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3 years ago

What are systematic errors when reacting sodium thiosulfate with hydrochloride acid?

Chemistry

2 answers:

irakobra [83]3 years ago

6 0

Answer:

¿Cuáles son los errores sistemáticos al reaccionar tiosulfato de sodio con ácido clorhídrico?

Explanation:El tiosulfato de sodio, antes denominado hiposulfito sódico, es un compuesto inorgánico cristalino que se encuentra con mayor frecuencia en forma de pentahidrato Na2S2O3·5H2O. Su estructura cristalina es de tipo monoclínica.

El anión tiosulfato se presenta con forma tetraédrica, en la que se produce la sustitución de un átomo de oxígeno por otro de azufre. La distancia entre los átomos de azufre (S-S) indica que entre ellos existe un enlace simple, lo que implica una carga negativa en las interacciones con el oxígeno (S-O) ya que estas presentan un enlace doble.

Zigmanuir [339]3 years ago

6 0

Explanation:

the solution turn cloudy due to the precipitation of sulphure

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Aldehydes may oxidize to form carboxylic acid exception to oxidizing formaldehyde with strong oxidizing agent.

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4 years ago

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Sauron [17]

Answer:

The wood

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3 years ago

118g of o2 gas is held at STP what is the volume of gas

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Answer:

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Explanation:

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4 years ago

What part of the atom do the different 35 and 37 represent?

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4 years ago

The thiocyanate polyatomic ion, SCN-, is commanly called a pseudohalogen because it acts very much like halide ions. For example

harina [27]

Answer:

A) $\frac{1mol(SCN)_{2}}{2molNaSCN}$

B) 0.025 mol (SCN)₂

C) 2 mol (SCN)₂

D) $\frac{1molMnSO_{4}}{2molH_{2}SO_{4}}$

E) 3.5504 mol H₂SO₄

Explanation:

2NaSCN +2H₂SO₄ + MnO₂ → (SCN)₂ + 2H₂O + MnSO₄ + Na₂SO₄

A) A conversion factor could be $\frac{1mol(SCN)_{2}}{2molNaSCN}$ , as it has the units that we want to convert to in the numerator, and the units that we want to convert from in the denominator.

B) 0.05 mol NaSCN * $\frac{1mol(SCN)_{2}}{2molNaSCN}$ = 0.025 mol (SCN)₂

C) With 4 moles of NaSCN and 3 moles of MnSO₄, the reactant is NaSCN so we use that value to calculate the moles of product formed:

4 mol NaSCN * $\frac{1mol(SCN)_{2}}{2molNaSCN}$ = 2 mol (SCN)₂

D) $\frac{1molMnSO_{4}}{2molH_{2}SO_{4}}$

E) 1.7752 mol MnSO₄ * $\frac{2molH_{2}SO_{4}}{1molMnSO_{4}}$ = 3.5504 mol H₂SO₄

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3 years ago