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3 years ago

Is h2c2o4 an Arrhenius base or arrhenius acid

Chemistry

1 answer:

maksim [4K]3 years ago

6 0

Answer:
Oxalic Acid is and Arrhenius Acid.

Explanation:
According to Arrhenius Theory of acid and base, "Acid is any substance which when dissolved in water produces H⁺ Ions".

Therefore, Oxalic Acid is a diprotic substance, which is capable of donating protons in water. This acidity of oxalic acid can be dedicated to the stability of conjugate base, this stability comes from resonance of the negative charges on Oxalate ion. Below reaction shows the dissociation of Oxalic Acid into Protons and Oxalate Ion.

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What would be the water potential of pure water at atmospheric pressure?

ANTONII [103]

Zero is the value of water potential of pure water at atmospheric pressure.

4 0

2 years ago

If a hydrocarbon molecule contains a triple bond, its chemical name ends in

murzikaleks [220]

-yne. Hope this helps.

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3 years ago

A student places a 100.0°C piece of metal that weighs 85.5 g into 122 mL of 16.0°C water. If the final temperature is 20.2°C, wh

Musya8 [376]

Answer:

The specific heat of the metal is 0.314 J/g°C

Explanation:

Step 1: data given

Temperature of the piece of metal = 100.0 °C

Mass of the metal = 85.5 grams

Volume of water = 122 mL = 122 grams

Temperature of water = 16.0 °C

The final temperature of water = 20.2 °C

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

Step 2: Calculate the specific heat of metal

Heat gained= heat lost

Qgained = - Qlost

Qwater = -Qmetal

Q = m*c* ΔT

m(metal)*c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒m(metal) = mass of metal = 85.5 grams

⇒c(metal) = the specific heat of metal = TO BE DETERMINED

⇒ΔT(metal) = the change of temperature of metal = T2 - T1 = 20.2 - 100 °C = -79.8 °C

⇒m(water) = the mass of water = 122 grams

⇒c(water) = the specific heat of water = 4.184 J/g°C

⇒ΔT(water) = the change of temperature of metal = T2 - T1 = 20.2 - 16.0 °C = 4.2 °C

85.5 *c(metal) * -79.8 = -122 * 4.184 * 4.2

c(metal) * (-6822.9) = -2143.9

c(metal) = 0.314 J/g°C

The specific heat of the metal is 0.314 J/g°C

7 0

3 years ago

Given the following data:

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its $\Delta H$ can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with $\Delta H$ given be denoted as (1), (2), (3), and the last equation (4). Let $a$ , $b$ , and $c$ be letters such that $a \times (1) + b \times (2) + c \times (3) = (4)$ . This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, $3 + (-1) = 2$ shall resemble the number of $\text{H}_2$ left on the product side when the second equation is directly added to the third. Similarly

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

$a = -1/2\\b = 1/2\\c = -1/2$ and

$-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)$

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

$a + b = -1/2 + 1/2 = 0$

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

3 years ago

A 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, what is the volume? *

elena-14-01-66 [18.8K]

The new volume when pressure increases to 2,030 kPa is 0.8L

BOYLE'S LAW:

The new volume of a gas can be calculated using Boyle's law equation:

P1V1 = P2V2

Where;

P1 = initial pressure (kPa)
P2 = final pressure (kPa)
V1 = initial volume (L)
V2 = final volume (L)

According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:

406 × 4 = 2030 × V2

1624 = 2030V2

V2 = 1624 ÷ 2030

V2 = 0.8L

Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.

Learn more about Boyle's law calculations at: brainly.com/question/1437490?referrer=searchResults

3 0

2 years ago