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ValentinkaMS [17]
4 years ago
9

The elementary reaction 2H20(g)<--->2H2(g)+O2(g) proceeds at a certain temperature until the partial pressures of H2O, H2,

and O2 reach 0.0700 atm, 0.00200 atm, and 0.00600 atm respectively. What is the value of the equilibrium constant at this temperature?
Chemistry
1 answer:
olya-2409 [2.1K]4 years ago
3 0

Answer:

4.9 x 10⁻⁶.

Explanation:

  • For the reaction:

<em>2H₂O(g) ⇄ 2H₂(g) + O₂(g),</em>

'

P of H₂ = 0.002 atm, P of O₂ = 0.006 atm, and P of H₂O = 0.07 atm.

∴ The equilibrium constant (Kp) = (P of H₂)²(P of O₂)/(P of H₂O)² = (0.002 atm)²(0.006 atm)/(0.07 atm)² = 4.9 x 10⁻⁶.

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Write an equation that shows the formation of a selenide ion from a neutral selenium atom
Georgia [21]

Answer:

Se + 2e- => Se-2

Explanation:

Ions are electrically charged particles, formed due to the gain or loss of electrons by an atom.

There are two types of ions cations and anions. An element that loses their electrons and forms positive ions are cations while metals and element that gains one or more electrons and forms negative ions are anions.

Selenium is an atom with atomic number 34 and represented as Se. The electronic configuration of Se is  1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4. Selenium will gain 2 more electrons to complete it p orbital and form an ion Se-2.

The equation is as below:

Se + 2e- => Se-2

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3 years ago
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Ivahew [28]

Answer:

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3 years ago
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

3 0
3 years ago
Which block of elements do the halogens belong to?<br> a. p block<br> c. d block<br> b. s block?
kykrilka [37]
A. p block

from 3a-8a are in the p block including halogen
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3 years ago
As a cooking method, braising is valued for its A. ability to retain flavor. B. quickness and convenience. C. simplicity and ver
LiRa [457]

Answer:

i belevie its A

Explanation:

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