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12345 [234]
4 years ago
12

Given the balanced equation representing a reaction:

Chemistry
1 answer:
notsponge [240]4 years ago
4 0

Answer:

it is a base because they accept H+ ions

Explanation:

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The cathode reaction is NiO2+H2O+2e−→Ni(OH)2+2OH−.

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Why pie bond are not perticipate in hybridaization​
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The first bond between two atoms is always a sigma bond and the other bonds are always pi bonds and a hybridized orbital cannot be involved in a pi bond. Thus we need to leave one electron (in case of Carbon double bond) to let the Carbon have the second bond as a pi bond.
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How many moles are in 2.3 grams of phosphorus?
GarryVolchara [31]
<em>The Molecular Mass for Phosphorus is 31 grames / mole.

So, 31g. P.....................1 mole
2.3g. P ..........................x mole

</em>x= \frac{2,3}{31} = \boxed {0,07}  \hbox { moles }<em>
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Please help!
tino4ka555 [31]
True- strong acids completely ionize
3 0
3 years ago
How many grams of barium sulfate are produced if 25.34 mL of 0.113 M BaCl2 completely react given the reaction: BaCl2 (aq) + Na2
jeyben [28]

<u>Answer:</u> The amount of barium sulfate produced in the given reaction is 0.667 grams.

<u>Explanation:</u>

To calculate the number of moles from molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of barium chloride = 0.113 M

Volume of barium chloride = 25.34 mL = 0.02534 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol

For the given chemical reaction:

BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = \frac{1}{1}\times 0.00286mol=0.00286mol of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

4 0
4 years ago
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