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Vlad [161]
3 years ago
5

If 500.0 mL of 0.450 M sodium phosphate is reacted with an excess of iron (II) nitrate solution, how many grams of iron (II) pho

sphate are produced?
Chemistry
1 answer:
Maurinko [17]3 years ago
8 0

Answer:

If 500.0 mL of 0.450 M sodium phosphate is reacted with an excess of iron (II) nitrate solution, how many grams of iron (II) phosphate are produced?

idk

Explanation:

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Trava [24]

Answer: 2.52 M

Explanation:

The product of molarity (moles/litre) and volume in litres yields moles, and the numbers of moles in two solutions means dilute and concentrated are equal, which is expressed by the following equation: 

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$$Given that\\M $1=12.0 \mathrm{M}$ or mole $/ \mathrm{L}$\\$\mathrm{V} 1=420 \mathrm{ml}$\\$\mathrm{M} 2=$ ?\\$\mathrm{V} 2=2.0 \mathrm{~L}$ or $2000 \mathrm{ml}$\\\\$\mathrm{M} 1 \mathrm{~V} 1=\mathrm{M} 2 \mathrm{~V} 2$\\$\mathrm{M} 2=\mathrm{M} 1 \mathrm{~V} 1 / \mathrm{V} 2$\\$=12.0 * 420 / 2000$\\$=2.52 \ \mathrm{M}$

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Varvara68 [4.7K]

Answer:- Cl_2+2KBr\rightarrow 2KCl+Br_2

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