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3 years ago

8

A mixture of hydrogen and oxygen gases has a total pressure of 5.50 atm. What is the partial pressure of the hydrogen if the P(O

2) = 1250 mm Hg?

1 answer:

Sloan [31]3 years ago

6 0

Answer:

P O 2 = 5.21 atm P C O 2 = 4.79 atm

Explanation:

Hope it helps!

You might be interested in

How many grams of oxygen would be needed to completely react with 18 grams of methane (CH4)?

S_A_V [24]

Answer:

35.91 grams of oxygen

Explanation:

6 0

3 years ago

Earth's axis is tilted. Tell what would happen if it was vertical instead.

slamgirl [31]

I think the like term is 6 hope this is right

7 0

3 years ago

Read 2 more answers

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

There is water on the pan of the scale as you measure the mass of an object. If you were to ignore the water, what would be the

mihalych1998 [28]

Remember that density refers to the "mass per unit volume" of an object.

So, if an object had a mass of 100 grams and a volume of 100 milliliters, the density would be 100 grams / 100 ml.

In the question, water on the surface of the scale would add weight, so the mass of the object that you're weighing would appear to be heavier than it really is. If that happens, you'll incorrectly assume that the density is GREATER than it really is

As an example, suppose that there was 5 ml of water on the surface of the scale. Water has a density of 1 gram per milliliter (1 g/ml) so the water would add 5 grams to the object's weight. If we use the example above, the mass of the object would seem to be 105 grams, rather than 100 grams. So, you would calculate:

density = mass / volume
density = 105 grams / 100 ml
density = 1.05 g/ml

The effect on density would be that it would erroneously appear to be greater

Hope this helps!

Good luck

6 0

4 years ago

A sample of methane gas, CH4, occupies 3.25 L at temperature of 19.0 o C. If the pressure is held constant, what will be the tem

Artemon [7]

Answer:

625.46 °C

Explanation:

We'll begin by converting 19 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

T(°C) = 19 °C

T(K) = 19 °C + 273

T(K) = 292 K

Next, we shall determine the Final temperature. This can be obtained as follow:

Initial volume (V₁) = 3.25 L

Initial temperature (T₁) = 292 K

Final volume (V₂) = 10 L

Final temperature (T₂) =?

V₁/T₁ = V₂/T₂

3.25 / 292 = 10 / T₂

Cross multiply

3.25 × T₂ = 292 × 10

3.25 × T₂ = 2920

Divide both side by 3.25

T₂ = 2920 / 3.25

T₂ = 898.46 K

Finally, we shall convert 898.46 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 898.46 K

T(°C) = 898.46 – 273

T(°C) = 625.46 °C

Therefore the final temperature of the gas is 625.46 °C

4 0

3 years ago