The number of moles in each sample will be 0.391 moles, 30.7 moles, 0.456 moles, and 1350 moles
<h3>What is the number of moles?</h3>
The number of moles of a substance is the ratio of the mass of the substance to the molar mass.
In other words; mole = mass/molar mass.
Thus:
- moles of 18.0 g
= 18.0/46
= 0.391 moles
- moles of 1.35 kg
= 1350/44
= 30.7 moles
- moles of 46.1 g
= 46.1/101.1
= 0.456 moles
- moles of 191.8 kg
= 191800/142
= 1350 moles
More on the number of moles of substances can be found here: brainly.com/question/1445383
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The percent yield of the reaction : 89.14%
<h3>Further explanation</h3>
Reaction of Ammonia and Oxygen in a lab :
<em>4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)</em>
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):

mass O₂ = 120 g
mol O₂(MW=32 g/mol) :

Mol ratio of reactants(to find limiting reatants) :

mol of H₂O based on O₂ as limiting reactants :
mol H₂O :

mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :

Answer:
The correct approach is Option B (Peer Review).
Explanation:
- Rather made reference to someone as a scientific peer-review, it encourages the specialist who has not been essential to the study team to analyze the study objectively and pointed out everyone's mistakes. It serves as major self-regulation for scholars and aims to make the publishing process somewhat credible. Hence, the solution to this issue is Peer Examination.
- Funding organizations rarely have the capabilities to recognize out mistakes, whereas definitive analysis is a method of study that helps to make a definitive statement. The gathering of data is simply a process of scientific study.
Other approaches do not apply to the example mentioned. Although the one mentioned is right.
Answer:
% (COOK)2H2O = 37.826 %
Explanation:
mix: (COOK)2H2O + Ca(OH)2 → CaC2O4 + H2O
∴ mass mix = 4.00 g
∴ mass (CaC2O4)H2O = 1.20 g
∴ Mw (COOK)2H2O = 184.24 g/mol
∴ Mw (CaC2O4)H2O = 146.12 g/mol
∴ r = mol (COOK)2H2O / mol (CaC2O4)H2O = 1
- % (COOK)2H2O = (mass (COOK)2H2O / mass Mix) × 100
⇒ mass (COOK)2H2O = (1.20 g (CaC2O4)H2O)×(mol (CaC2O4)H2O / 146.12 g (CaC2O4)H2O)×(mol (COOK)2H2O/mol (CaC2O4)H2O)×(184.24 g (COOK)2H2O/mol (COOK)2H2O)
⇒ mass (COOK)2H2O = 1.513 g
⇒ % (COOK)2H2O = ( 1.513 g / 4 g )×100
⇒ % (COOK)2H2O = 37.826 %