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3 years ago

Is the sky a solid liquid or gas

Chemistry

2 answers:

hram777 [196]3 years ago

8 0

Answer:

gas

Explanation:

The sky is not a solid body, nor does it have all of the same the physical properties of a solid body. In terms of its behavior, physically a gas is much closer to a liquid than a solid. If you drop the temperature of the atmosphere quick enough and low enough it can become a solid.

s2008m [1.1K]3 years ago

6 0

Gas.
The sky is not solid, not liquid but of gas. It is also called the atmosphere, one of the important spheres on earth that influences the climate –temperature, humidity, precipitation, atmospheric pressure, and the biotic community as a whole. The atmosphere or the sky is composed of many layers and is responsible to protecting the living organisms of the earth against the deadly ultraviolet rays and other strong radiation coming from the sun.

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The value of h for the reaction below is -73 kj how many kj of heat are released

jarptica [38.1K]

If the value of H is positive, it means you have to add that much heat to complete the reaction. If H is negative, it means that much heat is released during the chemical process. Because it is -73 kJ, 73 kJ of heat are released in the reaction.

7 0

3 years ago

Calculate the second volumes. <br> 7.03 Liters at 31 C and 111 Torr to STP

DIA [1.3K]

The second volume : V₂= 0.922 L

<h3> Further explanation </h3><h3>Given </h3>

7.03 Liters at 31 C and 111 Torr

Required

The second volume

Solution

T₁ = 31 + 273 = 304 K

P₁ = 111 torr = 0,146 atm

V₁ = 7.03 L

At STP :

P₂ = 1 atm

T₂ = 273 K

Use combine gas law :

P₁V₁/T₁ = P₂V₂/T₂

Input the value :

0.146 x 7.03 / 304 = 1 x V₂/273

V₂= 0.922 L

4 0

3 years ago

1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en

Maslowich

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

$n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles$

$n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles$

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

$n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles$

$m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g$

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

$\% = \frac{R_{r}}{R_{T}}*100$

<u>Donde</u>:

$R_{r}$ : es el rendimiento real

$R_{T}$ : es el rendimiento teórico

$\% = \frac{3,5}{5,043}*100 = 69,4$

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!

5 0

3 years ago

White phosphorus melts and then vaporizes at high temperatures. The gas effuses at a rate that is 0.404 times that of neon in th

Soloha48 [4]

White phosphorus melts and then vaporizes at high temperatures. The gas effuses at a rate that is 0.404 times that of neon in the same apparatus under the same conditions-There are 4 atoms of P in the molecule

Explanation:

Ar=30,97g/mol

$R_{Px$ / $R_Ne}$ = $\sqrt{Mr(Ne)/Mr(Px)}$ =0,404

0,404= $\sqrt{20,18/30,97*x}$

$0,404^{2}$ =20,18/30,97*x

X=20,18/30,97*0,163

X=4

There are 4 atoms of P in the molecule

8 0

3 years ago

What volume (in milliliters) of oxygen gas is required to react with 4.03 g of Mg at STP?

BigorU [14]

Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).

The first step is to find the number of moles of Mg in 4.03g of Mg. You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg. Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg. To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂. From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP. Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.

I hope this helps. Let me know in the comments if anything is unclear.

6 0

4 years ago

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