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Naddika [18.5K]
2 years ago
8

List the salts in order from weakest to strongest lattice energy (most positive to most negative). magnesium chloride, barium ch

loride, calcium chloride, strontium chloride Group of answer choices magnesium chloride < calcium chloride < strontium chloride < barium chloride calcium chloride < barium chloride < magnesium chloride < strontium chloride magnesium chloride < barium chloride < calcium chloride < strontium chloride strontium chloride < calcium chloride < barium chloride < magnesium chloride barium chloride < strontium chloride < magnesium chloride < calcium chloride barium chloride < strontium chloride < calcium chloride < magnesium chloride calcium chloride < magnesium chloride < strontium chloride < barium chloride
Chemistry
1 answer:
almond37 [142]2 years ago
6 0

Answer: Option (6) is the correct answer.

Explanation:

Lattice energy is described as the energy released when formation of 1 mole of an ionic compound occurs due to the combination of its constituent ions.

Also, lattice energy is inversely proportional to the distance between the cation and anion. And, when we move down a group then there occurs an increase in the atomic radii of the atoms.

This means that smaller is the ionic distance between the cation and anion, more will be the lattice energy between the atoms.

Therefore, order from weakest to strongest lattice energy (most positive to most negative) for the given compounds is as follows.

 barium chloride < strontium chloride < calcium chloride < magnesium chloride

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Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ
kirza4 [7]

<u>Answer:</u> The molecular formula for the given organic compound is C_{18}H_{20}O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=39.61g

Mass of H_2O=9.01g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, \frac{12}{44}\times 39.61=10.80g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, \frac{2}{18}\times 9.01=1.00g of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = \frac{0.9}{0.10}=9

For Hydrogen = \frac{1}{0.10}=10

For Oxygen = \frac{0.10}{0.10}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is C_9H_{10}O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

n=\frac{268.34g/mol}{134g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2

Thus, the molecular formula for the given organic compound is C_{18}H_{20}O_2.

3 0
2 years ago
Consider the following incomplete reaction. Mg + 2Y ---&gt; MgCl₂ + H₂ Choose the formula for the missing substance Y.
vesna_86 [32]
It would be B as the answer
7 0
2 years ago
3. Scott and James work at a grocery store. After the grocery store closed, they were playing a game with a shopping cart and Sc
weqwewe [10]

Answer: B) Crash 2; the force on the cart was stronger in this crash, so the force on the skateboard was also stronger.

Explanation:

6 0
1 year ago
At higher elevations, the boiling point of water decreases, due to the decrease in atmospheric pressure. As a result, what could
amid [387]
When we say decrease in boiling point, that means, we achieve boiling at a more lower temperature (lower than 100deg C). This is due to the lower atmospheric pressure. Boiling happens when the vapor pressure is equal the atmospheric pressure. Lower atmospheric pressure takes lower temperature for vapor pressure to equate with the atmospheric pressure. The answer here is letter B. 
At higher elevations, it would take longer to hard boil an egg, because there is a lower boiling point, so the egg is boiling in water at a lower temperature.

8 0
2 years ago
When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capac
STatiana [176]

Answer:

\Delta _{comb}H=-2,093\frac{kJ}{mol}

Explanation:

Hello!

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

Q_{rxn}+Q_{cal}=0

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\  \Delta _{comb}H=-2,093\frac{kJ}{mol}

Best regards!

4 0
1 year ago
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