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2 years ago

Why does magma composition change during fractional crystallization?.

1 answer:

6 0

Why does magma composition change during fractional crystallization? Different elements in the magma form crystals at different rates, leaving behind more of the unused elements. ... The crystals are denser than the magma.

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The breaking of _____________________ bonds in food releases energy for your body to use.

olya-2409 [2.1K]

Answer:

Chemical bonds

Explanation:

the chemicals in your food are what give you energy when they break down, that's how it releases energy Into your body.

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2 years ago

A weather map shows red semicircles on a line. What type of weather would one expect?

gtnhenbr [62]

A warm front brings drizzly weather.

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3 years ago

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Ultimately the carbon molecules in pyruvate end up as what molecules

jeyben [28]

CO2 Thats your answer

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2 years ago

Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under eac

valentinak56 [21]

Answer : The value of $\Delta G_{rxn}$ is, $8.867kJ/mole$

Explanation :

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction

$\Delta G_^o$ = standard Gibbs free energy

R = gas constant = 8.314 J/mole.K

T = temperature = $25^oC=273+25=298K$

Q = reaction quotient

First we have to calculate the $\Delta G_^o$ .

Formula used :

$\Delta G^o=-RT\times \ln K_p$

Now put all the given values in this formula, we get:

$\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})$

$\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole$

Now we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

$CO(g)+2H_2(g)\rightarrow CH_3OH(g)$

The expression for reaction quotient will be :

$Q=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

$Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}$

$Q=8.1\times 10^{5}$

Now we have to calculate the value of $\Delta G_{rxn}$ by using relation (1).

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

Now put all the given values in this formula, we get:

$\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})$

$\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole$

Therefore, the value of $\Delta G_{rxn}$ is, $8.867kJ/mole$

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2 years ago

_______ is the application of scientific knowledge to solve problems or create new products.

astraxan [27]

Having resources? i guess...
maybe it is critical thinking?

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3 years ago

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