2(g) → 2HCl(g)
We use the molar masses of hydrogen chloride gas HCl and hydrogen gas H2 to calculate for the mass of hydrogen gas H2 needed:
mass of H2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol H2 / 2 mol HCl) *
(2.02 g H2 / 1 mol H2)
= 4.056 g H2
We also use the molar masses of hydrogen chloride gas HCl and chlorine gas CL2 to calculate for the mass of hydrogen gas H2:
mass of CL2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol Cl2 / 2 mol HCl) *
(70.91 g Cl2 / 1 mol Cl2)
= 142.4 g Cl2
Therefore, we need 4.056 grams of hydrogen gas and 142.4 grams of chlorine gas to produce 146.4 grams of hydrogen chloride gas.
