Answer:
The balanced equation is:
2 HNO3 + Mg ---> Mg(NO3)2 + H2
From the equation, we can see that we need twice the moles of HNO3 than the moles of Mg
Moles of Mg:
Molar mass of Mg = 24 g/mol
Moles = Given mass / Molar Mass
Moles of Mg = 4.47 / 24 = 0.18 moles (approx)
Hence, 2(moles of Mg) = 0.36 moles of HNO3 will be consumed
Number of moles of HNO3 after the reaction is finished is the number of unreacted moles of HNO3
Unreacted moles of HNO3 = Total Moles - Moles consumed
Unreacted moles of HNO3 = 0.64 moles (approx)
Since we approximated the value of moles of Mg, the value of remaining moles of HNO3 will also be approximate
From the given options, we can see that 0.632 moles is the closest value to our answer
Therefore, 0.632 moles will remain after the reaction
Answer:
Explanation:
The steps of the Ostwald process:
Combinning the equations:
+
+
=
Simplifying:
The overall reaction is endothermic becuase the formation of new chemical bonds requires energy consumption.
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First, we need to calculate moles of hydrazoic acid NH3:
moles NH3 = molarity * volume
= 0.15 m * 0.025 L
= 0.00375 moles
moles NaOH = molarity * volume
= 0.15 m * 0.015 L
= 0.00225 moles
after that we shoul get the total volume = 0.025L + 0.015L
= 0.04 L
So we can get the concentration of NH3 & NaOH by:
∴[NH3] = moles NH3 / total volume
= 0.00375 moles / 0.04 L
= 0.09375 M
∴[NaOH] = moles NaOH / total volume
= 0.00225 moles / 0.04 L
= 0.05625 M
then, when we have the value of Ka of NH3 so we can get the Pka value from:
Pka = -㏒Ka
= - ㏒ 1.9 x10^-5
= 4.7
finally, by using H-H equation we can get PH:
PH = Pka + ㏒[salt/ basic]
PH = 4.7 +㏒[0.05625/0.09375]
∴ PH = 4.48
M = 2 . 8 . 2
Valence Electron of M = 2
M ==> M⁺² + 2 e⁻
a. M⁺² + OH⁻ ==> M(OH)₂
b. M⁺² + PO₄⁻³ ==> M₃(PO₄)₂
