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slega [8]
2 years ago
13

How many grams of NaOH are needed to neutralize 50 grams of H2SO4

Chemistry
2 answers:
slavikrds [6]2 years ago
8 0

Answer:

H2SO4 + 2NaOH ==> Na2SO4 + 2H2O

Explanation:

Ede4ka [16]2 years ago
8 0
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O

mols H2SO4 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols H2SO4 to mols NaOH.
2Naoh+h2so4 na2so4+2h2o
50g/100g=x/98g
X=98g×50g/100g
X=49g
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Use calc to determine whether it is possible to remove 99.99% Cu2 by converting it to Cu(s) in a solution mixture containing 0.1
inessss [21]

Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Explanation:

So, from the question/problem above we are given the following ionic or REDOX equations of reactions;

Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V

Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V

In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:

Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.

Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.

Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.

Thus, ΔG° = - 92640.

This is less than zero[0]. Therefore,  it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.  

7 0
3 years ago
Based on the electromagnetic spectrum what is the relationship between wave length and frequency
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Convert to moles 3.45 g Na
GenaCL600 [577]

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The free energy of formation of nitric oxide, NO, at 1000 K (roughly the temperature in an automobile engine during ignition) is
Assoli18 [71]

<u>Answer:</u> The value of K_p for the chemical equation is 8.341\times 10^{-5}

<u>Explanation:</u>

For the given chemical equation:

N_2(g)+O_2(g)\rightarrow 2NO(g)

To calculate the K_p for given value of Gibbs free energy, we use the relation:

\Delta G=-RT\ln K_p

where,

\Delta G = Gibbs free energy = 78 kJ/mol = 78000 J/mol  (Conversion factor: 1kJ = 1000J)

R = Gas constant = 8.314J/K mol

T = temperature = 1000 K

K_p = equilibrium constant in terms of partial pressure = ?

Putting values in above equation, we get:

78000J/mol=-(8.314J/Kmol)\times 1000K\times \ln K_p\\\\Kp=8.341\times 10^{-5}

Hence, the value of K_p for the chemical equation is 8.341\times 10^{-5}

4 0
3 years ago
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