Answer:
32.5g of sodium carbonate
Explanation:
Reaction of sodium carbonate (Na₂CO₃) with Mg²⁺ and Ca²⁺ as follows:
Na₂CO₃(aq) + Ca²⁺(aq) → CaCO₃(s)
Na₂CO₃(aq) + Mg²⁺(aq) → MgCO₃(s)
<em>1 mole of carbonate reacts per mole of the cations.</em>
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To know the mass of sodium carbonate we must know the moles of carbonate we need to add based on the moles of the cations:
<em>Moles Mg²⁺:</em>
2.91L * (0.0661 moles MgCl₂ / 1L) = 0.192 moles MgCl₂ = Moles Mg²⁺
<em>Moles Ca²⁺:</em>
2.91L * (0.0396mol Ca(NO₃)₂ / 1L) = 0.115 moles Ca(NO₃)₂ = Moles Ca²⁺
That means moles of sodium carbonate you must add are:
0.192 moles + 0.115 moles = 0.307 moles sodium carbonate.
In grams (Using molar mass Na₂CO₃ = 105.99g/mol):
0.307 moles Na₂CO₃ * (105.99g / mol) =
<h3>32.5g of sodium carbonate</h3>
Answer:
nombres uno
number 1
Explanation:
it is what it is according to physicodeferigolical
Answer: nn
Explanation:
The nanometre (international spelling as used by the International Bureau of Weights and Measures; SI symbol: nm) or nanometer (US spelling) is a unit of length in the metric system, equal to one billionth (short scale) of a metre (0.000000001 m).
A whiteout occurs when strong updrafts and downdrafts combine
The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.
<h3>What is empirical formula?</h3>
The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.
<h3>
How to find the empirical formula?</h3>
Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.
Moles phosphorus = 0.903 g phosphorus = 0.0293 mol
Moles bromine 6.99 g bromine=0.0875 mol
The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3
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