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Sav [38]
3 years ago
8

If I have a molar concentration of Na2S2O3 of .02M and a volume of .001L. As well as a Molar Concentration of KI .3M and H2O2 of

.1M. The equation of the reaction is 2S2O3+I3 ->3I+S4O6. The question is "How many moles of S2O3 are consumed". Do I just use the Molar concentration and the volume of the Na2S2O3?
Chemistry
1 answer:
Aleonysh [2.5K]3 years ago
4 0
You have to use everything that is given since you have to know which is the limiting reactant. We find the limiting reactant by calculating the number of moles of each reactant and compare the number of moles. The limiting reactant would be the one that is consumed fully by the reaction.


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What two structures in plant leaves help prevent the loss of water?.
Gelneren [198K]

Answer:

Upper cuticle and guard cells

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2 years ago
______ is a section of DNA that has information about a specific trait
stepan [7]
The answer is:
a gene
6 0
3 years ago
luminum and oxygen react according to the following equation: 4Al(s) +3O2(g) --> 2Al2O3(s) What mass of Al2O3, in grams, can
Slav-nsk [51]

Answer: 8.7 grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.  

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{4.6g}{27g/mol}=0.17moles

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

As oxygen is in excess, Aluminium is the limiting reagent and limits the formation of products.

According to stoichiometry:

4 moles of aluminium give = 2 moles of Al_2O_3(s)

Thus 0.17 moles of aluminium give=\frac{2}{4}\times 0.17=0.085mol

Mass of Al_2O_3=moles\times {\text {molar mass}}=0.085\times 102g/mol=8.7g

Thus the mass of Al_2O_3(s)  is 8.7 grams

8 0
3 years ago
Help ASAP!!!!!!!!!!!!!!!!!!!!!!!!!
BartSMP [9]

Answer:

ok      .

Explanation:            

3 0
3 years ago
The volume of a gas is 4.0 L when the pressure is 1.00 atm. At the same temperature, what is the pressure at which the volume of
Leokris [45]

<u>Answer:</u> 2.00 atm

<u>Explanation:</u>

The gas is kept under the same temperature in this problem. Assuming the amount of gas is constant, we can apply the Boyle's law.

The Boyle's law equation,

P₁V₁ =  P₂V  ₂

Plug in the values,

1.00 atm x 4.0 L =  P₂  x 2.0 L

Simplify,

4.00 atm L = 2  P₂ L

Now flip the equation,

2  P₂ L = 4.00 atm L

Dividing both sides by 2 we get,

P₂ = 2.00 atm

8 0
3 years ago
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