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3 years ago

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

Chemistry

2 answers:

Mila [183]3 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]3 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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A patient arrives in the emergency with a burn caused by steam. Calculate the heat that released when 17.7 g of steam

MaRussiya [10]

The steam releases 39.9 kJ when it condenses..

The steam condenses and transfers its energy to the skin.

q = mΔH_cond = 17.7 g × (-2257 J/1 g) = -39 900 kJ = -39.9 kJ

The negative sign shows that the steam is releasing energy

3 0

3 years ago

Consider two solutions: solution x has a ph of 4; solution y has a ph of 7. from this information, we can reasonably conclude th

KIM [24]

Consider two solutions: solution X has a pH of 4; solution Y has a pH of 7. From this information, we can reasonably conclude that the concentration of hydrogen ions (H⁺) or hydronium ions (H₃O⁺) in solution X is thousand times as great as the concentration of hydrogen ions or hydronium ions in solution Y.
Solution X: c(H⁺) = 10∧-pH = 10⁻⁴ mol/L = 0,0001 mol/L.
Solution Y: c(H⁺) = 10⁻⁷ mol/L = 0,0000001 mol/L.
0,0001 mol/L / 0,0000001 mol/L = 1000.

3 0

3 years ago

The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?

Vitek1552 [10]

The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
[OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
NaA → Na+ + A-
hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
A- + H2O ⇌ HA + OH-
Initial 0.0100 0 0
Change -x +x +x
Equilibrium 0.0100-x x x

We can now calculate the Kb for A-:
Kb = [HA][OH-] / [A-]
= x² / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
Ka = Kw / Kb
= 1.00x10^-14 / 1.00x10^-4
= 1.00x10^-10

7 0

3 years ago

True or false: The number of protons an element has can vary.

Lerok [7]

Answer:

true

Explanation:

5 0

3 years ago

Read 2 more answers

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

Answer: The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

5 0

3 years ago