The reaction Pb(NO3)2(aq) + NiCl2(aq)---------> PbCl2(s) + Ni(NO3)2(aq) is a precipitation reaction.
A chemical reaction is said to occur when two or more substances are combined to produce new substances. Chemical reactions are classified based on the kind of change taking place in the reaction.
In the reaction;
Pb(NO3)2(aq) + NiCl2(aq)---------> PbCl2(s) + Ni(NO3)2(aq)
We can see that a solid is formed when lead II nitrate reacts with nickel II chloride. This is a precipitation reaction.
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Answer:
= 3.78 g H₂O
Explanation:
2C₂H₆ + 3O₂ => 4CO₂ + 6H₂O
2.1g C₂H₆ = 2.1g/30.0 g/mol = 0.07 mole ethane
3.68g O₂ = 3.68g/32 g/mol = 0.115 mole oxygen
Limiting Reactant:
A quick way to determine limiting reactant is to divide moles of reactant by its respective coefficient in the balanced molecular equation. The smaller value is the limiting reactant.
moles ethane = 0.07 mole / 2 (the coefficient in balanced equation) = 0.035
moles oxygen = 0.115 mole / 3 (the coefficient in balanced equation) = 0.038
Since the smaller value is associated with ethane, then ethane is the limiting reactant and the problem is worked from the 0.07 moles of ethane in an excess of O₂.
From the equation stoichiometry ...
2 moles C₂H₆ in an excess of O₂ => 6 moles H₂O
then 0.07 mole C₂H₆ in an excess of O₂ => 6/2(0.07 moles H₂O = 0.21 mole
Converting to grams of water produced
= 0.21 mole H₂O X 18 g/mol = 3.78 g H₂O
1cm^3 = 1L would be the correct answer. One cubic centimeter equals .001 liter, so this equality above is not correct.
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Answer: B. Metal
Explanation: The catalyst used in the converter is mostly a precious metal such as platinum, palladium and rhodium Hope This Helped (=^･-･^=)
We are tasked to find the amount of O2 in grams given only the number of moles of oxygen gas. To solve the problem, we need first to calculate the molecular weight of oxygen. Based on the periodic table, elemental oxygen has a molecular weight of 16 g/mol. Thus its molecular weight is,
O2=16.0g/mol (2)=32 g/mol
To solve for the amount of compound oxygen in grams, we have,
O2 (g)=5 mol x 32 g/mol =160 g.
This cancel out the mols both in the numerator and denominator leaving only the g as a unit. Therefore, 5 mols of oxygen is equal to 160 g of oxygen.
