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3 years ago

What happens to the heat energy that leaves the milk?

Chemistry

1 answer:

bekas [8.4K]3 years ago

8 0

Answer:

it gets cold or warm

Explanation:

pleas get me 53 likes

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Please help! How many grams is 4.5 x 10^27

Eva8 [605]

Answer: The answer is 4.5e+27 grams, I hope I helped!

Explanation: N/A (I threw away my paper ^^")

4 0

3 years ago

`as much as we would like for it to be the case, we never get all i the product we are supposed to get when we run a real reacti

zavuch27 [327]

Need more information

3 0

3 years ago

At 25 ∘C , the equilibrium partial pressures for the reaction were found to be PA=5.16 bar, PB=5.04 bar, PC=4.11 bar, and PD=4.8

erastova [34]

Answer: 5.85kJ/Kmol.

Explanation:

The balanced equilibrium reaction is

$A(g)+2B(g)\rightleftharpoons 4C(g)+D(g)$

The expression for equilibrium reaction will be,

$K_p=\frac{[p_{D}]\times [p_{C}]}^4{[p_{B}]^2\times [p_{A}]}$

Now put all the given values in this expression, we get the concentration of methane.

$K_p=\frac{(4.85)\times [(4.11)^4}{(5.04)^2\times (5.16)}$

$K_p=10.6$

Relation of standard change in Gibbs free energy and equilibrium constant is given by:

$\Delta G^o=-2.303\times RT\times \log K_c$

where,

R = universal gas constant = 8.314 J/K/mole

T = temperature = $25^0C=(25+273)K=298 K$

$K_c$ = equilibrium constant = 10.6

$\Delta G^o=-2.303\times 8.314\times 298\times \log (10.6)$

$\Delta G^o=5850.23J/Kmol$

$\Delta G^o=5.85kJ/Kmol$

Thus standard change in Gibbs free energy of this reaction is 5.85kJ/Kmol.

3 0

3 years ago

Heating 2.40 g of the oxide of metal X (molar mass of X = 55.9 g/mol) in carbon monoxide (CO) yields the pure metal and carbon d

Korolek [52]

Answer:The molecular formula of the oxide of metal be $X_2O_3$ . The balanced equation for the reaction is given by:

$X_2O_3+3CO\rightarrow 3CO_2+2X$

Explanation:

Let the molecular formula of the oxide of metal be $X_2O_y$

$X_2O_y+yCO\rightarrrow yCO_2+2X$

Mass of metal product = 1.68 g

Moles of metal X = $\frac{1.68 g}{55.9 g/mol}=0.03005 mol$

1 mol of metal oxide produces 2 moles of metal X.

Then 0.03005 moles of metal X will be produced by:

$\frac{1}{2}\times 0.03005 mol=0.01502 mol$ of metal oxide

Mass of 0.01502 mol of metal oxide = 2.40 g (given)

$0.01502 mol\times (2\times 55.9 g/mol+y\times 16 g/mol)=2.40 g$

y = 2.999 ≈ 3

The molecular formula of the oxide of metal be $X_2O_3$ . The balanced equation for the reaction is given by:

$X_2O_3+3CO\rightarrow 3CO_2+2X$

8 0

3 years ago

How does the vegetation surface type affect the amount of runoff

maxonik [38]

Answer:

<em>The type of vegetation a surface does affect the </em><em>water coming from above to sink in or runoff. </em>

Explanation:

This is how the vegetation affects the runoff:-

The leaves and stems present in the vegetation do not let the water fall directly on the soil and makes the process rather slow which makes the water to get to the ground slowly and sink in properly inside the soil rather than running off.

If the vegetation present is dense with there was being hairy then also the water would not run out and will get absorbed by the roots letting the soil intact

5 0

3 years ago