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3 years ago

13

_ Ba(OH)2 + _ H3PO4 ----> _ BaHPO4 + _ H2O can someone please balance this?

1 answer:

Colt1911 [192]3 years ago

8 0

Answer:

3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O

Step-by-step explanation:

Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O

There are 3 atoms of Ba on the right side and 1atom on the left side. It can be balance by putting 3 in front of Ba(OH)2 as shown below:

3Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O

There are 2 atoms of P on the right side and 1atom on the left. It can be balance by putting 2 in front of H3PO4 as shown below:

3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + H2O

Now, there are a total of 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:

3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O

Now the equation is balanced as the numbers of the atoms of the different elements present on both sides are equal

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One more question on this test guys. please answer as fast as possible. If you do,

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1. The given rectangular equation is $x=2$ .

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$r=2}\sec \theta$

$\boxed{x=2\to r=2\sec \theta}$

2. The given rectangular equation is:

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This is the same as:

$x^2+y^2=6^2$

We use the relation $r^2=x^2+y^2$

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3. The given rectangular equation is:

$x^2+y^2=2y$

This is the same as:

We use the relation $r^2=x^2+y^2$ and $y=r\sin \theta$

This implies that:

$r^2=2r\sin \theta$

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$\boxed{x^2+y^2=2y\to r=2\sin \theta}$

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We substitute $y=r\sin \theta$ and $x=r\cos \theta$

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$\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}$

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