The reaction uses B) 9.0 g Br₂.
iron + bromine ⟶ product
2.0 g + <em>x</em> g ⟶ 11.0 g
According to the <em>Law of Conservation of Mass</em>, the total mass of the reactants must equal the total mass of the products.
∴2.0 + <em>x</em> = 11.0
<em>x</em> = 11.0 – 2.0 = 9.0
The reaction uses 9.0 g Br₂.
Answer:
5.83 mol.
Explanation:
- From the balanced reaction:
<em>2Al + 3Ag₂S → 6Ag + Al₂S₃,</em>
It is clear that 2 mol of Al react with 3 mol of Ag₂S to produce 1 mol of Ag and 1 mol of Al₂S₃.
Al reacts with Ag₂S with (2: 3) molar ratio.
<em>So, 2.27 mol of Al reacts completely with 3.4 mol of Ag₂S with (2: 3) molar ratio.</em>
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- The limiting reactant is Ag₂S.
- The excess "left over" reactant is Al.
The reamining moles of excess reactant "Al" = 8.1 mol - 2.27 mol = 5.83 mol.
Answer:
1. NaN₃(s) → Na(s) + 1.5 N₂(g)
2. 79.3g
Explanation:
<em>1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid sodium azide (NaN₃) into solid sodium and gaseous dinitrogen.</em>
NaN₃(s) → Na(s) + 1.5 N₂(g)
<em>2. Suppose 43.0L of dinitrogen gas are produced by this reaction, at a temperature of 13.0°C and pressure of exactly 1atm. Calculate the mass of sodium azide that must have reacted. Round your answer to 3 significant digits.</em>
First, we have to calculate the moles of N₂ from the ideal gas equation.
The moles of NaN₃ are:
The molar mass of NaN₃ is 65.01 g/mol. The mass of NaN₃ is:
Explanation:
1.
Cu(NO3)2 + 2NaCl(aq) --> CuCl2(aq) + 2NaNO3(aq)
2.
Cu(NO3)2 + 2NaOH(aq) --> Cu(OH)2(s) + 2NaNO3(aq)
A light blue precipitate of Cu(OH)2 is formed and NaNO3 in solution.
3.
Cu(NO3)2(aq) --> Cu2+(aq) + 2NO3^-2(aq)
2NaOH(aq) --> 2Na+(aq) + 2OH-(aq)
Cu2+(aq) + 2OH-(aq) --> Cu(OH)2(aq)
2Na+(aq) + 2NO3^-2(aq) --> 2NaNO3(aq)
4.
The reaction in both Questions 1 and 2 is called Double displacement reaction. A double-replacement reaction exchanges the cations and/or or the anions of two ionic compounds. A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate (precipitated) while the other in solution.
Since the cation and anions in Qustion 1 were exchanged, the same was done for Question 2, hence the identity of the precipitate in Question 2 was got.
Answer: i think it is Habitat but i am not 100%
Explanation:
