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slega [8]
3 years ago
13

Aqueous lithium sulfate was mixed with aqueous strontium chlorate, and a crystallized strontium sulfate product was formed. cons

ider the other product and its phase, and then write the balanced molecular equation for this precipitation reaction
Chemistry
1 answer:
Dmitriy789 [7]3 years ago
4 0

Lithium sulfate is Li2SO4

Strontium chlorate is Sr(ClO3)2

Strontium sulfate is SrSO4

 

So the complete balanced chemical reaction for this is:

 

Li2SO4 (aq) +  Sr(ClO3)2 (aq) -->  SrSO4 (s)  +  2 LiClO3 (aq)

 

 

This is a type of double replacement reaction since there is an exchange of ions.

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How many grams of Sg is required to produce 83.10 g SF6? S: +24F-->8SF
ozzi

Answer : The mass of S_8 required is 18.238 grams.

Explanation : Given,

Mass of SF_6 = 83.10 g

Molar mass of SF_6 = 146 g/mole

Molar mass of S_8 = 256.52 g/mole

The balanced chemical reaction is,

S_8+24F_2\rightarrow 8SF_6

First we have to determine the moles of SF_6.

\text{Moles of }SF_6=\frac{\text{Mass of }SF_6}{\text{Molar mass of }SF_6}=\frac{83.10g}{146g/mole}=0.569moles

Now we have to determine the moles of S_8.

From the balanced chemical reaction we conclude that,

As, 8 moles of SF_6 produced from 1 mole of S_8

So, 0.569 moles of SF_6 produced from \frac{0.569}{8}=0.0711 mole of S_8

Now we have to determine the mass of S_8.

\text{Mass of }S_8=\text{Moles of }S_8\times \text{Molar mass of }S_8

\text{Mass of }S_8=(0.0711mole)\times (256.52g/mole)=18.238g

Therefore, the mass of S_8 required is 18.238 grams.

7 0
3 years ago
A balloon is filled with 0.250 mole of air at 35°C. If the volume of the balloon is 6.23 liters, what is the absolute pressure o
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What is the pH if the [H+] concentration is 3 x10^-13​
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Answer:

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Answer:Reaccionan 54gr de Nitrato de plata al 39 % de pureza, con 72gr de ácido clorhídrico al 83% de pureza, en un proceso donde se obtienen 93 gr de cloruro de - 154… ... 93 gr de cloruro de plata. El otro producto es el ácido nítrico: Calcular el porcentaje de rendimiento de la reacción y balancearlo.

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