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4 years ago

What is the formula of Potassium Bicarbonate?

Chemistry

2 answers:

Nimfa-mama [501]4 years ago

8 0

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ It is KHCO3

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

shepuryov [24]4 years ago

7 0

The formula of Potassium Bicarbonate is KHCO3.

In the laboratory, baking soda can be prepared by passing excess of carbon dioxide gas through the saturated solution of potassium carbonate. As a result, potassium bicarbonate is formed. Being sparingly soluble in water, it gets precipitated. The precipitate is separated, washed and dried without heating.

K2CO3 + CO2 + H2O -----> 2KHCO3

Here, KHCO3 is a precipitate. It contains one atom of Potassium one atom of Hydrogen one atom of Carbon and three atoms of Oxygen.

Potassium Bicarbonate is also Potassium Hydrogen Carbonate.

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Answer: Im not sure if you are doing this in your class.

Explanation:

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3 years ago

Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or

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Answer :

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The oxidation number of a free element is always zero.
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The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of oxygen (O) in compounds is usually -2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
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Now we have to determine the oxidation state of the elements in the compound.

(a) $H_2SO_4$

Let the oxidation state of 'S' be, 'x'

$2(+1)+x+4(-2)=0\\\\x=+6$

Hence, the oxidation state of 'S' is, (+6)

(b) $Ca(OH)_2$

Let the oxidation state of 'Ca' be, 'x'

$x+2(-2+1)=0\\\\x=+2$

Hence, the oxidation state of 'Ca' is, (+2)

(c) $BrOH$

Let the oxidation state of 'Br' be, 'x'

$x+(-2)+1=0\\\\x=+1$

Hence, the oxidation state of 'Br' is, (+1)

(d) $ClNO_2$

Let the oxidation state of 'N' be, 'x'

$-1+x+2(-2)=0\\\\x=+5$

Hence, the oxidation state of 'N' is, (+5)

(e) $TiCl_4$

Let the oxidation state of 'Ti' be, 'x'

$x+4(-1)=0\\\\x=+4$

Hence, the oxidation state of 'Ti' is, (+4)

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Let the oxidation state of 'Na' be, 'x'

$x+(-1)=0\\\\x=+1$

Hence, the oxidation state of 'Na' is, (+1)

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3 years ago

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Explanation:

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3 years ago

Read 2 more answers

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ira [324]

Answer:

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3 years ago

Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2

Svet_ta [14]

Answer:

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Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)

This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

In this case:

ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol

4 0

3 years ago

What is the formula of Potassium Bicarbonate?​

What is the formula of Potassium Bicarbonate?