16 grams.
Looking at the reaction equation, it's evident that for each mole of HBr
consumed, 1 mole of CO2 will be produced. So let's calculate the molar mass
of HBr and CO2 first.
Atomic weight hydrogen = 1.00794
Atomic weight bromine = 79.904
Atomic weight carbon = 12.0107
Atomic weight oxygen = 15.999
Molar mass HBr = 1.00794 + 79.904 = 80.91194 g/mol
Molar mass CO2 = 12.0107 + 2*15.999 = 44.0087 g/mol
Moles HBr = 30 g / 80.91194 g/mol = 0.370773461 mol
Grams CO2 = 0.370773461 mol * 44.0087 g/mol = 16.317258 g
Rounding to 2 significant digits gives 16 grams.
Atomic mass unit is the answer
Explanation:
This situation is analogous to a double replacement reaction. A double-replacement reaction occurs when parts of two ionic composites are exchanged, making two new composites. One characteristic of this kind of reaction is that there are two compounds as reactants and two different compounds as products. So, in this case we can relate with it because there are two new couples with the same members that formed the previous couples.
The balanced chemical reaction is given as:
2H2S (g) + 3O2 (g) 2SO2 (g) + 2H2O (g)
We are given the volume at STP of H2S to be used for the reaction. This would be the starting point for the calculations. We do as follows:
45.0 L H2S ( 1 mol / 22.4 L) ( 3 mol O2 / 2 mol H2S ) ( 22.4 L / 1 mol ) = 67.5 L O2 needed
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Answer:
1.09 x 10⁻⁴ M
Explanation:
The equation of the reaction in given by
Ni²⁺ (aq) + 6NH₃ (aq) ⇔ Ni(NH₃)₆
At the beginning o the reaction, we have 0.18M concentration of Ni and 1.2M concentration of aqueous NH₃ and zero concentration of the product
As the reaction proceeds towards equilibrium, the concentration of the reactants decrease as the concentration of the product starts to increase
From the equation,
1 mole of Ni²⁺ reacts with 6 moles of aqueous NH₃ to give 1mole of Ni(NH3)
therefore
0.18 M of Ni would react with 1.08M (6 x 0.18M) aqueous NH₃ to give 0.18M of Ni(NH₃)6
At equilibrium,
1.08M of NH3 would have reacted to form the product leaving
(1.2 - 1.08)M = 0.12M of aqueous NH₃ left as reactant.
Therefore, formation constant K which is the ratio of the concentration of the product to that of the reactant is given by
K = [Ni(NH₃)₆} / [Ni²⁺] 6[NH₃]
5.5 x 10⁸ = 0.18 M / [Ni²⁺] [0.12]⁶
[Ni²⁺]= 0.18 M / (5.5 x 10⁸) (2.986 x 10⁻⁶)
=0.18 M / 0.00001642
= 1.09 x 10⁻⁴ M
[Ni]²⁺ = 1.09 x 10⁻⁴ M
Hence the concentration of Ni²⁺ is 1.09 x 10⁻⁴ M