Insoluble compounds in a chemical reaction are called?

A student is studying the rate of the following reaction: C2H4O NaOH --> H20 NaC2H3O Knowing that this is an exothermic react

emmasim [6.3K]

Answer:

HCl is not a catalyst because these are not used up during the chemical reactions.

Explanation:

Hello there!

In this case, according to the performed experiments, it is possible for us to realize that HCl cannot be a catalyst for this reaction because it is used up during the reaction. This is explained by the fact that catalyst are able to return to the original form once the reaction has gone to completion; this is the example of palladium in the hydrogenation or dehydrogenation of hydrocarbons depending on the case. Moreover, we know that the catalysts increase the reaction rate because they decrease the activation energy of the reaction and therefore the student observed such increase.

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8 0

3 years ago

An electron charge e mass m and a positron charge e mass m revolve around their common center of mass under the influence of the

Katen [24]

Answer: v = 2π2 Kme2 Z / nh

Explanation:

The formula for velocity of an electron in the nth orbit is given as,

v = 2π2 Kme2 Z / nh

v = velocity

K = 1/(4πε0)

m= mass of an electron

e = Charge on an electron

Z= atomic number

h= Planck’s constant

n is a positive integer.

3 0

4 years ago

Describe how the pH of a solution can be calculated

hjlf

Answer:

The pH of an aqueous solution is the measure of how acidic or basic it is. The pH of an aqueous solution can be determined and calculated by using the concentration of hydronium ion concentration in the solution.

Explanation:

6 0

3 years ago

If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),

garri49 [273]

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] (1)

Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻

pH = 7.40

pKa = 7.20

[A-] + [HA] = 0.10moles (2)

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

Then, you need to dilute both solutions to 2.00L with distilled water.

4 0

3 years ago

What is the product of the unbalanced equation below?

Stolb23 [73]

Answer:

D

Explanation:

The answer is D. I'm not sure that it is a solid. I don't think it is a ppte, which is the only way it can be a true solid. It is ionic if the reaction is taking place in water and there is someway to start the reaction. Be that as it may, the internal balace numbers of the chemical produced is the only possible answer. The balanced eq;uatioon is

2Al + 3Br2 ==> 2AlBr3

3 0

3 years ago