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3 years ago

Istinguish between ionic and molecular compounds in terms of the basic

Chemistry

1 answer:

Murrr4er [49]3 years ago

6 0

Answer:

In molecular compound atoms are held together through covalent bond. It is formed between the non metals.

In ionic compounds positive and negative ions held together through electrostatic attraction. It is formed between metal and non metal.

Explanation:

Ionic compound:

In ionic compounds positive and negative ions held together through electrostatic attraction. It is formed between metal and non metal.

For example:

Sodium chloride is ionic compound. Te sodium metal lose it electron and this electron is accept by the chlorine atom. Sodium became positive ion while chlorine became negative ion. The forces develop between them are electrostatic forces.

Molecular compound:

In molecular compound atoms are held together through covalent bond. It is formed between the non metals.

For example:

Carbon dioxide is molecular compound. It is made up of carbon and oxygen atoms. Both atoms share the electrons with each other and form two double bond O=C=O.

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3 years ago

A student dissolves of urea in of a solvent with a density of . The student notices that the volume of the solvent does not chan

Dimas [21]

The question incomplete , the complete question is:

A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.

Answer:

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

Explanation:

Moles of urea = $\frac{18.0 g}{60 g/mol}=0.3 mol$

Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

$Molarity(M)=\frac{\text{Moles of compound}}{\text{Volume of solution in L}}$

Molarity of the urea solution ;

$M=\frac{0.3 mol}{0.200 L}=1.50 M$

Mass of solvent = m

Volume of solvent = V = 200.0 mL

Density of the urea = d = 0.95 g/mL

$m=d\times V=0.95 g/mL\times 200.0 mL=190 g$

$m = 190 g = 190 \times 0.001 kg = 0.19 kg$

(1 g = 0.001 kg)

Molality of the urea solution ;

$Molality(m)=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}$

$m=\frac{0.3 mol}{0.19 kg}=1.58 m$

The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.

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3 years ago

Iodine 131, cesium 137, and strontium 90 are all considered isotopes. What are isotopes?

Nitella [24]

Answer:

Isotopes are variants of an element, where the number of neutrons is different but its number of protons stay the same.

Explanation:

If the number of protons had changed then it would just be a different element.

Electron number changing simply adds charge to the atom, and doesn't change it or anything.

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3 years ago

Read 2 more answers

Equation for questions 1-4:

Karolina [17]

Answer:

5.625 moles of oxygen, O₂.

Explanation:

The balanced equation for the reaction is given below:

4Al + 3O₂ —> 2Al₂O₃

From the balanced equation above,

4 moles of Al reacted with 3 moles of O₂.

Finally, we shall determine the number of mole of O₂ required to react with 7.5 moles of aluminum, Al. This can be obtained as illustrated below:

From the balanced equation above,

4 moles of Al reacted with 3 moles of O₂.

Therefore, 7.5 moles of Al will react with = (7.5 × 3)/4 = 5.625 moles of O₂.

Thus, 5.625 moles of O₂ is needed for the reaction.

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3 years ago

Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?

dlinn [17]

Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

C₄H₁₀: 2 moles
O₂: 13 moles
CO₂: 8 moles
H₂O: 10 moles

Being:

C: 12 g/mole
H: 1 g/mole
O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
O₂: 2*16 g/mole= 32 g/mole
CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

C₄H₁₀: 2 moles* 58 g/mole= 116 g
O₂: 13 moles* 32 g/mole= 416 g
CO₂: 8 moles* 44 g/mole= 352 g
H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

$mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10} }{116 grams of C_{4}H_{10}}$

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>

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3 years ago