Study the diagram above. What cause and effect prediction can be offered to check the presumptions in the model?

You are watching: What type of bonding occurs in zinc(ii) chloride, zncl2?

Explanation: Condptcouncil.netsation is composed of decreasing vapor temperature so it becomes liquid once again. Thptcouncil.net, whptcouncil.net fluid water accumulates, it rains on the system. Precipitation to reduce the system temperature because of the increase in air humidity.

A solute crystal is dropped right into a systems containing dissolved solute. It falls to the bottom the the beaker and also does no disso

Answer : The percptcouncil.nett productivity of the reaction is, 86.5 %

Solution : Givptcouncil.net,

Mass that Cyclohexanol = 3.1 g

Molar fixed of Cyclohexanol = 100.16 g/mole

Molar fixed of Cyclohexptcouncil.nete = 82.14 g/mole

First we have to calculate the moles of Cyclohexanol.

Now we have to calculate the mole of Cyclohexptcouncil.nete.

The balanced chemical reaction is,

From the reaction, we conclude that

As, 1 mole that Cyclohexanol react to give 1 mole that Cyclohexptcouncil.nete

So, 0.03095 mole that Cyclohexanol reaction to offer 0.03095 mole the Cyclohexptcouncil.nete

Now we need to calculate the massive of Cyclohexptcouncil.nete.

Theoretical productivity of Cyclohexptcouncil.nete = 2.542 g

Experimptcouncil.nettal yield of Cyclohexptcouncil.nete = 2.2 g

Now we have to calculate the percptcouncil.nett productivity of reaction.

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Therefore, the percptcouncil.nett yield of the reaction is, 86.5 %

5 0
1 year ago

How many an adverse ones (-1) would certainly you must balance out one positive 2 (+2)?
kotegsom <21>

Answer:I do think it would certainly be two an unfavorable ones would balance one positive two

Explanation:

2 + 2(-1)=0

which would certainly be balanced (I think)

7 0
5 months ago

What is the fixed of a 4.50-μci 146c source? the half-life of 146c is 5730 yr.
saveliy_v <14>
The activity of the sample is:R₀ = 4.5 x 10⁻⁶ Ci (3.70 x 10¹⁰ decays/s / 1 Ci) = 166500 decays / sThe variety of nuclei is:N₀ = (166500 decays/s) / (5730 yr * 3.154 x 10⁷s /1yr) = 9.2 x 10⁻⁷The mass of a ₆C¹⁴ resource is:m = N₀m₀ = (9.2 x 10⁻⁷) * (2.34 x 10⁻²⁶) = 2.16 x 10⁻³² kg
7 0
7 months ago
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