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3 years ago

For the reaction 2H2+O2 2H2O how many liters of water can be made from 5.00 L of oxygen gas and an excess of Hydrogen at STP

Chemistry

1 answer:

Veseljchak [2.6K]3 years ago

8 0

5L O2 x 1 mol O2/ 22.4L O2 x 2 mol H20/ 1 mol O2 x 22.4L H20/ 1 mol H20 = 10 L H2O
For future reference though, since its at STP that means that the coefficient are in proportion. Since oxygen has a coefficient of 1 and water has a coefficient of 2 for every 1 liter of oxygen there is 2 liters of water. Hence you started with 5 liters and ended with 10

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Write the equilibrium constant expression for this reaction: CH3OH (aq)+Cl-(aq)-> CH3Cl(aq)+OH-(aq)

bulgar [2K]

Answer: $K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}$

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients. Pure solids are assumed to have a concentration of 1.

The given balanced equilibrium reaction is:

$CH_3OH(aq)+Cl^-(aq)\rightleftharpoons CH_3Cl(aq)+OH^-(aq)$

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}$

Thus the equilibrium constant expression for this reaction is $K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}$

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3 years ago

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stepan [7]

Answer:

i believe its 5.2

Explanation:

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3 years ago

Predict the polarity of 6 real molecules. First, draw the molecules and any bond dipoles. Then draw any molecular dipoles. Expl

il63 [147K]

Answer:

Explanation:

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3 0

3 years ago

At a certain temperature, the equilibrium constant, K c , for this reaction is 53.3. H 2 ( g ) + I 2 ( g ) − ⇀ ↽ − 2 HI ( g ) K

swat32

Answer:

$[HI]_{eq}=0.942M$

Explanation:

Hello,

In this case, the initial concentrations of hydrogen and iodine are the same:

$[H_2]_0=[I_2]_0=0.600M$

Thus, considering the given undergoing chemical reaction, one states the law of mass action in terms of the change $x$ due to the chemical change as shown below:

$Kc=\frac{(2x)^2}{(0.600M-x)(0.600M-x)}=53.3$

Therefore, solving for $x$ by quadratic equation one obtains:

$x_1=0.471M;x_2=0.826M$

Nevertheless, the feasible result is the first one as the second one results in negative concentrations, thus, the hydroiodic acid equilibrium concentration turns out:

$[HI]_{eq}=2*0.471M=0.942M$