How many hydrogen atoms are present in a hydrocarbon chain of five carbon atoms with one double bond and the rest single bonds?

krek1111 [17]

First, let's start off by finding the mass of this whole hydrate.
(Note: the unit of measurement for mass will be amu)

Let's find the molecular mass of each element.
$Co=58.933$
$Cl=35.45$
$H=1.008$
$O=15.999$

Now, let's find the mass of each compound.

$CoCl_2=58.933+2(35.45)=129.833$
$H_2O=2(1.008)+15.999=18.015$

We have 6 molecules of H2O, so multiply 18.015 by 6 then add that with the weight of CoCl2.

$6(18.015)=108.09$
$129.833+108.09=237.923$

Now divide 108.09 (mass of all the H2O in the hydrate) by 237.923 (total mass of hydrate).

$\dfrac{108.09}{237.923}$

$\approx0.45431$

Turn that into a percentage and you get 45.431%.
Hope this helps! :)

6 0

3 years ago

Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine

yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

7 0

3 years ago

What is a polyprotic buffer?

coldgirl [10]

Buffers - mixtures of conjugate acid and conjugate base at ±1 pH unit from pH = pKa. Resistant to changes in pH in response to small additions of H+ or OH-. ... Polyprotic acids - dissociation of each H+ can be treated separately if the pKa values are different

3 0

3 years ago

Hi! My responsibility is to add and protection to plant cells. what i am

liq [111]

Probably a cell wall

4 0

3 years ago

*please help*

mojhsa [17]

Answer:

What was the verdict?

Explanation: