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3 years ago

Fungi and bacteria release nitrogen stored in dead tissue through a decomposition process called:

Chemistry

1 answer:

kirill [66]3 years ago

7 0

Fungi and bacteria release nitrogen stored in dead tissue through a decomposition process called B) ammonification.

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago

I have been stuck on this for a while I know you work it out through Hess's law but I can't seem to have the equations balanced

katovenus [111]

Answer:

uzb-prwk-khj

indians and all other countries can join

7 0

2 years ago

Elaborate on the difference in natural occurrences between fission and fusion reactions. A) Neither fission nor fusion reactions

vazorg [7]

<span>Okay then I would go with choice B since fusion takes place in the sun which is a giant star.</span>

8 0

3 years ago

Read 2 more answers

how many grams of calcium sulfate are produced from 10 grams of calcium nitrate and how many grams of calcium sulfate are produc

AlekseyPX

Answer: 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Ca(NO_3)_2=\frac{10g}{164g/mol}=0.061moles$

$\text{Moles of} Li_2SO_4=\frac{10g}{110g/mol}=0.091moles$

$Ca(NO_3)_2+Li_2SO_4\rightarrow 2LiNO_3+CaSO_4$

According to stoichiometry :

1 mole of $Ca(NO_3)_2$ require = 1 mole of $Li_2SO_4$

Thus 0.061 moles of $Ca(NO_3)_2$ will require= $\frac{1}{1}\times 0.061=0.061moles$ of $Li_2SO_4$

Thus $Ca(NO_3)_2$ is the limiting reagent as it limits the formation of product and $Li_2SO_4$ is the excess reagent.

As 1 mole of $Ca(NO_3)_2$ give = 1 mole of $CaSO_4$

Thus 0.061 moles of $Ca(NO_3)_2$ give = $\frac{1}{1}\times 0.061=0.061moles$ of $CaSO_4$

Mass of $CaSO_4=moles\times {\text {Molar mass}}=0.061moles\times 136g/mol=8.30g$

Thus 8.30 g of calcium sulfate are produced from 10 grams of lithium sulfate.

6 0

3 years ago

Individual measurements introduce errors that may be large or small depending on the quality of the tools and processes used. Sc

kirill [66]

Answer:

B. accepted value x 0.1

Explanation:

in the equation provided

$Percentage Error=\frac{Error}{AcceptedValue} X100$

Maximum allowed value of percentage error = 10%

put this value in the equation in stead of percentage error we get,

$10 = \frac{error}{AcceptedValue} X100\\ \frac{10}{100} =\frac{error}{AcceptedValue}\\.1=\frac{error}{AcceptedValue}\\error = .1 X Accepted Value$

so maximum error = .1 x accepted value

10 % percentage error means the experimental value has 10 % error compared to accepted value.so error will be 10 % of the accepted value

or .1 times of accepted value

8 0

3 years ago