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3 years ago

What conditions must be met to compress a gas

1 answer:

3 0

In order to compress a gas, high pressure should be met. <span>There are three major groups of compressed gases stored in cylinders: liquefied, non-liquefied and dissolved gases. In each case, the pressure of the gas in the cylinder are high. Hope this helps.</span>

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Using the equation 2H2+O2 = 2H2I, if 192 g if oxygen reacts completely, how many grams of hydrogen must react with it

dimaraw [331]

I assume there is a typo in the equation. It is H2O instead of H2I.

The ratio of H2 and O2 that react with each other is 2:1. To find out the grams of H2, we need to first find out the moles of O2 and H2.

Moles of O2 = mass of O2/molar mass of O2 = 192g/32g/mol = 6 mol. Therefore, the moles of H2 = 6mol *2 = 12 mol.

So the mass of H2 that reacts with O2 = moles of H2 * molar mass of H2 = 12 mol * 2 g/mol = 24 g

7 0

4 years ago

Answer question number 6

Romashka [77]

It will be the first one

4 0

4 years ago

Read 2 more answers

How does the vapor pressure of water at 10°C compared with its vapor pressure at 50°C ?

Sliva [168]

Answer: The vapor pressure of water at 10°C will be less as compared with its vapor pressure at 50°C.

Explanation:

Vapor pressure of a liquid is defined as the pressure exerted by the vapors in equilibrium with the liquid/solution at a particular temperature.

As Kinetic energy is dependent on the absolute temperature of the gas.

$K.E=\frac{3}{2}RT$

where R = gas constant

T = temperature

On increase in temperature, the kinetic energy of the molecules increase and thus more liquid molecules can escape to form vapours and thus will exert more vapor pressure.

Thus the vapor pressure of water at 10°C will be less as compared with its vapor pressure at 50°C.

4 0

3 years ago

Which household items would be considered a base? Select all that apply. *

mrs_skeptik [129]

Answer:

Drain Cleaner, soap, bleach

i think thats it, sorry if i got something wrong

6 0

3 years ago

Calculate the enthalpy of the reaction

harkovskaia [24]

Answer : The enthalpy of the reaction is, -2552 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given enthalpy of reaction is,

$4B(s)+3O_2(g)\rightarrow 2B_2O_3(s)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $B_2O_3(s)+3H_2O(g)\rightarrow 3O_2(g)+B_2H_6(g)$ $\Delta H_A=+2035kJ$

(2) $2B(s)+3H_2(g)\rightarrow B_2H_6(g)$ $\Delta H_B=+36kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_C=-285kJ$

(4) $H_2O(l)\rightarrow H_2O(g)$ $\Delta H_D=+44kJ$

Now we have to revere the reactions 1 and multiple by 2, revere the reactions 3, 4 and multiple by 2 and multiply the reaction 2 by 2 and then adding all the equations, we get :

(when we are reversing the reaction then the sign of the enthalpy change will be change.)

The expression for enthalpy of the reaction will be,

$\Delta H=-2\times \Delta H_A+2\times \Delta H_B-6\times \Delta H_C-6\times \Delta H_D$

$\Delta H=-2(+2035kJ)+2(+36kJ)-6(-285kJ)-6(+44)$

$\Delta H=-2552kJ$

Therefore, the enthalpy of the reaction is, -2552 kJ/mole

4 0

3 years ago