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3 years ago

Organisms breakdown _ in order to produce?

Chemistry

2 answers:

Novay_Z [31]3 years ago

3 0

Answer: Organisms breakdown __into smaller particles___ in order to produce__new compounds__?

Explanation:

branliest pls :)

lara [203]3 years ago

3 0

Glucose/Food, Energy

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3 years ago

What minimum mass of iron (II) nitrate must be added to 10.0 of a 0.0699 M phosphate solution in order to completely precipitate

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<u>Answer:</u> The minimum mass of iron (II) nitrate that must be added is 0.188 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of phosphate solution = 0.0699 M

Volume of solution = 10 mL = 0.010 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.0699M=\frac{\text{Moles of phosphate solution}}{0.010L}\\\\\text{Moles of phosphate solution}=(0.0699mol/L\times 0.010L)=6.99\times 10^{-4}mol$

The given chemical equation follows:

$2PO_4^{3-}(aq.)+3Fe(NO_3)_2(aq.)\rightarrow Fe_3(PO_4)_2(s)+6NO_3^-(aq.)$

By Stoichiometry of the reaction:

2 moles of phosphate solution reacts with 3 moles of iron (II) nitrate

So, $6.99\times 10^{-4}mol$ of phosphate solution will react with = $\frac{3}{2}\times 6.99\times 10^{-4}=1.049\times 10^{-3}mol$ of iron (II) nitrate

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of iron (II) nitrate = 180 g/mol

Moles of iron (II) nitrate = $1.049\times 10^{-3}$ moles

Putting values in above equation, we get:

$1.049\times 10^{-3}mol=\frac{\text{Mass of iron (II) nitrate}}{180g/mol}\\\\\text{Mass of iron (II) nitrate}=(1.049\times 10^{-3}mol\times 180g/mol)=0.188g$

Hence, the minimum mass of iron (II) nitrate that must be added is 0.188 grams

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4 years ago

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Answer:

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Organisms breakdown _____ in order to produce____?

Organisms breakdown _ in order to produce?