What my happen in the ecosystem of the fox was removed

Korolek [52]

Carbonic acid.............................................

6 0

3 years ago

Compound a reacts with one equivalent of h2 in the presence of a catalyst to give methylcyclohexane. Compound a can be formed up

V125BC [204]

Answer:

The compound a is 1-methyl cyclohexene (see attachment for structure).

Explanation:

The reaction of 1-Bromo-1-methylcyclohexane with sodium methoxide is a second-order reaction since the methoxide ion is a strong base and also a strong nucleophile. This ion attacks the alkyl halide faster than the alkyl halide can ionize to produce a first-order reaction. However, we can not see the product of nucleophilic substitution. The SN₂ mechanism is blocked due to the impediment of the 1-Bromo-1-methylcyclohexane. The main product, according to the Zaitsev rule, is the 1-methyl cyclohexene, thus forming a double bond.

Then, this cyclohexene is hydrogenated to form the cyclohexane.

6 0

3 years ago

You take three compounds consisting of two elements and decompose them. To determine the relative masses of X, Y, and Z, you col

miss Akunina [59]

Answer:

a) LAW OF MULTIPLE PROPORTIONS

b) 0.095g, 0.71g, 0.285g respectively

c) X2Y, YZ15, X6Y

d) hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

Explanation:

a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.

for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.

b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y as our relative mass, this implies that the relative mass of Y = 1g

mass of X = 0.4g

mass of Y = 4.2g

amount of X in 1g of Y = 0.4 x 1 /4.2

= 0.095g

for compound 2;

mass of Y = 1.4g

mass of Z = 1.0g

amount of Z in 1g of Y =1.0 x 1 /1.4

= 0.71g

for compound 3;

mass of X = 2.0g

mass of Y = 7.0g

amount of X in 1g of Y = 1 x 2/7

= 0.285g

c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;

compound 1/compound 3 = 0.095/0.285

= 1/3

compound 1/compound 2 = 0.095/0.71

= 2/15

compound 2/ compound 3 = 0.71/0.285

= 5/2

formular for compound 1 = X2Y

formula for compound 2 = YZ15

formular for compound 3 = X6Y

d) from the formular X2Y, we can get the amount of each product in XY using the ratios

%of compound XY in X = mass of compound X / total Mass

= 0.2/4.4 = 4.5%

as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%

hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

5 0

3 years ago

2. Classify the following solutions as acidic, basic, or neutral at 25OC.

PilotLPTM [1.2K]

Answer: a) pH = 13.00 : basic

b) $[H_3O^+]=1.0\times 10^{-12}$ : basic

c) pOH = 5.00 : basic

d) $[OH^-]=1.0\times 10^{-9}$ : acidic

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H_3O^+]$

Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.

a) pH = 13.00

As pH is more than 7, the solution is basic.

b) $[H_3O^+]=1.0\times 10^{-12}$

Putting in the values:

$pH=-\log[1.0\times 10^{-12}]$

$pH=12$

As pH is more than 7, the solution is basic.

c) pOH = 5.00

$pH+pOH=14.0$

$pH=14.0-5.00=9.00$

As pH is more than 7, the solution is basic.

d) $[OH^-]=1.0\times 10^{-9}$

Putting in the values:

$pOH=-\log[1.0\times 10^{-9}]$

$pOH=9.00$

$pH+pOH=14.0$

$pH=14.0-9.00=5.00$

As pH is less than 7, the solution is acidic.

3 0

3 years ago

determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth m

mamaluj [8]

This is an incomplete question, here is a complete question.

Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.

(a) $1s^22s^22p^63s^23p^5$

(b) $1s^22s^22p^63s^23p^63d^74s^2$

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

(d) $1s^22s^22p^63s^23p^63d^4s^1$

Answer :

(a) $1s^22s^22p^63s^23p^5$ → Halogen

(b) $1s^22s^22p^63s^23p^63d^74s^2$ → Transition metal

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$ → Transition metal

(d) $1s^22s^22p^63s^23p^63d^4s^1$ → Transition metal

Explanation :

Inert gas : These are the gases which lie in group 18.

Their general electronic configuration is: $ns^2np^6$ where n is the outermost shell.

Halogen : These are the elements which lie in group 17.

Their general electronic configuration is: $ns^2np^5$ where n is the outermost shell.

An alkali metal : These are the elements which lie in group 1.

Their general electronic configuration is: $ns^1$ where n is the outermost shell.

An alkaline earth metal : These are the elements which lie in group 2.

Their general electronic configuration is: $ns^2$ where n is the outermost shell.

Transition elements : They are the elements which lie between 's' and 'p' block elements. These are the elements which lie in group 3 to 12. The valence electrons of these elements enter d-orbital.

Their general electronic configuration is: $(n-1)d^{1-10}ns^{0-2}$ where n is the outermost shell.

(a) $1s^22s^22p^63s^23p^5$

The element having this electronic configuration belongs to the halogen family.

(b) $1s^22s^22p^63s^23p^63d^74s^2$

The element having this electronic configuration belongs to the transition family.

(c) $1s^22s^22p^63s^23p^63d^{10}4s^24p^6$

The element having this electronic configuration belongs to the transition family.

(d) $1s^22s^22p^63s^23p^63d^4s^1$

The element having this electronic configuration belongs to the transition family.

4 0

2 years ago