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2 years ago

What is the mean rate of reaction 3.4g of copper sulphate was produced in 3 days?

Chemistry

1 answer:

kramer2 years ago

3 0

Answer:

8.1×10^-8 mols-1

Explanation:

Now we have the mass of copper sulphate produced after three days. Recall that the rate of reaction is given as;

Rate= change in the concentration of product/time

At the beginning of the reaction, there was 0 moles of copper sulphate

After 72 hours or 259200 seconds, there was 3.4g/160gmol-1 = 0.021 moles of copper sulphate.

Note that 160gmol-1 is the molar mass of copper sulphate.

Hence;

Rate of reaction= 0.021 moles /259200 seconds

Hence, the rate of reaction is 8.1×10^-8 mols-1

Rate of reaction= 8.1×10^-8 mols-1

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Explain why you cannot just take away a proton to make a halogen negative.

Ilia_Sergeevich [38]

The force that holds protons and neutrons together is too strong to overcome.

<h3>Explanation</h3>

Consider the location of the particles in an atom.

Electrons are found outside the nucleus.
Protons and neutrons are found within the nucleus.

Protons carry positive charges and repel each other. The nucleus will break apart without the strong force that holds the protons and neutrons together. This force is much stronger than the attraction between the nucleus and the electrons. X-rays are energetic enough for removing electrons from an atom. However, you'll need a collider to remove protons from a stable nucleus. You could well have ionized the atom with all that energy.

Also, changing the number of protons per nucleus will convert the halogen atom to an atom of a different element. Rather than making the halogen negative, removing a proton will convert the halogen atom to the negative ion of a different element.

6 0

2 years ago

) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H

Elanso [62]

Answer:

1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).

2. The difference in pH values is 4.95.

Explanation:

1. The pH of a compound can be found using the following equation:

$pH = -log([H_{3}O^{+}])$

First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.

<u>Trimethyl ammonium</u>:

We can calculate [H₃O⁺] using the Ka as follows:

(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺

1.0 - x x x

$Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}$

$10^{-pKa} = \frac{x*x}{1.0 - x}$

$10^{-9.80}(1.0 - x) - x^{2} = 0$

By solving the above equation for x we have:

x = 0.097 = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01$

<u>Phenol</u>:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x x x

$Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}$

$10^{-10} = \frac{x^{2}}{1.0 - x}$

$1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0$

Solving the above equation for x we have:

x = 9.96x10⁻⁶ = [H₃O⁺]

$pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00$

Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.

2. The difference in pH values for the two acids is:

$\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95$

Therefore, the difference in pH values is 4.95.

I hope it helps you!

7 0

2 years ago

Bonds between two atoms that are equally electronegative are _____. bonds between two atoms that are equally electronegative are

Anna [14]

Bonds of two atoms of equal electronegativity are nonpolar covalent bonds.

Your second sentence is identical to the first sentence; I'll bet the second sentence is "Bonds between two atoms that are unequally electronegative are polar covalent bonds."

4 0

2 years ago

Jaiden is writing a report about the structure of the atom. In her report, she says that the atom has three main parts and two s

andre [41]

No, because an atom consists of two main parts and three subatomic
particles, protons, neutrons, and electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - e.g. Hydrogen has 1 proton and 1 electron. Neutrons do not effect the overall charge of the atom, and only increase the atomic mass.

7 0

2 years ago

Read 2 more answers

When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at

STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

$1.22 * 10^{5} = - 8.314* 2400 * ln K$

$\\ 1.22 * 10^{5} = -19953.6 * ln K$

$ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022$

So, the equilibrium constant is 0.0022.

4 0

2 years ago