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prisoha [69]
3 years ago
9

For the reaction: 2 A (g) + B (s)= 2 C(s) + D (g)

Chemistry
1 answer:
jolli1 [7]3 years ago
8 0

Answer:

The value of the equilibrium constant = 5.213

Explanation:

Here K_p (equilibrium constant) is referred to as the  partial pressure of product divided by the  partial pressure of reactant with each pressure term raised to power that is equal to its stoichiometric coefficient in balanced equation .

As such only gas appear in K_p  expression as solids takes a value of 1;

SO ; in the  given equation from the question:

2 A (g) + B (s) ----> 2 C(s) + D (g)

K_p = \dfrac{[D]}{[A]^2}

K_p = \dfrac{2.71}{0.721^2}

K_p = 5.213

The value of the equilibrium constant = 5.213

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How many moles in 2 h2o molecules
alexandr1967 [171]

Answer:

2 molecules of water represents 3.32 x 10^-24 moles of water.

Explanation:

To find the solution to this problem, you have to use the concept of Avogadro´s number, that is in 1 mol of any element o compound there are                    6.022 x 10^23 molecules. Then,

1 mol H2O ------------- 6.022 x 10^23 molecules

x= 3.32 x 10^-24 ---- 2 molecules.

2 molecules of water represents 3.32 x 10^-24 moles of water.

5 0
3 years ago
Stoichionetry Quiz
Maru [420]

Answer:

a

Explanation:

first form your equation and it is C3H8+5O2-4H2O+3CO2 44g of C3H8 produce 3moles of CO2/400g of C3H8 produce x the answer is 27.27moles when you cross multiply

4 0
3 years ago
A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO2 to effuse
ludmilkaskok [199]

Answer:

\large \boxed{\text{5.9 s}}

Explanation:

Graham’s Law applies to the effusion of gases:

The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

r \propto \dfrac{1}{\sqrt{M}}

If you have two gases, the ratio of their rates of effusion is

\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}

The time for diffusion is inversely proportional to the rate.

\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}

Let CO₂ be Gas 1 and O₂ be Gas 2

Data:

M₁ = 44.01

M₂ = 32.00

Calculation

\begin{array}{rcl}\dfrac{t_{2}}{t_{1}} & = & \sqrt{\dfrac{M_{2}}{M_{1}}}\\\\\dfrac{t_{2}}{\text{5 s}}& = & \sqrt{\dfrac{44.01}{32.00}}\\\\& = & \sqrt{1.375}\\t_{2}& = & \text{5 s}\times 1.173\\& = & \mathbf{5.9 s} \\\end{array}\\\text{It will take $\large \boxed{\textbf{5.9 s}}$ for the carbon dioxide to effuse.}

4 0
3 years ago
Calculate the mass of water formed when 4g of methane is burnt in excess oxygen​
wariber [46]

Answer:

Mass of water formed: 9g

Explanation:

Hope it helps you!

6 0
2 years ago
10) The presence of nitrates in soil can be shown by warming the soil with
tresset_1 [31]

Answer:

ammonia

Explanation:

Nitrogen fertilizers contain N in the forms of ammonium, nitrate and urea. Upon application to the soil, urea-N rapidly hydrolyzes to ammonia, thus it shares similar characteristics as ammonia-based N fertilizers.

5 0
2 years ago
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