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The number of moles of carbonate (CO3) in sodium carbonate (Na2CO3) can be calculated by dividing the mass of carbonate in the compound by the molar mass of the compound.

no. of moles of CO3 = mass of CO3 ÷ molar mass of Na2CO3

Molar mass of Na2CO3 = 23(2) + 12 + 16(3)

= 46 + 12 + 48 = 106g/mol

mass of CO3 = 12 + 48 = 60g

no. of moles of CO3 = 60/106

no. of moles of CO3 = 0.566mol

Therefore, there are 0.566 moles of carbonate in sodium carbonate.

Learn more about number of moles at: brainly.com/question/1542846

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2 years ago

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The following diagrams represent mixtures of NO(g) and O2(g). These two substances react as follows: 2NO(g)+O2(g)→2NO2(g) It has

Alja [10]

This is an incomplete question, here is a complete question and an image is attached below.

The following diagrams represent mixtures of NO(g) and O₂(g). These two substances react as follows:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

It has been determined experimentally that the rate is second order in NO and first order in O₂.

Based on this fact, which of the following mixtures will have the fastest initial rate?

The mixture (1). The mixture (2). The mixture (3).

Answer : The mixture 1 has the fastest initial rate.

Explanation :

The given chemical reaction is:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate law expression is:

$Rate=k[NO]^2[O_2]$

Now we have to determine the number of molecules of $NO\text{ and }O_2$

In mixture 1 : There are 5 $NO$ and 4 $O_2$ molecules.

In mixture 2 : There are 7 $NO$ and 2 $O_2$ molecules.

In mixture 3 : There are 3 $NO$ and 5 $O_2$ molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(5)^2\times (4)$

$Rate=k(100)$

The rate law expression for mixture 2 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(7)^2\times (2)$

$Rate=k(98)$

The rate law expression for mixture 3 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(3)^2\times (5)$

$Rate=k(45)$

Hence, the mixture 1 has the fastest initial rate.

4 0

2 years ago

Vanillin is the compound containing carbon, hydrogen, and oxygen that gives vanilla beans their distinctive flavor. The combusti

nignag [31]

<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C 14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O molar mass of C=12 g/mole molar mass of H=1 g/mole 0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C 0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H mg O= 30.4 mg vanillin - 19.2 mg C â€“ 1.56 mg H = 9.64 mg O molar mass of O=16 g/mole 9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602 C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602 C2.66H2.66O1 is the empirical formula; to obtain whole numbers multiply by 3 3[C2.66H2.66O1] = C8H8O3 above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu The empirical formula weight and the molecular formula weight are the same . Molecular formula is C8H8O3.</span>

7 0

3 years ago