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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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In the simulation, open the Custom mode. The beaker will be filled to the 0.50 L mark with a neutral solution. Set the pH to 2.8

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3 years ago

A student mixes chemicals A and B together and records the amount of time it takes for a color change to occur. The student repe

k0ka [10]

Answer:

It is mentioned that the student is mixing chemicals A and B and observes the time taken for the color to change. However, in the experiment, it is noticed that the student has repeated the procedure five times and each time he or she is modifying the concentration of chemical B. Thus, it is clear that the concentration of chemical B is the independent variable in the experiment. An independent variable is illustrated as the variable, which is controlled or modified in the experiment.

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3 years ago

In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is SO42−(aq)+Sn2+(aq)→H2

allsm [11]

Answer:

The final balanced equation is :

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

Explanation:

$SO_4^{2-}(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+Sn^{4+}(aq)$

Balancing in acidic medium:

First we will determine the oxidation and reduction reaction from the givne reaction :

Oxidation:

$Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)$

Balance the charge by adding 2 electrons on product side:

$Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^-$ ....[1]

Reduction :

$SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)$

Balance O by adding water on required side:

$SO_4^{2-}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)$

Now, balance H by adding $H^+$ on the required side:

$SO_4^{2-}(aq)+4H^+(aq)\rightarrow H_2SO_3(aq)+H_2O(l)$

At last balance the charge by adding electrons on the side where positive charge is more:

$SO_4^{2-}(aq)+4H^+(aq)+2e^-\rightarrow H_2SO_3(aq)+H_2O(l)$ ..[2]

Adding [1] and [2]:

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

The final balanced equation is :

$SO_4^{2-}(aq)+4H^+(aq)+Sn^{2+}(aq)\rightarrow H_2SO_3(aq)+H_2O(l)+Sn^{4+}(aq)$

4 0

3 years ago

The student wants to test the conductivity of each solution. Prior to carrying out the investigation, the student needs to ident

puteri [66]

Answer:b

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3 0

2 years ago

Read 2 more answers

calculate pressure exerted by 1.255 mol of CI2 in a volume of 5.005 L at a temperature 273.5 k using ideal gas equation

balu736 [363]

Answer:

The pressure is 5.62 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

P= ?
V= 5.005 L
n= 1.255 mol
R= 0.082 $\frac{atm*L}{mol*K}$
T= 273.5 K

Replacing:

P* 5.005 L= 1.255 mol* 0.082 $\frac{atm*L}{mol*K}$ *273.5 K

Solving:

$P=\frac{1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K}{5.005 L}$

P= 5.62 atm

<u><em>The pressure is 5.62 atm.</em></u>

8 0

2 years ago