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2 years ago

Chlorine and bromine have very similar chemical properties. This is best explained by the fact that both elements:

Chemistry

1 answer:

34kurt2 years ago

7 0

What is the similarity between chlorine and bromine?

Chlorine and bromine have very similar chemical properties and are best explained by the fact that they both belong to the category of halogens.

Similarities shared by halogens:

Chemical elements are represented as periods and groups on a periodic table. Chemical elements with comparable properties make up each category. Halogens are group 17 elements that are primarily non-metallic.

In the case of halogens, each element's atom possesses seven valence electrons, $ns^{2} np^{5}$ , in which n stands for the energy level. The electron configuration of chlorine is $[Ne]3s^{2} 3p^{5}$ , and the electron configuration of bromine is $[Ar]3d^{10}4s^{2} 4p^{5}$ .

Because of comparable electron configurations, elements of the same group have similar properties. The number of valence electrons, which are located in the outermost (highest energy) s and p orbitals, is particularly similar.

Learn more about halogens here;

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How much heat energy, in kilojoules, is required to convert 72.0 gg of ice at −−18.0 ∘C∘C to water at 25.0 ∘C∘C ? Express your a

Ghella [55]

Answer: The enthalpy change is 34.3 kJ

Explanation:

The conversions involved in this process are :

$(1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

m = mass of water = 72.0 g

$c_{s}$ = specific heat of ice = $2.09J/g^0C$

$c_{l}$ = specific heat of liquid water = $4.184J/g^0C$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6010 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]$ $\Delta H=34279.8J=34.3kJ$ (1 KJ = 1000 J)

Therefore, the enthalpy change is 34.3 kJ

5 0

3 years ago

Consider the following reaction: 3Fe(s) + 4H₂O(g) ➞ Fe₃O₄(s) + 4H₂(g). To answer the following question: "How many moles of hot

Elodia [21]

Answer: 2

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Fe_3O_4=\frac{275g}{233.5g/mol}=1.18moles$

$3Fe(s)+4H_2O(g)\rightarrow Fe_3O_4(s)+4H_2(g)$

According to stoichiometry :

1 mole of $Fe_3O_4$ are produced by = 4 moles of $H_2O$

Thus 1.18 moles of $Fe_3O_4$ will be produced by= $\frac{4}{1}\times 1.18=4.72moles$ of $H_2O$

Mass of $H_2O=moles\times {\text {Molar mass}}=4.72moles\times 18g/mol=85.0g$

Thus 85.0 g of $H_2O$ will be required and 2 steps are required to get the answer.

6 0

3 years ago

3. Unlike global winds, local winds (select the two answers that are correct)

Viktor [21]

Answer:

Small-scale convection currents arise from uneven heating on a smaller scale. This kind of heating occurs along a coast and in the mountains. Small-scale convection currents cause local winds. Local winds blow over a much smaller area and change direction and speed over a shorter period of time than global winds.

Maybe that will help you answer the question.

6 0

3 years ago

When octane (C8H18) is burned in the presence of oxygen, the yield of products (carbon dioxide and water) is 87%. What mass of c

ahrayia [7]

Answer:

14.5g of CO₂ are produced

Explanation:

The reaction of octane with oxygen is:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

<em>Where 1 mole of octane (Molar mass: 114.23g/mol) reacts with 25/2 moles of O₂ (Molar mass 32g/mol) to produce 8 moles of CO₂ and 9 moles of water.</em>

When 21.0 g of octane is burned with 19.0 g of oxygen gas you need to find <em>limiting reactant </em>to find how many moles of products are formed:

Octane: 21.0g ₓ (1mol / 114.23g) = 0.184 moles octane

Oxygen: 19.0g ₓ (1 mol / 32g) = 0.594 moles oxygen

For a complete reaction of 0.184 moles of octane you will need:

0.184 moles C₈H₁₈ ₓ (25/2 moles O₂ / 1 mole C₈H₁₈) = <em>2.3 moles of oxygen</em>

As you have just 0.594 moles of oxygen, <em>Oxygen is limiting reactant.</em>

Based on chemical equation, 25/2 of O₂ produce 8 moles of CO₂, that means theoretical yield of CO₂ with 0.594 moles of O₂ is:

0.594 moles O₂ ₓ (8 moles CO₂ / 25/2 moles O₂) = 0.380 moles of CO₂

But, as yield of products is 87%, moles produced of CO₂ are:

0.380 moles of CO₂ ₓ 87% = 0.331 moles CO₂ are produced.

As molar mass of CO₂ is 44g/mol, mass of CO₂ in 0.331 moles is:

0.331 moles CO₂ ₓ (44g / mol) =

<h3>14.5g of CO₂ are produced</h3>

6 0

3 years ago

Which of the four models represents a solid

ICE Princess25 [194]

Answer:

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Explanation:

6 0

3 years ago