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3 years ago

Why are state symbols used in chemical equations

Chemistry

2 answers:

DedPeter [7]3 years ago

4 0

Answer:

state symbol meaning

[s] solid

liquid

gas

aq] aqueous dissolved in water

Explanation:

often chemical equations are written showing the state that each substance is in. the [s] sign means that the compound is a sold. the l sign means the substance is a liquid. the aq sign stands for aqueous in water and means the compound is dissolved in water

thank me later.

galina1969 [7]3 years ago

4 0

State symbols are used in chemical equations bc they identify what phase the substances are in.

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Sedbober [7]

Answer:

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7 0

3 years ago

Determine whether or not baco3 will be more soluble in acidic solution than in pure water.

Volgvan

Barium carbonate (BaCO₃) <span>will be more soluble in acidic solution than in pure water, because Ksp (solubility constant) in water for this salt is very low.
In acid (for example hydrochloric acid) barium carbonate dissolves more because it forms weak electrolyte carbonic acid:
BaCO</span>₃(s) + 2HCl(aq) → BaCl₂(aq) + H₂CO₃(aq).

4 0

3 years ago

This question involves two calculations. The answer to the first part will be

ozzi

Answer :

(a) The mass of $Al_2O_3$ produced is, 15.2 grams.

(b) The percent yield of the reaction is, 72.5 %

Explanation :

Part (a) :

Given,

Mass of $Al$ = 85.1 g

Molar mass of $Al$ = 27 g/mol

First we have to calculate the moles of $Al$

$\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{85.1g}{27g/mol}=3.15mol$

Now we have to calculate the moles of $Al_2O_3$

The balanced chemical equation is:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the reaction, we conclude that

As, 4 moles of $Al$ react to give 2 moles of $Al_2O_3$

So, 3.15 moles of $Al$ react to give $\frac{2}{4}\times 3.15=1.58$ mole of $Al_2O_3$

Now we have to calculate the mass of $Al_2O_3$

$\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3$

Molar mass of $Al_2O_3$ = 102 g/mole

$\text{ Mass of }Al_2O_3=(1.58moles)\times (102g/mole)=161.2g$

Therefore, the mass of $Al_2O_3$ produced is, 161.2 grams.

Part (b) :

Now we have to calculate the percent yield of the reaction.

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 116.9 g

Theoretical yield = 161.2 g

Now put all the given values in this formula, we get:

$\text{Percent yield}=\frac{116.9g}{161.2g}\times 100=72.5\%$

Therefore, the percent yield of the reaction is, 72.5 %

8 0

3 years ago

2+2= ? 3+3=? 2x2? 5867 cats + 800 dogs?

sukhopar [10]

Answer:

6667 animals

Explanation:

4 0

2 years ago

A 90.00 mL solution of NaCl and ethyl alcohol has 7.83 g of salt dissolved in the alcohol. Calculate the molar concentration of

nalin [4]

The molar concentration of the solution is calculated as follows

find the moles of NaCl used to make the solution

moles = mass/molar mass
mass =7.83 g
molar mass =58.5 g/mol
moles = 7.83 g/ 58.5 g/mol = 0.134 moles

molarity ( concentration in mol/l) = number of moles/volume in liters

volume in liters = 90ml/1000 =0.09 liters

molarity is therefore = 0.134 moles/0.09 L =1.49 M

8 0

3 years ago

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