- #1
tobyguy
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If you mix BaNO(3)2 + NaOH what would be the chemical chemical reaction in formula?
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.
- #2
stunner5000pt
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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
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
Last edited:
- #3
tobyguy
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Thanks that really helped. 
- #4
dextercioby
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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.
Ba\left(NO_{3}\right)_{2}+NaOH\rightarrow NaNO_{3}+Ba(OH)_{2}\downarrow
Do you see the difference?
Daniel.
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19.
For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Figure 1 Overall Structure Diagram
Figure 2: Top view of the piston when it is cylindrical
A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N.
Figure 3: Modifying the structure to incorporate a fixed internal piston
When I modify the piston...
Let's declare that for the cylinder,
mass = M = 10 kg
Radius = R = 4 m
For the wall and the floor,
Friction coeff = ##\mu## = 0.5
For the hanging mass,
mass = m = 11 kg
First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on.
Force on the hanging mass
$$mg - T = ma$$
Force(Cylinder) on y
$$N_f + f_w - Mg = 0$$
Force(Cylinder) on x
$$T + f_f - N_w = Ma$$
There's also...
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