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kramer
3 years ago
11

Write the formula for the conjugate acid of each of the following bases. 1. NH₃

Chemistry
1 answer:
tino4ka555 [31]3 years ago
8 0

Answer:

According to Bronsted-Lowry theory, an<u> acid</u> is any species that can<em> donate</em> H+ (proton), while a<u> base</u> is any species that can<em> accept</em> a H+ ion.

  1. NH_{3}  

                          NH_{3}(aq) + H_{2}O(aq) → NH_{4}^{+}(aq)+ OH^{-}(aq)

NH_{4}^{+} is the conjugate acid of NH_{3}

      2. ClO_{3}^{-}

                           ClO_{3}^{-}(aq) + H_{2}O(aq) → HClO_{3}(aq) + OH^{-}(aq)

HClO_{3} is the conjugate acid of ClO_{3}^{-}

       3. SO_{4}^{2-}

                             SO_{4}^{2-}(aq) + H_{2}O(aq) → HSO_{4}^{-}(aq) + OH^{-}(aq)

HSO_{4}^{-} is the conjugate acid of  SO_{4}^{2-}  

         4. OH^{-}

                              OH^{-}(aq) + H^{+}(aq) → H_{2}O(aq)

H_{2}O is the conjugate acid of OH^{-}

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This reaction was at equilibrium when 0.2 atm of iodine gas was pumped into the container, what happened to the equilibrium and
n200080 [17]

Answer:

Q was < K. Partial pressure of hydrogen decreased, iodine increased

Explanation:

<em>After iodine was added the Q was [Select] K so the reaction shifted toward the Products [Select] ,The partial pressure of hydrogen [Select], Iodine [Select] |,and hydrogen iodide Decreased</em>

Based on the equilibrium:

H2(g) + I2(g) ⇄ 2HI(g)

K of equilibrium is:

K = [HI]² / [H2] [I2]

<em>Where [] are concentrations at equilibrium</em>

And Q is:

Q = [HI]² / [H2] [I2]

<em>Where [] are actual concentrations of the reactants.</em>

<em />

When the reaction is in equilibrium, K=Q.

But as [I2] is increased, Q decreases and Q was < K

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7 0
3 years ago
How many grams of water are produced when 2.50 mol oxygen reacts with hydrogen?
storchak [24]

Given:

2.5 mol O2

Required:

Grams of water

Solution:

2H2 + O2 -> 2H2O

Moles of H2O = (2.5 mol O2)(2 moles H2O/1 mole O2) = 5 moles H2O

Molar mass of H2O = 18.02 g/mol

Mass of H2O = (5 moles H2O)( 18.02 g/mol)

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3 years ago
How many atoms are in 52.3 g of lithium hypochlorite (LiClO)?​
garri49 [273]

Answer:

1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.

Explanation:

To find the amount of atoms that are in 52.3 g of lithium hypochlorite, we must first find the amount of moles. We do this by dividing by the molar mass of lithium hypochlorite.

52.3 g ÷ 58.4 g/mol = 0.896 mol

Next we must find the amount of formula units, we do this be multiplying by Avagadro's number.

0.896 mol × 6.02 × 10²³ = 5.39 × 10²³ f.u.

Now to get the amount of atoms we can multiply the amount of formula units by the amout of atoms in one formula unit.

5.39 × 10²³ f.u. × 3 atom/f.u. = 1.62 × 10²⁴ atoms

1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.

7 0
2 years ago
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