Calculate the number of moles in a 14.5 gram sample of C4H10.

Answers

Answer 1

Answer: Moles= mass divided by molar mass
Molar mass= 12.01(4) + 1.01(10)
= 58.14g/mol

Moles=14.5g / 58.14g/mol
=0.249

Therefore there are approx 0.249 moles in a 14.5g sample of C4H10

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The atomic mass unit (amu) isa. the mass of a single atom of carbon
b. one millionth of a gram
c. approximately the mass of a proton
d. approximately the mass of an electron

Answers

That would be c. approximately the mass of a proton

The three components essential for a fire are:a. Oxygen, fuel, ignition source.
b. Oxygen, fuel, primer.
c. Oxygen, fuel, oxygen enriched environment.
d. Oxygen, fuel, a liquid.

Answers

Hey there

The answer should be : Oxygen, fuel , heat

I hope that's help !

Ammonia is produced by the following reaction. 3H2(g) + N2(g) ---> 2NH3(g) When 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because

Answers

To determine the limiting reactant, we first calculate the number of moles of the reactant.

7.00 g H2 ( 1 mol / 2.02 g ) = 3.47 mol H2
70.00 g N2 ( 1 mol / 28.02 g) = 2.50 mol N2

From the balanced reaction, we see that there is a 3 is to 1 ratio of the reactants. So, for every 3 moles of H2 we need 1 mole of N2. Given the amount of reactant, for 2.50 moles of N2 we need 7.5 moles of H2 which obviously we have a lesser amount. Hydrogen is the limiting reactant.

The correct answer is A

Consider the solutions, 0.04 m urea [(NH2)2C=O)], 0.04 m AgNO3 and 0.04 m CaCl2. Which has (i) the highest osmotic pressure, (ii) the lowest vapor pressure, (iii) the highest boiling point?

Answers

Answer:

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

Explanation:

The colligative properties depend upon the number of solute particles in a solution.

The following four are the colligative properties:

a) osmotic pressure : more the concentration of the solute, more the osmotic pressure

b) vapor pressure: more the concentration of the solute, lesser the vapor pressure.

c) elevation in boiling point: more the concentration of the solute, more the boiling point.

d) depression in freezing point: more the concentration of the solute, lesser the freezing point.

the number of particle produced by urea = 1

the number of particle produced by AgNO3 = 2

the number of particle produced by CaCl2 = 3

As concentrations are same, CaCl2 will have more number of solute particles and urea will have least

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

Final answer:

The solution with the highest number of particles in solution (CaCl2 in this case), experiences the highest osmotic pressure, lowest vapor pressure and highest boiling point due to the principles of colligative properties.

Explanation:

The question pertains to the colligative properties of solutions, which would be governed by the number of particles in the solution. The solutions are 0.04 m urea [(NH2)2C=O)], 0.04 m AgNO3, and 0.04 m CaCl2. For (i) Highest osmotic pressure, the solution with the highest ion count would yield the highest osmotic pressure. CaCl2 dissociates into three ions (Ca²+, and 2 Cl¯), therefore, it would exhibit the highest osmotic pressure. For (ii) Lowest vapor pressure, this would coincide with the solution with the highest osmotic pressure, again making it CaCl2, due to the greatest decrease in vapor pressure. For (iii) the highest boiling point, this too would be CaCl2 for the reasons stated above. The presence of more particles in a solution interferes more with the evaporation process, requiring more energy (higher temperature) to achieve boiling.

Learn more about Colligative Properties here:

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What is the correct electron configuration for carbon?a. 1s22s22p2
b. 1s22s22p4
c. 1s22s22p5
d. 1s22s22p63s23p2

Answers

It is A; [He] 2s² 2p² is another way to put it

Your answer would be A.

Cuantas moléculas de oxigeno se producen por la descomposición de 28.5 g de H2O2 (masa molecular = 34.0g/mol) de acuerdo a la ecuación 2H2O2(l) → 2H2O(l)+O2(g)

Answers

The question is as follows: How many oxygen molecules are produced by the decomposition of 28.5 g of H2O2 (molecular mass = 34.0g / mol) according to the equation

2H2O2 (l) → 2H2O (l) + O2 (g)

Answer: There are $2.52 * 10^(23)$ molecules are produced by the decomposition of 28.5 g of $H_(2)O_(2)$ according to the equation $2H_(2)O(l) \rightarrow 2H_(2)O(l) + O_(2)(g)$ .

Explanation:

Given: Mass of $H_(2)O_(2)$ = 28.5 g

As moles is the mass of a substance divided by its molar mass. Hence, moles of $H_(2)O_(2)$ is calculated as follow.

$Moles = (mass)/(molarmass)\n= (28.5 g)/(34.0 g/mol)\n= 0.838 mol$

According to the given equation, 2 moles of $H_(2)O_(2)$ gives 1 mole of $O_(2)$ . So, moles of $O_(2)$ produced by 0.838 moles of $H_(2)O_(2)$ will be calculated as follows.

$Moles of O_(2) = (0.838 mol)/(2)\n= 0.419 mol$

This means that moles of $O_(2)$ produced is 0.419 mol.

As per the mole concept, 1 mole of every substance has $6.022 * 10^(23)$ molecules.

So, molecules of $O_(2)$ present in 0.419 mole are as follows.

$0.419 * 6.022 * 10^(23)\n= 2.52 * 10^(23)$

Thus, we can conclude that there are $2.52 * 10^(23)$ molecules are produced by the decomposition of 28.5 g of $H_(2)O_(2)$ according to the equation $2H_(2)O(l) \rightarrow 2H_(2)O(l) + O_(2)(g)$ .

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