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4 years ago

When an alkane reacts with an element from group 7a, the reaction is referred to as?

Chemistry

2 answers:

Mila [183]4 years ago

8 0

Answer:

Halogenation.

Explanation:

Hello,

In this case, we must remember that group VIIA elements are known as halogens which include F, Cl, Br, I and At. Now, when an alkane reacts with the diatomic form of one of those halogen a haloganation chemical reaction occurs forming an alkyl halide and a hydrogen halide. For instance, for ethane, if it reacts with iodine, ethyl iodide and hydrogen iodide are yielded based on the given example:

$CH_3-CH_3+I_2\rightarrow CH_3-CH_2-I+HI$

Best regards

Elena-2011 [213]4 years ago

7 0

Alkanes are saturated hydrocarbon, that is they contain hydrogen and carbon without a double or triple bond between the carbon atoms, e.g. ethane, propane. Group 7a in the periodic table are called halogens e.g chlorine, bromine. Alkanes react with halogens in a reaction called substitution, where halogens replace hydrogen atoms in alkanes.

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3 years ago

If 4.12 l of a 0.850 m-h3po4 solution is be diluted to a volume of 10.00 l, what is the concentration of the resulting solution?

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The process in which the concentration of the solution is lessened by the addition of water is said to be dilution and equation of dilution relates the initial concentration and volume of stock solution with the final concentration and volume of the solution.

Formula is given by:

$C_{1}V_{1}=C_{2}V_{2}$ (1)

where,

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$V_{2}$ = 10.00 L

Substitute the give values in formula (1),

$0.850 M\times 4.12L=C_{2}\times 10.00 L$

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3 years ago

Determine the pH of 0.050 M HCN solution. HCN is a weak acid with a Ka equal to 4.9 x 10-10<br> DONE

nadezda [96]

Answer:

The pH of the solution is 5.31.

Explanation:

Let " $\alpha$ is the dissociation of weak acid - HCN.

The dissociation reaction of HCN is as follows.

$HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}$

Initial C 0 0

Equilibrium c(1- $\alpha$ ) c $\alpha$ c $\alpha$

Dissociation constant = $Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}$

$=\frac{c\alpha^{2}}{(1-\alpha)}$

In this case weak acids $\alpha$ is very small so, (1- $\alpha$ ) is taken as 1.

$Ka=C\alpha^{2}$

$\alpha=\sqrt\frac{ka}{c}$

From the given the concentration = 0.050 M

Substitute the given value.

$\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}$

$[H_{3}O^{+}]=c\alpha$

$[H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}$

$pH= -log[H_{3}O^{+}]$

$=-log[4.9\times10^{-6}]$

$=6-log 4.9= 5.31$

Therefore, The pH of the solution is 5.31.

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