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3 years ago

The net ionic equation for the reaction between aqueous sulfuric acid and aqueous sodium hydroxide is ________.

1 answer:

5 0

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

2H⁺ + SO₄²⁻ + 2Na⁺ + 2OH⁻ → 2Na⁺ + SO₄²⁻ + 2H₂O

H⁺ + OH⁻ → H₂O (the net ionic equation)

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tiny-mole [99]

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b) 590

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3 years ago

What is boiling point

frutty [35]

Answer:

Its the temperature at which the molecules of a matter especially a liquid attain higher kinetic energy and the rate of collision becomes increased

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2 years ago

10.0 grams of water are heated during the preparation of a cup of coffee 1.0x 103 j of the heat are added to the water. which is

katovenus [111]

<u>Answer:</u> The final temperature of the coffee is 43.9°C

<u>Explanation:</u>

To calculate the final temperature, we use the equation:

$q=mC(T_2-T_1)$

where,

q = heat released = $1.0\times 10^3J=1000J$

m = mass of water = 10.0 grams

C = specific heat capacity of water = 4.184 J/g°C

$T_2$ = final temperature = ?

$T_1$ = initial temperature = 20°C

Putting values in above equation, we get:

$1000J=10.0g\times 4.184J/g^oC\times (T_2-20)\\\\T_2=43.9^oC$

Hence, the final temperature of the coffee is 43.9°C

6 0

3 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.