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swat32
3 years ago
15

Is the equation Ag2O ----> Ag+ O2 balanced

Chemistry
2 answers:
maw [93]3 years ago
4 0
Ok so simple 1Ag2O ---> 1Ag+1O2 do u understand how I got that????
natita [175]3 years ago
3 0
This is a decomposition reaction.
Silver has a charge of plus one, and oxygen has a charge of minus two. So this equation is not balanced. 
The first part is right. The two should be a subscript. The second part is not right and you should add the 2 to the Silver instead of the oxygen.
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Answer:

Higher the frequency, the higher the energy

Explanation:

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6 0
3 years ago
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What is the concentration of a potassium iodate solution after you complete the following porcedure? Pipette 10 mL of a 0.31 M p
zubka84 [21]

<u>Given:</u>

Initial concentration of potassium iodate (KIO3) M1 = 0.31 M

Initial volume of KIO3 (stock solution) V1 = 10 ml

Final volume of KIO3 V2 = 100 ml

<u>To determine:</u>

The final concentration of KIO3 i.e. M2

<u>Explanation:</u>

Use the relation-

M1V1 = M2V2

M2 = M1V1/V2 = 0.31 M * 10 ml/100 ml = 0.031 M

Ans: The concentration of KIO3 after dilution is 0.031 M

4 0
3 years ago
It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta
Zigmanuir [339]

Answer:

\lambda=241.9\ nm

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that  

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,  

1 mole = 6.023\times 10^{23}\ electrons

So,  

6.023\times 10^{23} electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of \frac {495.0}{6.023\times 10^{23}}\ kJ of energy.

Energy required = 82.18\times 10^{-23}\ kJ

Also,  

1 kJ = 1000 J

So,  

Energy required = 82.18\times 10^{-20}\ J

Also, E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}

\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}

\lambda=\frac{19.878}{10^6\times \:82.18}

\lambda=2.4188\times 10^{-7}\ m

Also,  

1 m = 10⁻⁹ nm

So,  

\lambda=241.9\ nm

6 0
3 years ago
What is the mass of 0.45 moles of Sodium?
ch4aika [34]

Answer:

10.35

Explanation:

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