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marissa [1.9K]
3 years ago
9

There are certain trends with which you should become very familar (recognizing these trends will save you time!) This part of t

he question is dedicated to that task.
a) Whenever you see 9 or a multiple of 9 in the integration ratio, which group should you first consider for being responsible for that signal.
b) Whenever you see a quartet and triplet on a spectrum, which group should you first consider for being responsible for those signals?
c) Whenever you see septet and a doublet on a spectrum, which group should you first consider for being responsible for those signals?
d) Whenever you see 6 or a multiple of 6 in the integration ratio, which group should you first consider for being responsible for that signal
e) Whenever you see 3 as the actual number of protons for a given signal, which group should you first consider for being responsible for that signal

Chemistry
1 answer:
Gemiola [76]3 years ago
6 0

Answer:

Explanation:

The objective of this  question is all about identifying the phenomena that holds true for the statement being said in each instance. Let; walk through them.

a) Whenever you see 9 or a multiple of 9 in the integration ratio, which group should you first consider for being responsible for that signal.

( C₄H₉ )Tert. Butyl group

b) Whenever you see a quartet and triplet on a spectrum, which group should you first consider for being responsible for those signals?

(CH₃CH₂)  ethyl group

c) Whenever you see septet and a doublet on a spectrum, which group should you first consider for being responsible for those signals?

(CH(CH₃)₂)  isoproply group

d) Whenever you see 6 or a multiple of 6 in the integration ratio, which group should you first consider for being responsible for that signal

(CH(CH₃)₂)  isoproply group

e)  Whenever you see 3 as the actual number of protons for a given signal, which group should you first consider for being responsible for that signal.

CH₃- methyl group

A perfect description showing  the explanation of each answers chosen is explained with an aid of diagram below.

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The table shows columns that brenda uses for her notes on the properties of elements. her notes state that some elements can rea
AleksandrR [38]

The element should be placed in the column for metalloids. Option A

<h3>What is the property of metals?</h3>

We know that the properties of metals can be used to identify an unknown substances as a metal. We know that metals are able to react with acids to liberate the metal salt and hydrogen gas. Metals do not react with bases because they can not accept electrons.

Metalloids are generally amphoteric in nature. They can react with both the acid and the basses to form compounds. The nonmetals do not react with acid or base.

Looking at the result, we are told that the entry in the second row have the entries  nothing, nothing, nothing. If this is the outcome, then it follows that the element should be placed in the column for metalloids.

Learn more about metalloids:brainly.com/question/2548493

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4 0
1 year ago
Use Hess's Law to determine the enthalpy change (∆H) for the reaction: ClF + F2 → ClF3 Given: 2ClF + O2 → Cl2O + F2O. ∆H=167.4kJ
emmasim [6.3K]

Answer:

The enthalpy change (∆H) for the reaction is -108.7 kJ

Explanation:

Hess's law can be stated as: when the reactants are converted to products, the enthalpy change is the same, regardless of whether the reaction is carried out in one step or in a series of steps. Then, Hess's Law states that the enthalpy of one reaction can be achieved by algebraically adding the enthalpies of other reactions.

So,  to calculate the ∆H (heat of reaction) of the combustion reaction, that is, the heat that accompanies the entire reaction, you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient ( number of molecules of each compound participating in the reaction) and finally subtract them.

Enthalpy of combustion = ΔH = ∑Hproducts - ∑Hreactants

2 ClF + O₂ → Cl₂O + F₂O ∆H=167.4kJ

Cl₂O + 3 F₂O → 2 ClF₃ + 2 O₂ ∆H= -341.4kJ  

The previous equation must be inverted, and the enthalpy value is also inverted, that is, the sign is changed.

2 F₂ + O₂ →2 F₂O ∆H=-43.4kJ

Reactants and products are added or canceled, taking into account that certain substances sometimes appear as a reagent and others as a product, so they are totally eliminated (there is nothing left of them anywhere in the reaction, if the same amount in reagents and products) or partially (this substance remains, in less quantity, only on one side), obtaining:

2 ClF + 2 F₂ → 2 ClF₃

Then, as all the reactants and products have a stoichiometric coefficient of 2, dividend by that number is obtained:

ClF + F₂ → ClF₃

Adding the enthalpies algebraically, and dividing by 2, because to get the "data" reaction you had to multiply by two, you get:

ΔH= [167.4 kJ - 341.4 kJ - 43.3 kJ]÷2

ΔH= -108.7 kJ

<u><em>The enthalpy change (∆H) for the reaction is -108.7 kJ</em></u>

3 0
3 years ago
Help for this practice work
allochka39001 [22]

Answer:

It's the third option.

Explanation:

In order for the chemical equation to be correctly it needs the same number of atoms of each element on both sides of the equal sign

8 0
2 years ago
How are you! :D I hope your doing good
Tanya [424]

Answer:

Yeah, Fine..

Explanation:

5 0
2 years ago
An enzyme with molecular weight of 310 kDa undergoes a change in shape when the substrate binds. This change can be characterize
oee [108]

Answer:

(a) r = 6.26 * 10⁻⁷cm

(b) r₂ = 6.05 * 10⁻⁷cm

Explanation:

Using the sedimentation coefficient formula;

s =  M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle

s = M ( 1 - Vρ) / N*6πnr

making r sbjct of formula, r =  M (1 - Vρ) / N*6πnrs

Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s

r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)

r = 6.26 * 10⁻⁷cm

b. Using the formula r₂/r₁ = s₁/s₂

s₂ = 0.035 + 1s₁ = 1.035s₁

making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁

r₂ = 6.3 * 10⁻⁷cm / 1.035

r₂ = 6.05 * 10⁻⁷cm

8 0
3 years ago
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