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2 years ago

What is the suffix for -✯ Alcohols✯ Ketones✯ Halogensthankyou ~

Chemistry

2 answers:

MA_775_DIABLO [31]2 years ago

8 0

Answer:

Kinematics is the study of motion of a system of bodies without directly considering the forces or potential fields affecting the motion. In other words, kinematics examines how the momentum and energy are shared among interacting bodies.

antoniya [11.8K]2 years ago

7 0

Answer:

ane, al, keto

ol, al, keto

ol, al, one

ol, ane, one.

Explanation:

The suffix –ol is used in organic chemistry principally to form names of organic compounds containing the hydroxyl (–OH) group, mainly alcohols (also phenol). The suffix was extracted from the word alcohol. The suffix also appears in some trivial names with reference to oils (from Latin oleum, oil).

Functional group is a ketone, therefore suffix = -one

Hydrocarbon structure is an alkane therefore -ane

The longest continuous chain is C5 therefore root = pent

The first point of difference rule requires numbering from the left as drawn to make the ketone group locant 2-

pentan-2-one or 2-pentanone

CH3CH2CH2C(=O)CH3

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A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na

s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f: after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

$Normality=molarity \times n-factor$

but in case of NaOH and HCl n-factor is 1 for each.

hence

normality=molarity;

At equivalence point: $N_1V_1=N_2V_2$

Before equivalence point : $N_1V_1>N_2V_2$

After the equivalence point: $N_1V_1$

$N_1V_1=100\times1=100$

case a: 5.00 mL of 1.00 M NaOH

$N_2V_2=5\times1=5$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

$N_2V_2=100\times1=100$

$N_1V_1=N_2V_2$ hence it is equivalence point

case c: 10.0 mL of 1.00 M NaOH

$N_2V_2=10\times1=10$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

$N_2V_2=150\times1=150$

$N_1V_1$ hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

$N_2V_2=50\times1=50$

$N_1V_1>N_2V_2$ hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

$N_2V_2=200\times1=200$

$N_1V_1$ hence it is after the eqivalence point

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3 years ago

The message that a neuron carries

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3 years ago

Which of the following statements best explains why Native Americans were overtaken so easily by European settlers?

MaRussiya [10]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

The equilibrium constant for the reaction

Hitman42 [59]

The question is incomplete, here is the complete question:

The equilibrium constant for the reaction

N₂O₄(g)⇌2NO₂ at 2°C is Kc = 2.0

If each yellow sphere represents 1 mol of N₂O₄(g) and each gray sphere 1 mol of NO₂ which of the following 1.0 L containers represents the equilibrium mixture at 2°C?

The image is attached below.

Answer: The system which represents the equilibrium having value of $K_c=2.0$ is system (b)

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$