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miss Akunina [59]
2 years ago
15

How many degrees of unsaturation are lost or gained in the base-induced transformation from mito-ph to its photoactive ring-open

ed form?
Chemistry
1 answer:
Fudgin [204]2 years ago
6 0

Unsaturation (IHD) 2 hydrogen Needed

IHD = [(2n+2) -H]/2
(H: X=1, N=-1, O= zero)

Unsaturation:
Double bonds = 1
Rings = 1
Triple Bonds = 2

The degrees of unsaturation in a molecule are additive — a molecule with one double bond has one degree of unsaturation, a molecule with two double bonds has two degrees of unsaturation, and so forth.



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If you have ever peeled the label off a glass jar, you may have noticed that the glue does not easily wash off with water. Howev
shepuryov [24]

Answer:

It can be removed by acidic chemicals

Explanation:

8 0
2 years ago
Calculate the heat of reaction, ΔH°rxn, for overall reaction for the production of methane, CH4.
Lesechka [4]

<u>Answer:</u> The enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

For the given chemical reaction:

C(s)+2H_2(g)\rightarrow CH_4(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CH_4(g))})]-[(1\times \Delta H^o_f_{(C(s))})+(2\times \Delta H^o_f_{(H_2(g))})]

We are given:

\Delta H^o_f_{(C(s))}=0kJ/mol\\\Delta H^o_f_{(H_2)}=0kJ/mol\\\Delta H^o_f_{CH_4}=-74.9kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-74.9))]-[1\times 0)+(2\times 0)]\\\\\Delta H^o_{rxn}=-74.9kJ

Hence, the enthalpy of the reaction for the production of CH_4 is coming out to be -74.9 kJ

3 0
2 years ago
In 200 g of a concentrated solution of 70.4 wt% nitric acid (r = 1.41 g/mL, FW(HNO3) = 63.01 g/mol), how many grams of water are
mars1129 [50]

Answer:

59.2 grams

Explanation:

We are given that 70.4% of the weight of the total 200 g of the concentration is made up of nitric acid, the remaining information is not required to solve the problem. Since water and nitric acid are the only components of the solution, the total weight of water is given by:

W = 200*(1-0.704)\\W=59.2\ g

There are 59.2 grams of water in this solution.

5 0
2 years ago
During an experiment, the percent yield of calcium chloride from a reaction was 85.22%. Theoretically, the expected amount shoul
devlian [24]

The formula to find yield is

(Actual Yield)/(Theorectical Yield) x100

Just do the math.

85.22% x 113 = 96.2986

Convert it to 3 significant figures

Ans: 96.3g

7 0
2 years ago
Read 2 more answers
What would the % p, on the npk ratio, be reported as for a 100g sample containing 7.5g of phosphorous?
kati45 [8]
When the concentration is expressed in percentage, you simply have to divide the amount of substance to the total amount of the mixture, then multiply it by 100. In this case, when you want to find the percentage of P in the sample, the solution is as follows:

%P = 7.5 g/100 g * 100 =<em> 7.5%</em>
6 0
1 year ago
Read 2 more answers
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