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Fed [463]
3 years ago
15

Write the expression for the equilibrium constant Kp for the following reaction. Enclose pressures in parentheses and do NOT wri

te the chemical formula as a subscript. For example, enter (PNH3 )2 as (P NH3)2. If either the numerator or denominator is 1, please enter 1 2 MoO3(s) ↔ 2 MoO2(s) + O2(g)
Chemistry
1 answer:
madam [21]3 years ago
6 0

Answer: K_p={(p_{O_2})}

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K

K_p is the constant of a certain reaction at equilibrium for gaseous reactants and products.

For the given chemical reaction:

MoO_3(s)\rightleftharpoons 2MoO_2(s)+O_2(g)

The expression of  for above equation follows:

K_p={(p_{O_2})}

As solids do not exert pressure, MoO_3 and MoO_2  are not involved.

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Some antacid tables contain aluminum hydroxide. The aluminum hydroxide reacts with stomach acid according to the equation: Al(OH
MatroZZZ [7]

Answer:

26.67 mol HCl

Explanation:

Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O

In order to solve this problem, we need to c<u>onvert Al(OH)₃ moles to HCl moles</u>.

To do so we use the<em> stoichiometric ratios</em> of the balanced reaction:

  • 8.89 mol Al(OH)₃ * \frac{3molHCl}{1molAl(OH)_{3}} = 26.67 mol HCl

Thus 26.67 moles of HCl would react completely with 8.89 moles of Al(OH)₃.

4 0
2 years ago
PLEASE ANSWER QUICK I WILL MARK BRAINLIEST A spreadsheet is a _____. A. way to organize data on a computer B. place to list obse
Oxana [17]

Answer:

The answer is D part of your lab notebook used for writing vocabulary

Hope it helps!

3 0
3 years ago
If 75 grams of oxygen react, how many grams of aluminum are required?
german

Answer:

84.24 g

Explanation:

Given data:

Mass of oxygen = 75 g

Mass of Al required to react = ?

Solution:

Chemical equation:

4Al + 3O₂     →   2Al₂O₃

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 75 g/ 32 g/mol

Number of moles = 2.34 mol

Now we will compare the moles of oxygen with Al.

                          O₂         :          Al

                           3          :             4

                        2.34        :         4/3×2.34 = 3.12 mol

Mass of Al required:

Mass = number of moles × molar mass

Mass = 3.12 mol × 27 g/mol

Mass = 84.24 g

5 0
3 years ago
A student found that the titration had taken 10.00 ml of 0.1002 m naoh to titration 0.132 g of aspirin, a monoprotic acid. calcu
Harman [31]
The balanced equation for the reaction between NaOH and aspirin is as follows;
NaOH + C₉H₈O₄ --> C₉H₇O₄Na + H₂O
stoichiometry of NaOH to C₉H₈O₄ is 1:1
The number of NaOH moles reacted - 0.1002 M / 1000 mL/L x 10.00 mL 
Number of NaOH  moles - 0.001002 mol
Therefore number of moles of aspirin - 0.001002 mol
Mass of aspirin reacted - 0.001002 mol x 180.2 g/mol = 0.18 g
However the mass of the aspirin sample is 0.132 g but 0.18 g of aspirin has reacted, therefore this question is not correct.

4 0
3 years ago
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stepladder [879]

Answer:

Chemical

Explanation:

7 0
3 years ago
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