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ozzi
3 years ago
6

In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions)

, which affects lattice energy (a measure of the attractive force holding those ions together). Based on ion sizes, arrange these compounds by their expected lattice energy. Note that many sources define lattice energies as negative values. Please arrange by magnitude and ignore the sign. |lattice energy|=absolute value of the lattice energy Greatest|lattice energy|. RbCl ,RbBr ,Rbl ,RbF.
Chemistry
1 answer:
Elena L [17]3 years ago
3 0

Answer:

RbI<RbBr<RbCl<RbF

Explanation:

As stated in the question, the latice energy depends on the relative size of the ions. When the action size is constant as in the question, the lattice energy now depends on the relative of the anions. The order of increase in ionic sizes among the halide ions is fluoride<Chloride<Bromide<Iodide. This order of increasing size means that the lattice energy will decrease accordingly as shown in the answer.

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A balloon contains 269.7 L of helium at 6.12ºC and 1.00 atm. What is the temperature (in ºC) of the gas if the volume has increa
MatroZZZ [7]

Answer:

T2= 7.3°C

Explanation:

To solve this problem we will use Charles law equation i.e,

V1/T1 = V2/T2    

Given data

V1 = 269.7 L

T1 = 6.12 °C

V2= 320.4 L

T2=?

Solution:

Now we will put the values in equation

269.7 L / 6.12°C  = 320.4 L / T2

T2=  320.4 L × 6.12°C/ 269.7 L

T2= 1960.85 °C. L /269.7 L

T2= 7.3°C

3 0
3 years ago
How many mole of oxygen are present in 3 mole of sodium sulfate, Na2SO4
GarryVolchara [31]
N(Na₂SO₄)=3 mol

n(O)=4n(Na₂SO₄)

n(O)=4*3=12 mol
3 0
3 years ago
When 125 grams of FeO react with 25.0 grams of Al, how many grams of Fe can be produced? FeO + Al → Fe + Al2O3 25.9 g Fe 38.7 g
Serga [27]

<u>Answer:</u> The mass of iron produced will be 77.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For FeO:</u>

Given mass of FeO = 125 g

Molar mass of FeO = 71.8 g/mol

Putting values in equation 1, we get:

\text{Moles of FeO}=\frac{125g}{71.8g/mol}=1.74mol

  • <u>For aluminium:</u>

Given mass of aluminium = 25.0 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{25.0g}{27g/mol}=0.93mol

The given chemical reaction follows:

3FeO+2Al\rightarrow 3Fe+Al_2O_3

By Stoichiometry of the reaction:

2 moles of aluminium metal reacts with 3 mole of FeO

So, 0.93 moles of aluminium metal will react with = \frac{3}{2}\times 0.93=1.395mol of FeO

As, given amount of FeO is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium metal is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 mole of iron metal

So, 0.93 moles of aluminium metal will produce = \frac{3}{2}\times 0.93=1.395moles of iron metal

  • Now, calculating the mass of iron metal from equation 1, we get:

Molar mass of iron = 55.85 g/mol

Moles of iron = 1.395 moles

Putting values in equation 1, we get:

1.395mol=\frac{\text{Mass of iron}}{55.85g/mol}\\\\\text{Mass of iron}=(1.395mol\times 55.85g/mol)=77.6g

Hence, the mass of iron produced will be 77.6 grams

4 0
3 years ago
11. What are transition metals?
kolezko [41]

Answer:

Elements that fall between those on the left and right sides of the periodic table

Explanation:

Transition metals:

These are present at the center of periodic table.

These are d-block elements.

They include the elements of group 3 to 12 in periodic table.

They have large charge to radius ratio.

They mostly form paramagnetic compounds.

They shoes more than one oxidation state.

They form colored compounds.

They all have high melting and boiling point.

They have high densities.

They form stable complexes.

The elements of f-block are also transition but they are called inner transition.These are consist of two series lanthanide and actinides.

8 0
3 years ago
When 1000 C of charge is passed through CuSO4 solution, x g of copper is deposited. How much charge should be passed through the
Stels [109]

Number of charge = 5018 C

<h3>Further explanation</h3>

Given

1000 C of charge for x grams of copper

Required

Number of charge

Solution

Faraday's Law :

\tt W=\dfrac{e.i.t}{96500}\\\\W=\dfrac{e.Q}{96500}

For 1000 C, W = x grams

\tt x=\dfrac{1000.e}{96500}=0.0104e

For 5x grams :

\tt 5x=\dfrac{e.Q}{96500}\\\\5\times 0.0104e=\dfrac{e.Q}{96500}\\\\Q=\dfrac{96500\times 5\times 0.0104e}{e}=5018~C

6 0
3 years ago
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