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malfutka [58]
3 years ago
8

List the three requirements for a correctly written chemical equation.2. Give one example of a word equation, one example of a f

ormula equation, and one example of a chemical equation.3. Write formulas for each of the following compounds.a. potassium hydroxide b. calcium nitrate
Chemistry
1 answer:
erma4kov [3.2K]3 years ago
8 0

Answer:

Requirements for a correctly written chemical equation are reactants and products, their formula and valency

Explanation:

Formula of the given compound are -

1 - Potassium Hydroxide - KOH

2 - Calcium Nitrate - Ca(NO_3)_2

The requirements for a correctly written chemical equation are -

  • Identifying reactants and products
  • Formula of reactants and products
  • Valency of elements

Example of word equation, formula equation, and chemical equation is as follows -

Aluminium + iron9(III)oxide ⇒ aluminium oxide + iron (word equation)

Al_(_s_)  + Fe_2O_3_(_s_)    ⇒   Al_2O_3_(_s_)  + Fe_(s) (formula equation)              

2Al_(_s_) + Fe_2O_3_(_s_)    ⇒   Al_2O_3_(_s_)   + 2Fe_(_s_) (chemical equation)

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A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr
umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

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The correct answer is 3.
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