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3 years ago

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1 answer:

6 0

Saturated oils are oils where every carbon is sp3 hybridized and attached to two other carbons and two hydrogen. An unsaturated oil features a pi bond and a sigma bond between one or more carbons. This sigma bond interacts with Br2 by way of an addition reaction, the double bond is broken and two bromine are added to the carbon chain. The resulting structure is colorless so in a way the oil absorbs the colored Br2 into a colorless molecule. So, the more Saturated an oil is, the more Br2 it can accept and that's why Br2 can be used to detect the presence of satiated oils. While adding Br2, the solution will stay colorless as long as there are double bonds to accept it.

Hope this makes sense, if you've talked about reaction mechanisms this should be pretty straightforward.

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Consider the chemical equation for the ionization of CH3NH2 in water. Estimate the percent ionization of CH3NH2 in a 0.050 M CH3

anyanavicka [17]

Answer: The percent ionization of $CH_3NH_2$ in a 0.050 M $CH_3NH_2(aq)$ solution is 8.9 %

Explanation:

$CH_3NH_2+H_2O\rightarrow OH^-+CH_3NH_3^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_b=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= concentration = 0.050 M and $\alpha$ = degree of ionisation = ?

$K_b=4.4\times 10^{-4}$

Putting in the values we get:

$4.4\times 10^{-4}=\frac{(0.050\times \alpha)^2}{(0.050-0.050\times \alpha)}$

$(\alpha)=0.089$

percent ionisation = $0.089\times 100=8.9\%$

8 0

3 years ago

Write the ions present in a solution of na3po4. express your answers as chemical formulas separated by a comma. offset subscript

dexar [7]

Sodium/natrium is a metal from first column group so it should have one 1+ charge. Phosphate ion has 3- charge. That is why there 3 natrium ion for 1 phosphate ion when this molecule is dissolved in water. The ion formula would be:

(Na) $_{3}$ (PO $_{4}$ ) ==> 3 Na $^{+}$ + PO $_{4} ^{3-}$

5 0

3 years ago

you find that 7.36g of a compound has decomposed to give 6.93g of oxygen. the only other element in the compound is hydrogen. if

slavikrds [6]

First solve the moles of oxgen present in the compound

mol O = 6.93 g O ( 1 mol O / 16 g O )
mol O = 0.43 mol H

then solve the moles of hydrogen present
mol H = ( 7.36 - 6.93) g H ( 1 mol H / 1 g H)
mol H = 0.43 mol H
so the O and H are in the same mole content so the molecular formula would be OH, but the molar mass will not satisfy. so the answer would be
H2O2

8 0

3 years ago

Read 2 more answers

Please help! Thanks :D

Digiron [165]

Species shown in bold are precipitates.

Ca(NO₃)₂ + 2 KOH → Ca(OH)₂ + 2 KNO₃
Ca(NO₃)₂ + Na₂C₂O₄ → CaC₂O₄ + 2 NaNO₃

Cu(NO₃)₂ + 2 KI → CuI₂ + 2 KI
Cu(NO₃)₂ + 2 KOH → Cu(OH)₂ + 2 KNO₃
Cu(NO₃)₂ + Na₂C₂O₄ → CuC₂O₄ + 2 NaNO₃

Ni(NO₃)₂ + 2 KOH → Ni(OH)₂ + 2 KNO₃
Ni(NO₃)₂ + Na₂C₂O₄ → NiC₂O₄ + 2 NaNO₃

Zn(NO₃)₂ + 2 KOH → Zn(OH)₂ + 2 KNO₃
Zn(NO₃)₂ + Na₂C₂O₄ → ZnC₂O₄ + 2 NaNO₃

A double replacement reaction takes place only if it reduces in the concentration of ions in the solution. For example, the reaction between Ca(NO₃)₂ and KOH produces Ca(OH)₂. Ca(OH)₂ barely dissolves. The reaction has removed Ca²⁺ and OH⁻ ions from the solution.

Some of the reactions lead to neither precipitates nor gases. They will not take place since they are not energetically favored.

Compare the first and last row:

Both Ca(NO₃)₂ and Zn(NO₃)₂ react with KOH. However, between the two precipitates formed, Ca(OH)₂ is more soluble than Zn(OH)₂.

As a result, add the same amount of KOH to two Ca(NO₃)₂ and Zn(NO₃)₂ of equal concentration. The solution that end up with more precipitate shall belong to Zn(NO₃)₂.

Compare the second and third row:

Cu(NO₃)₂ reacts with KI, but Ni(NO₃)₂ does not. Thus, add equal amount of KI to the two unknowns. The solution that forms precipitate shall belong to Cu(NO₃)₂.

8 0

3 years ago

Write the chemical formula for disilicon hexabromide.

Rasek [7]

Answer:

Br6Si2

Explanation:

5 0

3 years ago