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Reika [66]
3 years ago
12

Hydrogen ions cannot exist alone but they exist after combining with water molecules true or false​

Chemistry
1 answer:
juin [17]3 years ago
4 0
True because hydrogen ions combines with h2o to make hydronium ion, so in a sense h2o is acting like a base.
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Whoever awser correct I will give you Brainly
vivado [14]

Answer:

Option A.

2Na + 2H2O —> 2NaOH + H2

Explanation:

To know which option is correct, we shall do a head count of the number of atoms present on both side to see which of them is balanced. This is illustrated below below:

For Option A:

2Na + 2H2O —> 2NaOH + H2

Reactant >>>>>>> Product

2 Na >>>>>>>>>>> 2 Na

4 H >>>>>>>>>>>> 4 H

2 O >>>>>>>>>>>> 2 O

Thus, the above equation is balanced.

For Option B:

2Na + 2H2O —> NaOH + H2

Reactant >>>>>>> Product

2 Na >>>>>>>>>>> 1 Na

4 H >>>>>>>>>>>> 3 H

2 O >>>>>>>>>>>> 1 O

Thus, the above equation is not balanced.

For Option C:

2Na + H2O —> 2NaOH + H2

Reactant >>>>>>> Product

2 Na >>>>>>>>>>> 2 Na

2 H >>>>>>>>>>>> 4 H

1 O >>>>>>>>>>>> 2 O

Thus, the above equation is not balanced.

For Option D:

Na + 2H2O —> NaOH + 2H2

Reactant >>>>>>> Product

1 Na >>>>>>>>>>> 1 Na

4 H >>>>>>>>>>>> 5 H

2 O >>>>>>>>>>>> 1 O

Thus, the above equation is not balanced.

From the illustrations made above, only option A is balanced.

7 0
3 years ago
Base your answer to this question on the information below.During a bread-making process, glucose is converted to ethanol and ca
Damm [24]

Answer: 1 C6H12O6===> 2 C2H5OH + 2 CO2

75 In the space in your answer booklet, draw a structural formula for the alcohol formed in this reaction. [1]

Explanation:

6 0
3 years ago
A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?
Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles

Now we have to calculate the moles of Br_2

{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles

{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles

The balanced chemical reaction is,

2Na(s)+Br_2(g)\rightarrow 2NaBr

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = \frac{2}{1}\times 0.189=0.378 moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = \frac{2}{1}\times 0.189=0.378 moles of Sodium bromide

Now we have to calculate the percent yield of reaction

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%

Therefore, the percent yield is, 93.4%

3 0
3 years ago
Use chemical formulas to write the equation for the reaction between carbon and oxygen
Mandarinka [93]
C=Carbon
O=Oxygen
Oxygen is a diatomic molecule, so the subscript two needs to be used after the O in the formula.
C+O2->CO2
Hope this helps!
7 0
3 years ago
PLEASE HELP it would be really nice of you to answer this question i would be very grateful <3
Aleksandr-060686 [28]

Answer: D (The abundance percentage of each isotope)

Explanation: hope this helps!

7 0
3 years ago
Read 2 more answers
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