All categories

3 years ago

HNO3 + H2O → H3O^+ + NO3 which ones are acids and which ones are bases

Chemistry

1 answer:

nlexa [21]3 years ago

4 0

Answer:

HNO₃ and H₃O⁺ are acids

H₂O and NO₃⁻ are bases

Explanation:

The chemical equation is:

HNO₃ + H₂O → H₃O⁺ + NO₃⁻

There are several definitions of acid and bases: Arrhenius', Bronsted-Lowry's and Lewis'.

Bronsted-Lowry model defines and acid as a donor of protons, H⁺.

In the given equation HNO₃ is such substance: it releases an donates its hdyrogen to form the H₃O⁺ ion.

On the other hand, a base is a substance that accepts protons.

In the reaction shown, H₂O accepts the proton from HNO₃ to form H₃O⁺.

Thus, H₂O is a base.

In turn, on the reactant sides the substances can be classified as acids or bases.

H₃O⁺ contain an hydrogen that can be donated and form H₂O; thus, it is an acid (the conjugated acid), and NO₃⁻ can accept a proton to form HNO₃; thus it is a base (the conjugated base).

You might be interested in

When 50.0 mL of 1.27 M of HCl(aq) is combined with 50.0 mL of 1.32 M of NaOH(aq) in a coffee-cup calorimeter, the temperature of

sergij07 [2.7K]

Answer:

-55.9kJ/mol is the change in enthalpy of the reaction

Explanation:

In the reaction:

HCl(aq) + NaOH(aq) → H₂O(l) + NaCl

Some heat is released per mole of reaction.

To know how many moles reacts we need to find limiting reactant:

Moles HCl = 0.050L ₓ (1.27mol / L) = 0.0635 moles HCl

Moles NaOH = 0.050L ₓ (1.32mol / L) = 0.066 moles NaOH

As there are more moles of NaOH than moles of HCl, HCl is limiting reactant and moles of reaction are moles of limiting reactant, 0.0635 moles

Using the coffee-cup calorimeter equation we can find how many heat was released thus:

Q = C×m×ΔT

Where Q is heat released, C is specific heat of the solution (4.18J/g°C), m is mass of solution (100g because there are 100mL of solution -50.0mL of HCl and 50.0mL of NaOH- and density is 1g/mL) and ΔT is change in temperature (8.49°C)

Replacing:

Q = 4.18J/g°C×100g×8.49°C

Q = 3548.8J of heat are released in the reaction

Now, change in enthalpy, ΔH, is equal to change in heat (As is released heat ΔH < 0) per mole of reaction, that is:

ΔH = Heat / mol of reaction

ΔH = -3548.8J / 0.0635 moles of reaction

Negative because is released heat.

ΔH = -55887J / mol

ΔH =

<h3>-55.9kJ/mol is the change in enthalpy of the reaction</h3>

3 0

3 years ago

A gas that was heated to 150 Celsius has a new volume of 1587.4 L. What was its volume when its temperature was 100 Celsius?

sdas [7]

Answer:

$\boxed {\boxed {\sf 1058.3 \ L}}$

Explanation:

We are asked to find the new volume of a gas after a change in temperature. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula for this law is:

$\frac {V_1}{T_1}= \frac{V_2}{T_2}$

The gas was heated to 150 degrees Celsius and had a volume of 1587.4 liters.

$\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{T_2}$

The temperature was 100 degrees Celsius, but the volume is unknown.

$\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{100 \textdegree C}$

We are solving for the volume at 100 degrees Celsius, so we must isolate the variable V₂. It is being divided by 100°C and the inverse of division is multiplication. Multiply both sides of the equation by 100°C.

$100 \textdegree C *\frac {1587.4 \ L }{150 \textdegree C} = \frac {V_2}{100 \textdegree C} * 100 \textdegree C$

$100 \textdegree C *\frac {1587.4 \ L }{150 \textdegree C} = V_2$

The units of degrees Celsius cancel.

$100 *\frac {1587.4 \ L }{150 } = V_2$

$100 *10.58266667 \ L = V_2$

$1058.266667 \ L = V_2$

The original measurement of volume has 5 significant figures, so our answer must have the same. For the number we calculated, that is the tenth place. The 6 in the hundredth place to the right tells us to round to 2 up to a 3.

$1058.3 \ L = V_2$

The volume of the gas at 100 degrees Celsius is approximately 1058.3 liters.

6 0

3 years ago

Please help! Why is cooler, drier air related to High Pressure

azamat

Due to density differences among two air masses. Due to high air pressure in cooler, drier gases, the air is supposed tobe rather dense and it does, mind you, tend to move towards warm or moist areas.

7 0

3 years ago

Shadow club!!!!!!!!!!

pychu [463]

WHATS THATT ???????????????

7 0

3 years ago

Read 2 more answers

How many moles are in 3.4×10^23 molecules of H2SO4

Kazeer [188]

1 mole ----------- 6.02x10²³ molecules
? moles ---------- 3.4x10²³ molecules

(3.4x10²³) x 1 / 6.02x10²³ =

3.4x10²³ / 6.02x10²³ => 0.564 moles of H2SO4

6 0

3 years ago