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Katarina [22]
3 years ago
6

Will mark brainliest. Anyone willing to do this?

Chemistry
1 answer:
Norma-Jean [14]3 years ago
7 0

Answer:

ACTIVITY 1

Sample 1 has a stronger taste of lemon, and is more sour.

Sample 2 has a sweeter taste, my guess is because there's more sugar:lemon juice ratio.

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What type of reactive intermediate is formed in the reaction of an alkene with aqueous acid to give an alcohol?
Mariana [72]

Answer:

The intermediate is carbocation.

Explanation:

The reaction of alkene with an acid to give alcohol is an electrophilic addition reaction.

In electrophilic reaction, the alkene is being attacked by an alkene.

in case of aqueous acid reaction with alkene , the electrophile is [H⁺].

When the proton attacks the alkene it generates carbocation.

The carbocation generated is formed based on stability of carbocation.

The mechanism is shown in the figure.

5 0
3 years ago
Rationalize the difference in boiling points between the members of the following pairs of substances. Part APart complete HF (2
Mrrafil [7]

Answer:

HF has the higher boiling point because HF molecules are more polar. Part B: CHBr3 molecules possess stronger intermolecular interaction due to higher molar mass than CHCl3

Explanation:

Fluorine is more electronegative than chlorine. This implies that HF is more polar and possess stronger hydrogen bonds than HCl molecules.

In part B, the magnitude of dispersion forces depend on molar mass, the greater the molar mass, the greater the magnitude of dispersion forces between molecules, hence CHBr3 has a greater boiling point than CHCl3

4 0
3 years ago
6.0 L of oxygen gas is at a temperature of 5K. If the temperature of the gas is lowered to 1K at constant pressure, what is the
DerKrebs [107]

Answer:

THE NEW VOLUME AT 1 K IS 1.2 L

Explanation:

Using Charles' law which states that the volume of a given gas is directly proportional to its temperature provided the pressure remains constant.

Mathematically written as;

V1/T1 = V2/T2   at constant pressure

V1 = Initial volume = 6L

T1 = initial temperature = 5K

T2 = Final temperature = 1K

V2 = final volume = unknown

Re-arranging the equation by making V2 the subject of the equation, we obtain;

V2 = V1 T2 / T1

V2 = 6 * 1 / 5

V2 = 1.2 L

The new volume of the gas sample at 1 K is 1,2 L

6 0
3 years ago
Read 2 more answers
Consider the reaction described by the following chemical equation.
oee [108]

Answer:

Q = -897 kJ/mol

Explanation:

From the given information:

The heat released Q = -65.9 kJ

To start with the molar mass of H_2O_2 = 2 × (molar mass of H) + 2 × (molar mass of O)

= (2 × 1.008) + (2 × 16.0 )

= 34.016 g/mol

However, given that:

mass of H_2O_2  2.50 g

The number of moles of H_2O_2  = \dfrac{mass}{molar \ mass}

= \dfrac{2.5}{34.016}

= 7.349 \times 10^{-2} \ mol

Finally; Using the formula:

\Delta H = \dfrac{Q}{number \ of \ moles}\\ \\ Q = \dfrac{-65.9 \ kJ}{7.349 \times 10^{-2} \ mol}

Q = -897 kJ/mol

6 0
2 years ago
What kind of particles make up a compound
kap26 [50]
Two or more elements make up a compound
5 0
3 years ago
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