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3 years ago

Saturated hydrocarbons contain carbon and hydrogen atoms that have ____ bond(s). \ 1 or 2?

Chemistry

2 answers:

Tom [10]3 years ago

7 0

Saturated has only 1 c to c bond while unsaturated has 2 or 3 c to c bond

jasenka [17]3 years ago

5 0

Answer : Saturated hydrocarbons contain carbon and hydrogen atoms that have 1 bond.

Explanation :

Hydrocarbon : It is an organic compound that contains only hydrogen and the carbon atoms.

They are of two types which are saturated hydrocarbon and unsaturated hydrocarbon.

Saturated hydrocarbons : It is defined as the hydrocarbons in which a single bond is present between the carbon-carbon atoms. The general formula for these hydrocarbons is $C_nH_{2n+2}$

Unsaturated hydrocarbons : It is defined as the hydrocarbons which have double or triple covalent C-C bonds. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is $C_nH_{2n}$ and $C_nH_{2n-2}$

Hence, from this we conclude that the saturated hydrocarbon contains 1 bond between the carbon and hydrogen atoms.

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arlik [135]

Answer:

Explanation:

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3 years ago

Please help, i will mark you as brainliest

Mrrafil [7]

Answer:

The order of reactivity of metals is as follows, Potassium > Sodium > Lithium > Calcium > Magnesium > Aluminium > Zinc > Iron > Copper > Silver > Gold.

Explanation:

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Iron<Zinc<Magnesium<Sodium

4 0

2 years ago

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A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?

Viktor [21]

The answer for the following problem is mentioned below.

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.

Explanation:

Given:

mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 125.3 grams

We know;

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 + 3(95)

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 +285 = 405 grams

We also know;

No of molecules at STP conditions( $N_{A}$ ) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

$\frac{m}{M} =\frac{N} }{}$ N÷ $N_{A}$

$\frac{405}{125.3}$ = $\frac{N}{6.023*10^23}$

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules

3 0

3 years ago

At sea level, where the pressure was 104 kPa and temperature 21.1 ºC, a certain mass of air occupies 2.0 m3 . To what volume wil

Romashka [77]

Answer:

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 104 kPa

$P_2$ = final pressure of gas = 52 kPa

$V_1$ = initial volume of gas = $2.0m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $21.1^oC=273+21.1=294.1K$

$T_2$ = final temperature of gas = $-5.0^oC=273+(-5.0)=268 K$

Now put all the given values in the above equation, we get:

$\frac{104 kPa\times 2.0m^3}{294.1 K}=\frac{52 kPa\times V_2}{268 K}$

$V_2=3.64 m^3$

The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is $3.64 m^3$ .

3 0

3 years ago

g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron

Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O Oxidation

BrO₃⁻ → Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ Reduction

In order to balance the main reaction and balance the electrons we multiply (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0

3 years ago