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Aleksandr
18 days ago
7

When Z-4,5-dimethyloct-4-ene is treated with hydrogen chloride, HCl, the result is:________.

Chemistry
1 answer:
alisha [2.9K]18 days ago
6 0
The proper choice is c.
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How many iodide ions are present in 65.5ml of .210 m AlI3 solution
castortr0y [3046]

Answer:

2.48\times 10^{22} ions are present in solution.

Explanation:

Molarity of the solution = 0.210 M

Volume of the solution = 65.5 ml = 0.0655 L

Moles of aluminum iodide = n

Molarity=\frac{\text{Moles of compound}}{\text{Volume of the solution(L)}}

0.210M=\frac{n}{0.0655 L}

n = 0.013755 moles of aluminum iodide

Each mole of aluminum iodide yields 3 moles of iodide ions:

Thus, 0.013755 moles of aluminum iodide will provide:

3\times 0.013755 moles=0.041265 mol moles of iodide ions

The total number of iodide ions in 0.041265 moles:

0.041265 mol\times 6.022\times 10^{23}=2.48\times 10^{22} ions

2.48\times 10^{22} ions are present in solution.

7 0
11 days ago
How many grams of AlF3 are in 2.64 moles of AlF3?
Anarel [2989]
To determine the mass of AlF3 in 2.64 moles of AlF3, we use the formula: mass = moles x molar mass, which results in 221.76 grams of AlF3.
3 0
1 month ago
How many grams of mercury are present in a barometer that holds 6.5mL of mercury? ...?
alisha [2963]
<span>13.6
grams of mercury are present in a barometer that holds
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3 0
1 month ago
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2CH4(g)⟶C2H4(g)+2H2(g)
Alekssandra [3086]

Answer: The enthalpy change for the reaction is, 201.9 kJ

Explanation:

Based on Hess’s law of constant heat summation, the energy released or absorbed in a chemical reaction stays constant, regardless of whether the process unfolds in one step or multiple steps.

This principle implies, that chemical equations can be treated analogously to algebraic expressions, allowing addition or subtraction to create the needed equation. Thus, the overall enthalpy change corresponds to the summation of the individual enthalpy changes of the reactions occurring in between.

The balanced equation for CH_4 appears as follows,

2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)    \Delta H^o=?

The intermediate balanced reactions are outlined as follows,

(1) CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)     \Delta H_1=-890.3kJ

(2) C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)     \Delta H_2=-136.3kJ

(3) 2H_2(g)+O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=-571.6kJ

(4) 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)     \Delta H_4=-3120.8kJ

Next, we will multiply the first reaction by 2, reverse the second, and reverse and halve the third and fourth reactions before combining them. This gives us:

(1) 2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)     \Delta H_1=2\times (-890.3kJ)=-1780.6kJ

(2) C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)    \Delta H_2=-(-136.3kJ)=136.3kJ

(3) H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)    \Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ

(4) 2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)     \Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ

Therefore, the expression for the enthalpy of the reaction is,

\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4

\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)

\Delta H=201.9kJ

Hence, the enthalpy change for this reaction is, 201.9 kJ

5 0
1 month ago
Two pure elements react to form a compound. One element is an alkali metal, X, and the other element is a halogen, Z. Which of t
VMariaS [2998]
The answer is option c. It consists of ionic bonds.Explanation: For instance, sodium chloride (NaCl) is an example where a group 1 element (an alkali metal) combines with a halogen. Elements from group 1 produce +1 charged cations. This ionic compound cannot possess the formula XZ₂. Ionic compounds are also known to dissolve in water and do not feature covalent bonds.
4 0
24 days ago
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