Question 1 Points 3 23 and Louis immerses his left hand in a beaker containing cold water and immerses his right hand in a beaker containing warm water. Then, he immerses both his hands on a beaker containing water at room temperature. Which of the following statements are true? 1. The hand that was in hot water would feel cold. 2. The hand that was in cold water would feel hot. 3. His two hands will feel the same hotness. Que O2 and 3 0 1 and 2 o 1 and 3 1.2, and 3​

Answer:

Hydrogen and fluorine form only one bond while carbon forms four bonds to other atoms.

Explanation:

This question brings us to the idea of valency. Fluorine is univalent while carbon is tetravalent.

Univalent means that fluorine can only form one bond to hydrogen while carbon forms as much as four bonds because it is tetravalent.

Hence fluorine and hydrogen have only one peg while carbon has four.

Answer:

0.465

Explanation:

To find the volume of a substance, divide the mass by the density.

M/D = V

10.0 / 21.5 = 0.4651163

Then round to 3 significant figures: and the density is 0.465

Answer:

Arrange the following compounds in order of acidity (highest to lowest): H2O, CH3COOH , HCl

A. CH3COOH > HCl > H2O

B. H2O > CH3COOH > HCl

C. HCl > H2O > CH3COOH

D. HCl > CH3COOH > H2O

Explanation:

The given substances are acetic acid, hydrochloric acid, and water.

Since HCl is a strong acid and it undergoes complete ionization.

CH3COOH acetic acid is a weak acid and it undergoes partial dissociation in water.

Pure water is a neutral substance.

Hence, the order of acidity is shown below:

HCl > CH3COOH > H2O.

Among the given options, option D is the correct answer.

Answer:

The answer is 20

Explanation:

1) The order of decreasing priority would allow determining E/Z or R/S is "dbca".

2)  The order of decreasing priority would allow determining E/Z or R/S is "acbd".

Absolute configuration can be described as to the spatial arrangement of atoms within a chiral molecular entity. Absolute configuration in organic molecules, where carbon is bonded to four different substituents.

The absolute configuration has used a set of rules to describe the relative positions around the chiral center atom. The most common labeling method is the descriptors R or S where R and S refer to Rectus and Sinister.

The group with the highest atomic number will get the highest priority and the group with the lowest atomic number substituents will get the lowest priority. Therefore, the order of priority is -CH₂Br > -CH₂OH > -CH₂NH₂ > -CH₃.

Therefore, the order of priority for the second part is -F > -CHO > -CH₂OH > -CH₃.

Learn more about absolute configuration, here:

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Answer:

Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane

Explanation:

The reaction of (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) forms the following product:

The answer is (R)-2-chlorobutane.

The reaction take splace through [tex]S_{N} _2[/tex] mechansim and inversion in configuration happens.

Answer:

Water and Aluminium oxide

Explanation:

Have a nice day.

Answer:

dissolved in the solute

Explanation:

A solvent is the component that dissolves the solute and is present in larger amount. The type of solution is determined by the state of the solute and solvent. If you have NaCl, a solid, dissolved in water, a liquid, the type of solution is a solid/liquid solution.

Answer:

We know that the student was measuring a spectrum, and observed double yellow lines, he claimed that it was Sodium.

There are multiple elements with double yellow lines, like Mercury or Sodium, but Sodium has two bright yellow lines, so it is usually identified by them.

So when we look at a spectrum and we see a strong doublet in the yellow range, we can easily assume that it is Sodium.

Here we assume that the student only saw the yellow doublet, this would imply that the yellow doublet is way more intense than the other lines, that can't be seen (while for other elements with double yellow lines, we should see other lines with similar intensity) then we can conclude that it is Sodium.

The student is correct.

Solution :

a). The separation of 4-methylbenzoic acid from 1,4-dimethoxybenzene will work but it will result in lower recovery.

In the reaction of acid-base to form a sodium 4 - methoxy benzoate, that is soluble in the water, 4-methoxy benzoic acid reacts with the sodium bicarbonate to give sodium 4-methoxybenzoate as well as carbonic acid.

b). The pKa for the 4-methoxybenzoic acid is [tex]4.46[/tex], and that of carbonic acid is [tex]6.37[/tex]

c). The Keq for the reaction is [tex]10(6.37 - 4.46) = 101.91[/tex]

Therefore, the equilibrium lies to the right  and also the reaction favors the products and the separation works.

But the recovery will be low when compared to the extraction with Sodium hydroxide as the strong base will drive the equilibrium further to the right position, thus neutralizing all the acids virtually. And the weak base will partially neutralize the acid.

Answer:

The rate of the reverse reaction before any addition of CH3CO2- is zero

Explanation:

When the reaction is in equilibrium:

HCH3CO2(aq) + H2O ⇄ H3O+(l)(aq) + CH3CO2^-(aq)

The reaction quotient, Q = Ka and no more products or reactants are produced because their concentrations are in the right proportion.

Now, as no reaction occurs,

This question is incomplete, the complete question is;

A certain liquid X has a normal boiling point of 150.4 °C and a molar boiling point elevation constant kb is 0.60 °Ckgmol⁻¹.

A solution is prepared by dissolving some urea (NH22CO) in 750 g of X. This solution boils at 150.9 °C . Calculate the mass of urea that was dissolved. Round your answer to 3 significant digits.

Answer:

the mass of urea that was dissolved is 37.5 g

Explanation:

Given the data in the question;

normal boiling point of X; Tb⁰ = 150.4 °C

boiling point of solution Tb = 150.9 °C

Change in boiling point Δt = Tb - Tb⁰  = 150.9 °C - 150.4 °C = 0.5 °C

Kb = 0.6 °C.kg.mol⁻¹

V = 750 g

Now, we know that

Δt = Kb × molality

so

0.5 = 0.6 × molality

molality = 0.5 / 0.6

molality = 0.833

we know that molar mass of urea is 60 g/mol

so

molality = mass × 1000 / molar mass × volume( g )

we substitute

0.833 = ( mass × 1000 ) / ( 60 × 750 )

0.833 = ( mass × 1000 ) / 45000

0.833 × 45000 = mass × 1000

mass = ( 0.833 × 45000 ) / 1000

mass = 37485 / 1000

mass = 37.485 ≈ 37.5 g   { 3 significance figure }

Therefore, the mass of urea that was dissolved is 37.5 g

Answer:

118.06 mL

Explanation:

The neutralization reaction between HBr (acid) and Ba(OH)₂ (base) is the following:

2HBr + Ba(OH)₂ → BaBr₂ + 2H₂O

According to the equation, 2 moles of HBr react with 1 mol Ba(OH)₂. Thus, at the equivalence point the moles of acid and base react completely:

2 moles HBr = 1 mol Ba(OH)₂

We can replace the moles by the product of the molar concentration (M) and volume (V):

2 x (M HBr) x (V HBr) = M Ba(OH)₂ x V Ba(OH)₂

Now, we introduce the data in the equation to calculate the volume in mL of Ba(OH)₂:

V Ba(OH)₂ = (2 x (M HBr) x (V HBr))/M Ba(OH)₂

                 = (2 x 0.311 M x 57.7 mL)/(0.304 M)

                 = 118.06 mL

Therefore, 118 mL of Ba(OH)₂ are needed.

Answer:

Explanation:

H

Answer:

liquefied petroleum gas

LPG is prepared by refining natural gas. it is made by refining crude oil or from extracted natural gas streams as they emerge from the ground.

Answer:

True.

Explanation:

The information presented in the question above regarding linoleic acid is true. Linoleic acid is, in fact, found in many oils and fats in the ester form. In addition, linoleic acid is considered a polyunsaturated fatty acid, due to the presence of two unsaturations in its composition. Its chemical formula is CH3-(CH2)4-CH=CH-CH2-CH=CH-(CH2)7COOH and it is an essential fatty acid for the human body, as it is essential in the composition of arachidonic acid that is responsible for building muscle, managing body fat thermogenesis, and regulating core protein synthesis.

Answer:

The choice that is not an indicator of a chemical change is "State of matter changes". More common than not, chemical reactions produce energy in the form of light or heat. Along with energy, they also produce a new substance called the product that could be in any state of matter (solid, gas, or liquid).

Explanation:

Answer:

7.640 kg

Explanation:

Step 1: Write the balanced complete combustion equation for ethanol

C₂H₆O + 3 O₂ ⇒ 2 CO₂ + 3 H₂O

Step 2: Calculate the moles corresponding to 4 kg (4000 g) of C₂H₆O

The molar mass of C₂H₆O is 46.07 g/mol.

4000 g × 1 mol/46.07 g = 86.82 mol

Step 3: Calculate the moles of CO₂ released

86.82 mol C₂H₆O × 2 mol CO₂/1 mol C₂H₆O = 173.6 mol CO₂

Step 4: Calculate the mass corresponding to 173.6 moles of CO₂

The molar mass of CO₂ is 44.01 g/mol.

173.6 mol × 44.01 g/mol = 7640 g = 7.640 kg

anode is oxidation

so look at the reduction potential for Pt and Kl

the one with the smaller reduction potential will undergo oxidation

the one with the larger reduction potential will undergo reduction

you have to flip the equation that undergoes oxidation because the reduction table always gives reduction equations

Answer: There are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.

Explanation:

Given: [tex]M_{1}[/tex] = 1.0 M,    [tex]V_{1}[/tex] = ?

[tex]M_{2}[/tex] = 0.1 M,    [tex]V_{2}[/tex] = 200 mL

Formula used is as follows.

[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]

Substitute values into the above formula as follows.

[tex]M_{1}V_{1} = M_{2}V_{2}\\1.0 M \times V_{1} = 0.1 M \times 200 mL\\V_{1} = \frac{0.1 M \times 200 mL}{1.0 M}\\= 20 mL[/tex]

Thus, we can conclude that there are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.

Answer:

2

Explanation:

2 alkenes are formed by E2 elimination of HBr from 3-bromo -3, 4 - dimethylhexane using a strong base such as sodium methoxide

Answer:

a. [tex]Ag^0[/tex]

b. [tex]Ag^{+}[/tex]

c. [tex]Ca^{2+}F_2^-[/tex]

d. [tex]H_2^+S^{2-}[/tex]

e. [tex](C^{4+}O_3^{2-})^{-}[/tex]

f. [tex](Cr^{6+}O_4^{2-})^{2-}[/tex]

g. [tex]Cl_2^0[/tex]

h. [tex]Fe^0[/tex]

i. [tex]Cu^{2+}Cl_2^-[/tex]

j. [tex]C^{4-}H_4^+[/tex]

Explanation:

Hello there!

In this case, according to the concept of charge balance, which tell us that the overall charge is zero for any compound, except ions, it turns out possible to proceed as follows:

a. [tex]Ag^0[/tex]

b. [tex]Ag^{+}[/tex]

c. [tex]Ca^{2+}F_2^-[/tex]

d. [tex]H_2^+S^{2-}[/tex]

e. [tex](C^{4+}O_3^{2-})^{-}[/tex]

f. [tex](Cr^{6+}O_4^{2-})^{2-}[/tex]

g. [tex]Cl_2^0[/tex]

h. [tex]Fe^0[/tex]

i. [tex]Cu^{2+}Cl_2^-[/tex]

j. [tex]C^{4-}H_4^+[/tex]

Keep in mind lonely elements have 0 as their oxidation state.

Regards!

Answer:

oshfjidgshsjdh

Explanation:

918474828

Answer:

i have no answer for part A

part B

the one that has a 4 can be divided by 2 because reducing

part c

you can determine if an equation is written in the correct way by balancing the equation as if it had not been done already.

Answer:

a.draw 2,3 dicholoro octane

Explanation:

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Answer:

azertyuiopazertyuiiop

Answer:

The Rf value of ibuprofen increases

Explanation:

TLC involves the elution of a solute using a mobile phase(solvent). The stationary phase is made of an adsorbent such as silica.

The extent of interaction between the solute and the mobile phase affects the Rf value. The greater the interaction between the solute and the solvent, the greater the Rf value.

On the other hand, the polarity of the solvent and the solute also affects the Rf value. If the solvent is changed from t-butyl methyl ether to acetone, the Rf value for ibuprofen increases because ibuprofen is polar and acetone is also polar hence there is greater interaction between the solvent and solute.

Answer:

747.5 mL

Explanation:

Assuming ideal behaviour, we can solve this problem by using Boyle's law, which states that at constant temperature:

Where in this case:

We input the data given by the problem:

And solve for V₁: