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3 years ago

7

What is the similarity between man and woman

2 answers:

Travka [436]3 years ago

4 0

Answer:

both human,both make mistakes,both can be stup.id

Explanation:

hope this helps

<3

Mkey [24]3 years ago

3 0

Blood type, skin tone,hair colour, human, breathe oxygen

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How do you find moles

Marizza181 [45]

A mole is equivalent to 6.02 x 10^23 molecules at STP. Use a conversion factor like the following.
Example: The initial value x The value you are trying to convert to
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The value you are trying to get rid of

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3 years ago

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Convection currents are caused by...

eimsori [14]

Answer:

A

Explanation:

that should be the answer

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2 years ago

Give an example of a physical model of the circulatory system

GaryK [48]

Arteries carry blood away from the heart and veins carry blood back to the heart. Hope that helps (:

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3 years ago

Ammonia gas combines with excess oxygen gas to produce nitric oxide and water. How many grams of ammonia gas would have to react

Sedbober [7]

Answer:

11.6g of NH₃(g) have to react

Explanation:

For the reaction:

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g) ΔH = -905kJ

<em>4 moles of ammonia produce 905kJ</em>

Thus, if you want to produce 154kJ of energy you need:

154kJ × (4 mol NH₃ / 905kJ) = <em>0.681moles of NH₃. </em>In mass -Molar mass ammonia is 17.031g/mol-

0.681mol NH₃ × (17.031g / mol) = <em>11.6g of NH₃(g) have to react</em>

3 0

3 years ago

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