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3 years ago

12.00 moles of naclo3 will produce how many grams of O2? (Moles to Grams)

1 answer:

4 0

2NaClO₃ → 2NaCl + 3O₂

mole ratio of NaClO₃ to O₂ is 2 : 3

∴ if moles of NaClO₃ = 12 mol

then moles of O₂ = $\frac{12 mol * 3}{2}$

= 18 mol

Mass of O₂ = mol of O₂ × molar mass of O₂

= 18 mol × 16 g/mol

= 288 g

So I wasn't sure which equation to use since you did not specify so I just used the decomposition reaction. If you should have used another reaction then just follow the same steps and you'll get your answer.

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Which of the following solutions would have the lowest freezing point?

raketka [301]

I would like to help but what are the solutions?

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3 years ago

What evidence do you have that tpgs-750m increases the reaction rate? How does tpgs-750m act to increase the reaction rate?

notka56 [123]

TPGS-750 M conducts reaction under lower temperature then the reactions carried out in organic solvents.

It is a chemical molecule of the second generation known as an amphiphile, with both hydrophilic and lipophilic qualities.

In contrast to any other organic solvent, water is used for transition metal catalysed cross-coupling, making TPGS-750M a second generation surfactant with physical features that encourage this process.

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7 0

1 year ago

Which of the following is TRUE about when two atoms form a bond? a. Repulsion is greater than attraction b. Neither atoms will h

mixer [17]

Answer:

b: Neither atoms will have a full valence shell

Explanation:

I think sooo

8 0

3 years ago

Permanganate ion reacts in basic solution with oxalate ion to form carbonate ion and solid manganese dioxide. Balance the skelet

saw5 [17]

Answer:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

Explanation:

First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.

Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.

Reduction:

MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)

Oxidation:

C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-

Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.

2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)

3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-

Now combine both equations and eliminate repeating H+ and H2O.

2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)

turns into:

2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)

5 0

4 years ago

A student isolated 25 g of a compound following a procedure that would theoratically yield 81 g. Wht was his percentt yield?

Sati [7]

<u>Answer:</u> The percent yield of the compound is 30.86 %.

<u>Explanation:</u>

To calculate the percentage yield of a compound, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of compound = 25 g

Theoretical yield of compound = 81 g

Putting values in above equation, we get:

$\%\text{ yield of compound}=\frac{25g}{81g}\times 100\\\\\% \text{yield of compound}=30.86\%$

Hence, the percent yield of the compound is 30.86 %.

3 0

4 years ago