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2 years ago

A 2.36-gram sample of NaHCO3 was completely decomposed in an experiment.

Chemistry

1 answer:

enyata [817]2 years ago

8 0

Answer:

Explanation:

a) The mass of the reactants is 2.36 grams, and the mass of the products is 1.57 grams plus the mass of the carbonic acid. Thus, using the law of conservation of mass, we get the mass of the carbonic acid is 2.36 - 1.57 = 0.79 grams.

b) The gram-formula mass of sodium bicarbonate is 84.006 g/mol, meaning that 2.36/84.006 = 0.028 moles were consumed. Thus, this means that in theory, 0.014 moles of carbonic acid should have been produced, which would have a mass of (0.014)(62.024)=0.868 grams. Thus, the percentage yield is (0.79)/(0.868) * 100 = 91%

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3 years ago

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3 years ago

Read 2 more answers

Determine the number of ions produced in the dissociation of the compound listed. AlF3

scoray [572]

Answer is: the number of ions produced in the dissociation of aluminium fluoride is 4.

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AlF₃(aq) → Al³⁺(aq) + 3F⁻(aq).
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3 years ago

The ion MnOis often used to analyze for the Fe2+ content of an aqueous solution by using the (unbalanced) reaction Mno+Fe2+ + Fe

dusya [7]

Answer:

B) 0.230 M

Explanation:

The first step is to balance the reaction between the Ferrous ion and the permanganate ion:

$5~Fe^+^2~+~MnO_4^-^1~+~8H^+~->~5Fe^+^3~+~Mn^+^2~+4H_2O$

Then we have to calculate the moles of $MnO_4^-$ :

$M~=~\frac{mol}{L}$

$mol~=~M*L$

$mol~=~0.033~M*0.0633L=~0.002088~mol~MnO_4^-$

Then using the molar ratio we can find the moles of $Fe^+^2$ :

$0.002088~mol~MnO_4^-\frac{5~mol~Fe^+^2}{1~mol~MnO_4^-}=0.01044~mol~Fe^+^2$

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$M=\frac{0.01044~mol~Fe^+^2}{0.0455~L}=0.230~M$

5 0

3 years ago

PLEASE HELP!!!!!!

pogonyaev

Answer:

$m_{B}^{theoretical}=0.365gB$

$Y=87.1\%$

Explanation:

Hello there!

In this case, since the reaction (A->B) have an initial amount of pure 4-aminobenzoic acid, the first step to compute the theoretical yield is to solve the following stoichiometric setup:

$m_{B}^{theoretical}=0.303gA*\frac{1molA}{137.14gA}*\frac{1molB}{1molA}*\frac{165.19 gB}{1molB}\\\\ m_{B}^{theoretical}=0.365gB$

Whereas A stands for 4-aminobenzoic acid and B for the benzocaine. Moreover, we compute the percent yield by dividing the actual yield (0.318 g) by the theoretical one (0.365 g):

$Y=\frac{0.318g}{0.365g} *100\%\\\\Y=87.1\%$

Best regards!

4 0

2 years ago