What Happens When Ba(NO3)2 Reacts with NaOH?

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The reaction between barium nitrate (Ba(NO3)2) and sodium hydroxide (NaOH) is a double displacement reaction, resulting in the formation of sodium nitrate (NaNO3) and barium hydroxide (Ba(OH)2). The balanced chemical equation is Ba(NO3)2 + 2NaOH → 2NaNO3 + Ba(OH)2. It is noted that barium has a valence of 2, which means two hydroxide ions are needed to form barium hydroxide. A correction was made regarding the initial reaction formula provided, emphasizing the importance of proper balancing. The discussion highlights the significance of understanding valences in chemical reactions.
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If you mix BaNO(3)2 + NaOH what would be the chemical chemical reaction in formula?
 
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a simple double displacement reaction

Ba (NO_{3})_{2} + 2Na OH \rightarrow 2NaNO_{3} + Ba (OH)_{2}

balance away!

p.s. dexter is right since Ba has 2 valences there will be 2 OH ions in barium hydroxide
 
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Thanks that really helped. :-p
 
The reaction provided is incorrect.

Ba\left(NO_{3}\right)_{2}+NaOH\rightarrow NaNO_{3}+Ba(OH)_{2}\downarrow

Do you see the difference?

Daniel.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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