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N76 [4]
3 years ago
14

What is the simplest unit factor that relates the number of hydrogen in sucrose to the number of oxygens?

Chemistry
1 answer:
butalik [34]3 years ago
8 0

The simplest unit factor that relates the number of hydrogen in sucrose to the number of oxygens is 1 O atom/ 2 H atoms.

Sucrose is a diasaccharide having the formula; C12H22O11.

This implies that there are twenty two atoms of hydrogen and eleven atoms of oxygen in a molecule of sucrose.

This gives a ratio of 2 hydrogen atoms to one oxygen atom. Therefore, the simplest unit factor that relates the number of hydrogen in sucrose to the number of oxygens is 1 O atom/ 2 H atoms.

Learn more: brainly.com/question/11347582

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The standard cell potential, E°cell, for a reaction in which two electrons are transferred between the reactants is +1.33 V. Cal
Alinara [238K]

Answer:

ΔG° = 2.57 × 10² kJ

The reaction is spontaneous.

Explanation:

<em>The standard cell potential, E°cell, for a reaction in which two electrons are transferred between the reactants is +1.33 V. Calculate the standard free energy change, ΔG°, in kJ for this reaction and determine if it is spontaneous or nonspontaneous at 25°C.</em>

<em />

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = -n × F × E°cell

where,

n: moles of electrons transferred

F: Faraday's constant

E°cell: standard cell potential

ΔG° = - (2 mol) × (96468 J/V . mol) × 1.33 V

ΔG° = -2.57 × 10⁵ J = 2.57 × 10² kJ

ΔG° < 0 means that the reaction is spontaneous.

4 0
3 years ago
How many molecules does 88 grams CO2 contain? A) 1.2 X 1024 B) 1.8 X 1024 C) 3.0 X 1023 D) 6.0 X 1023
mylen [45]

Hey there!:

Molar mass CO2 = 44.01 g/mol

44.01 g CO2 ------------------- 6.02*10²² molecules CO2

88 g CO2 ------------------------ y

y = 88 * ( 6.02*10²² ) / 44.01

y = 5.29*10²⁵ / 44.01

y = 1.2 * 10²⁴ molecules of CO2

Answer A


Hope that helps!

7 0
3 years ago
Reaction of tert−butyl pentyl ether [CH3CH2CH2CH2CH2OC(CH3)3] with HBr forms 1−bromopentane (CH3CH2CH2CH2CH2Br) and compound H.
EleoNora [17]

Answer:The compound H is is 2-Methyl prop-1-ene and its structure can be found in attachment. The m/z value of 2-Methyl prop-1-ene is 56 which clearly matches with the mas spectrum data. The structure can also be ascertained using the provided IR data. The IR data has absorption stretching frequency in the region of 1650cm⁻¹ which is due to the C=C double bond. The stretching frequency at 3150-3000cm⁻¹ is due to the unsaturated C-H bond. The stretching frequency at 3000-2850cm⁻¹ is due to the saturated C-H bonds. Kindly refer attachment for the mechanism.

Explanation:

The reaction of tert-butyl ether with HBr leads to the formation of 1-bromopentane and tertbutyl alcohol.

The tert butyl alcohol formed undergoes E1 elimnation reaction to give 2-Methyl prop-1-ene.

The mass spectra and IR data available for the compound H completely matches with that of 2-Methyl prop-1-ene hence we can ascertain that the compound H is 2-Methyl prop-1-ene.

The m/Z value of 2-Methyl prop-1-ene  is 56 which is in compete accordance with the provided data for compound H .

The IR data also completely matches with the structure of 2-Methyl prop-1-ene as the following Infrared absorption peaks are provided which matches with that of 2-Methyl prop-1-ene :

The absorption stretching frequency in the region of 1650cm⁻¹ corresponds to the  C=C double bond which is clearly evident in 2-Methyl prop-1-ene.

The stretching frequency at 3150-3000cm⁻¹ corresponds to  unsaturated C-H bond.

The stretching frequency at 3000-2850cm⁻¹ is due to the saturated C-H bonds.

The mechanism of the reaction involves the following steps:

1. The oxygen atom in tert-butyl pentyl ether is protonated by treating it with Hydrogen bromide and Br⁻ is lost from hydrogen bromide.

2. Now since the oxygen atom is protonated it turns into a good leaving group and can leave as tertiary butyl alcohol. The eliminated Br⁻ now  attacks in a SN2 manner from the back side at the primary carbon center which leads to the formation of 1-Bromopentane and tertiary-butyl alcohol

3.The tert-butyl alcohol formed further reacts with HBr present to give elimiantion product 2-Methyl prop-1-ene  through E1 elimination mechanism. The OH is protonated  and further it gets eliminated  as H₂O leading to formation of  a tertiary carbocation. The tertiary carbocation formed gives an elimination product of 2-Methyl prop-1-ene .

Kindly refer the attachment for complete reaction mechanism.

8 0
3 years ago
Fill in the missing in information: (protons, electrons, atomic#, or
zloy xaker [14]

Answer:

sorry po Hindi ko po alam

3 0
3 years ago
Write a balanced chemical equation that describes an acid-base reaction that would allow you to determine the moles of aspirin p
Kryger [21]

A balanced equation representing the acid-base reaction that allows the calculation of the moles of aspirin in a sample is

C_{9} H_{8}O_{4} + NaOH⇒ C_{9} H_{7}O_{4} Na+ H_{2} O

Aspirin, also known as acetylsalicylic acid, is a nonsteroidal anti-inflammatory drug used to reduce pain, fever, and/or inflammation, and as an antithrombotic.

Aspirin is a benzoic acid with an ortho-substituted acylated alcohol function (actually a phenol). Therefore, two reactions can occur when aspirin and NaOH are combined: In fact, several aspirin formulations contain this ingredient.

Hence, A balanced equation representing the acid-base reaction that allows the calculation of the moles of aspirin in a sample is

C_{9} H_{8}O_{4} + NaOH⇒ C_{9} H_{7}O_{4} Na+ H_{2} O

To know more about Aspirin, here : brainly.com/question/23878261

#SPJ4

6 0
1 year ago
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