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2 years ago

What happens when K2O dissolves in water?

1 answer:

7 0

Firstly

the overall reaction is k2o + H2O --------> 2KOH

That above is the molecular equation

net ionic equation, as the specatator ion, the one that is not involved in the reation is in fact K+

THE NET IONIC EQUATION is therefore

O2- + H2O ------>2OH-

Acid is defined as the proton donar, in otherwards the one that donated a positve charge, here , the water donated a postive H+ ion to the O2-, SO the h2o is the acid while the O2- is the base.

Acid : h2o

Spectator ion : k+, pottasium ion

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2 years ago

What mass of Ca(OH)2 will be used to make 45.6g of NaOH?

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2 years ago

Which is a molecule? A. Ne B. NaCL C. Trail Mix D. graphene

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B) NaCl is a molecule.

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3 years ago

What is one of the names for a column in the periodic table?

zheka24 [161]

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Explanation:

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2 years ago

Read 2 more answers

A 1.2516 gram sample of a mixture of caco3 and na2so4 was analyzed by dissolving the sample and completely precipitating the ca

Dennis_Churaev [7]

Answer:

0.009725 moles of H2C2O4

0.009725 moles CaCO3

Mass percentage = 77.77%

Explanation:

Step 1: The balanced equation

2MnO4- +5C2H2O4+6H+ →2Mn2+ +10CO2+8H2O

We can see that for 2 moles of Mno4- consumed , there is 5 moles of C2H2O4 needed and 6 moles H+ to produce 2 moles Mn2+, 10 moles of CO2 and 8 moles of H2O

Step 2: Calculate moles of MnO4-

Molarity = Moles/volume

Moles of Mno4- = Molarity of MnO4- * Volume of Mno4-

Moles of Mno4- = 0.1092M * 35.62 *10^-3 L

Moles of MnO4- = 0.00389 moles

Step 3: Calculate moles of H2C2O4

Since there is needed 5 moles of C2H2O4 to consume 2 moles of MnO4-

then for 0.00389 moles of MnO4-, there is 5/2 *0.00389 = 0.009725 moles of H2C2O4

Step 4: Calculate moles of CaCO3

moles of H2C2O4 = moles CaCO3, therefore, 0.009725 moles H2C2O4 = 0.009725 moles CaCO3

Step 5: Calculate mass of CaCO3

Molar mass of CaCO3 = 100.09 g/mole

Mass of CaCO3 = moles of CaCO3 * Molar mass of CaCO3

Mass of CaCO3 = 0.009725 moles * 100.09 g/mole = 0.9734 g

Step 6: Calculate percentage by weight of CaCO3

Mass of CaCO3 = 0.9734g

Mass of original sample = 1.2516g

Mass percentage = 0.9734/1.2516 *100% = 77.77%

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3 years ago