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3 years ago

Determine whether or not each chemical formula is an empirical formula. C6H12O6 CaCO3 NaMnO4 Ba3(PO4)2 K2C2O4

Chemistry

1 answer:

zavuch27 [327]3 years ago

5 0

Answer:

- C6H12O6 is not an empirical formula because it can be simplified to CH2O.

- CaCO3 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- NaMnO4 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- Ba3(PO4)2 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- K2C2O4 is not an empirical formula because it can be simplified to KCO2.

Explanation:

Hello!

In this case, since the empirical formulas are referred to those formulas that have been simplified up to the smallest whole number for each subscript per atom, we can see that:

- C6H12O6 is not an empirical formula because it can be simplified to CH2O.

- CaCO3 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- NaMnO4 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- Ba3(PO4)2 is an empirical formula as well as molecular formula because it has been simplified up to the smallest subscript per atom.

- K2C2O4 is not an empirical formula because it can be simplified to KCO2.

Best regards!

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C. CO is added to the reaction mixture.

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3 years ago

After 20 min, a reactant has decomposed to 85 % of its original concentration. Which order would the reaction need to be to allo

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Answer : The value of rate constant is, $0.0949\text{ min}^{-1}$

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First we have to calculate the rate constant, we use the formula :

Expression for rate law for first order kinetics is given by:

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3 years ago

A sample of g of pure aluminum metal is added to mL of M hydrochloric acid. The volume of hydrogen gas produced at standard temp

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Answer:

V = 11.2L are produced

Explanation:

... Sample of 27g of pure aluminium, 3added to 333 mL of 3.0 M HCl..

Based on the chemical reaction:

2Al(s) + 6HCl(aq) → 2AlC₃(aq) + 3H₂(g)

Where 3 moles of hydrogen are produced when 6 moles of hydrochloric acid reacts with 2 moles of Al.

To solve this question, we need to determine limiting reactant converting each reactant to moles. With limiting reactant and the chemical reaction we can find moles of hydrogen and its volume at STP (T=273.15K; P=1atm), thus:

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27g * (1mol / 26.98g) = 1mol of Al

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333mL = 0.333L * (3mol/L) = 1mol HCl

For a complete reaction of 1 mole of HCl are required:

1mol HCl * (2mol Al / 6mol HCl) = 0.333 moles of Al. As there is 1 mole of Al, Al is in excess and HCl is limiting reactant.

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1mol HCl * (3 moles H₂ / 6 mol HCl) = 0.5moles H₂ are produced.

Using ideal gas law:

PV = nRT

V = nRT/P

Where V is volume

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T is absolute temperature: 273.15K

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V = 0.5mol*0.082atmL/molK*273.15K / 1atm

<h3>V = 11.2L are produced</h3>

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