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2 years ago

Calculate the percentage of sodium and carbon and oxygen in sodium carbonate

1 answer:

8 0

Percentage of Sodium [Na] in Sodium Carbonate [Na2CO3] = 46/106 x 100 = 43.40%. Percentage of Carbon [C] in Sodium Carbonate [Na2CO3] = 12/106 x 100 = 11.32%. Percentage of Oxygen [O] in Sodium Carbonate [Na2CO3] = 48/106 x 100 = 45.28% .....

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2 years ago

Which element(s) are not balanced in this equation ?

Alekssandra [29.7K]

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3 years ago

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rutherford's experiment indicated that matter was not as uniform as it appears what part of his experimental results implied thi

MaRussiya [10]

Rutherford theorized that atoms have their charge concentrated in a very small nucleus.

This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).

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According to Rutherford model of the atom:

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6 0

2 years ago

What volume (in mL) of a 0.200 MHNO3 solution is required to completely react with 27.6 mL of a 0.100 MNa2CO3 solution according

ladessa [460]

Answer:

There is 27.6 mL of a 0.200 M HNO3 solution required

Explanation:

Step 1: The balanced equation is:

Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)

This means for 1 mole Na2CO3 consumed, there is consumed 2 mole of HNO3 and there is produced 2 moles of NaNO3, 1 mole of CO2 and 1 mole of H2O

Step 2: Calculating moles of Na2CO3

moles of Na2CO3 =volume of Na2CO3 * Molarity of Na2CO3

moles of Na2CO3 = 27.6 *10^-3 * 0.1 M = 0.00276 moles

Step 3: Calculating moles of HNO3

In the balanced equation, we can see that for 1 mole of Na2CO3 consumed, there are consumed 2 moles of HNO3.

So for 0.00276 moles consumed of Na2CO3, there are consumed 0.00552 moles of HNO3.

This means 0.00276 moles of the base Na2CO3 would react with 0.00552 moles of the acid HNO3

Step 4: Calculating the volume of HNO3

volume of HNO3 = moles of HNO3 / Molarity of HNO3

volume of HNO3 = 0.00552 moles / 0.200 M = 0.0276 L

0.0276 L = 27.6 ml

There is 27.6 mL of a 0.200 M HNO3 solution required

4 0

2 years ago

The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?

Agata [3.3K]

Answer: The molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

Explanation:

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

$PbI_2\rightleftharpoons Pb^{2+}+2I^-$

s 2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Pb^{2+}][I^-]^2$

We are given:

$K_{sp}=7.9\times 10^{-9}$

Putting values in above equation, we get:

$7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L$

Hence, the molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

3 0

3 years ago

Calculate the percentage of sodium and carbon and oxygen in sodium carbonate​

Calculate the percentage of sodium and carbon and oxygen in sodium carbonate