M(NiS₂) = 11.2 g.
n(NiS₂) = m(NiS₂) ÷ M(NiS₂).
n(NiS₂) = 11.2 g ÷ 122.8 g/mol.
n(NiS₂) = 0.091 mol.
m(O₂) = 5.43 g.
n(O₂) = 5.43 g ÷ 32 g/mol.
n(O₂) = 0.17 mol; limiting reactant.
From chemical reaction: n(NiS₂) : n(O₂) = 2 : 5.
0.091 mol : n(O₂) = 2 : 5.
n(O₂) = 0.2275 mol, not enough.
n(NiO) = 4.89 g .
n(O₂) : n(NiS) = 5 : 2.
n(NiS) = 0.068 mol.
m(NiS) = 0.068 mol · 74.7 g/mol = 5.08 g.
percent yield = 4.89 g / 5.08 g · 100% = 96.2%.
Answer:
Micky Mo is suffering from respiratory acidosis.
Explanation:
The pCO2 level in micky"s body is higher than normal it means the excess amount of CO2 will reacts with water to generate carbonic acid(H2CO3).
On the other hand according to the question total HCO3- also higher than normal.As a result the excess HCO3- will react with proton to form carbonic acid which is in turn dissociate to generate CO2 and H2O to maintain normal acid base homeostasis.
From that point of view it can be said Micky Mo is suffering from respiratory acidosis.
Answer is: <span>unbalanced electronegativity of the hydrogens and oxygens as they share electrons.
Oxygen has greater electronegativity than hydrogen, because of that oxygen is partially negative and hydrogen is partially positive.
</span>Electronegativity<span> is a </span>chemical property<span> that describes the tendency of an </span>atom<span> to attract a shared pair of </span>electrons<span> towards itself.</span>
First, multiply the mass by the molar mass of neon to find out how many moles of neon there are. Then, multiply by 22.4 to find out how many liters there are.
6.745g Ne x 1 mole Ne/20 g Ne x 22.4 L/1 mole Ne = 7.5544 L
So solve
this we must know the amount of sugar per gram of coffee
<span>i.
</span>10 g sugar /100 g coffee = 0.1 g sugar/ 1 g coffee
<span>ii.
</span>10 g sugar / 200 g coffee = 0.05 g sugar / 1 g coffee
<span>iii.
</span>4 g sugar / 200 g coffee = 0.02 g sugar / 1 g coffee
<span>iv.
</span>4 g sugar / 100 g coffee = 0.04 g sugar / 1 g coffee
So the ranking from sweetest so the least is:
<span> i, ii, iv, iii</span>
