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netineya [11]
3 years ago
5

Un buzo que nada en las profundidades despide pequeñas burbujas de aire¿Aumenta disminuye o permanece igual el tamaño de las bur

bujas conforme suben a la superficie?
Chemistry
1 answer:
alekssr [168]3 years ago
5 0

Answer:

Al ascender las burbujas van aumentando de tamaño.

Explanation:

Las burbujas que produce el buzo debajo del agua son pequeñas moléculas de dióxido de carbono gaseoso producto de la respiración del mismo.

Ahora, a medida que las burbujas suben a la superficie, la presión que sufren estas (Presión debido al agua), es menor conforme van ascendiendo debido a la ley de Boyle: A medida que la presión aumenta, el volumen va disminuyendo.

Esto significa que al ascender las burbujas van aumentando de tamaño debido a que la presión que sufren estas es menor que cuando están a mayores profundidades.

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How many moles are in 17.2g k2s
Natalija [7]

Answer: 0.156 mol

Explanation:

To find the moles of 17.2 g K₂S, we need to know the molar mass to convert.

17.2g*\frac{mol}{110.256 g} =0.156 mol

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3 years ago
Some states authorize a to make healthcare decisions ------- for patients who can't make them because they're incapacitated. O a
Tom [10]

Explanation:

A student flips four different coins at the same time. Which choice shows all of the possibilities for the results of the coin toss? (H = heads, T = tails)

A HH, TT, HT, TH

B HHHT, HTTT, HHTT

C HHH, TTT, HTH, HTT

D HHHH, HTHH, HTTH, HTTT, TTTT

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3 years ago
1. Write the balanced chemical equation for the synthesis of magnesium phosphide from its elements, including the word “energy”
Evgen [1.6K]

Answer:

3Mg(s) +2P(s) -------> Mg3P2(s) + energy

Keq= [Mg3P2]/[Mg]^3 [P]^2

Explanation:

The equation for the formation of magnesium phosphide from its elements is;

3Mg(s) +2P(s) -------> Mg3P2(s) + energy

Hence we can see that three moles of magnesium atoms combines with two moles of phosphorus atoms to yield one mole of magnesium phosphide. The equation written above is the balanced chemical reaction equation for the formation of the magnesium phosphide.

The equilibrium expression for the reaction K(eq) will be given by;

Keq= [Mg3P2]/[Mg]^3 [P]^2

8 0
3 years ago
The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th
MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

\frac{K_1}{K_2}=?

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}

where,

K_2 = rate constant reaction -1

K_1 = rate constant reaction -2

Activation energy of the reaction 1 ,Ea_1 = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2,Ea_1  = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

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

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

\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340

0.4284 is the ratio of the rate constants.

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B answer is b I repeat it’s b
4 0
3 years ago
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