Arrange the following atoms in order of increasing electronegativity.

Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a plat

Mila [183]

Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values of the atomic mass of each element that form the compounds:

N: 14 g/mol
H: 1 g/mol
O: 16 g/mol

So, the molar mass of the compounds in the reaction is:

NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
O₂: 2*16 g/mol= 32 g/mol
NO: 14 g/mol + 16 g/mol= 30 g/mol
H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol

By stoichiometry, they react and occur in moles:

NH₃: 4 moles
O₂: 5 moles
NO: 4 moles
H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

NH₃: 4 moles*17 g/mol= 68 g
O₂: 5 moles*32 g/mol= 160 g
NO: 4 moles*30 g/mol= 120 g
H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

$mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }$

mass of O₂≅17.65 g

17.65 grams of O2 are needed for a complete reaction.

3 0

3 years ago

Burning 1 kg of coal releases about 3 million joules of energy. If you could use all of the chemical energy to lift another kilo

Katarina [22]

I could lift 3.06 x 10⁵ m high

<h3>Further explanation </h3>

Energy is the ability to do work. Energy can change from one energy to another

Potential energy is the energy that an object has because of its position

The potential energy can be formulated:

Ep = m. g. h

E = potential energy of an object, joule

m = object mass, kg

g = gravity acceleration, m / s²

h = height of an object, m

energy of coal = 3.10⁶ J

mass = 1 kg

g = 9.8 m/s²

$\tt h=\dfrac{E}{m.g}\\\\h=\dfrac{3.10^6}{1\times 9.8}=3.06\times 10^5`m$

5 0

3 years ago

How many particles are there in 1.43 g of a molecular compound with a gram molecular mass of 233g

dalvyx [7]

1.43g x (1 mole / 233g) x (6.022x10^23 particles / mole) = 3.70 x 10^ 21 particles

7 0

3 years ago

A 1.40 L sample of O2 at 645 Torr and 25 °C, and a 0.751 L sample of N2 at 1.13 atm and 25 °C, are both transferred to the same

anyanavicka [17]

Answer:

P(O₂) = 0.595 atm

P(N₂) = 0.424 atm

Total Pressure = 1.019 atm

Explanation:

To solve this problem we use PV=nRT for both gases in their containers, in order to calculate the moles of each one:

O₂:

645 Torr ⇒ 645 /760 = 0.85 atm

25°C ⇒ 25 + 273.16 = 298.16 K

0.85 atm * 1.40 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0487 mol O₂

N₂:

1.13 atm * 0.751 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

n = 0.0347 mol N₂

Now we can calculate the partial pressure for each gas in the new container, because the number of moles did not change:

O₂:

P(O₂) * 2.00 L = 0.0487 mol O₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(O₂) = 0.595 atm

N₂:

P(N₂) * 2.00 L = 0.0347 mol N₂ * 0.082 atm·L·mol⁻¹·K⁻¹ *298.16 K

P(N₂) = 0.424 atm

Finally we add the partial pressures of all gases to calculate the total pressure:

Pt = 0.595 atm+ 0.424 atm = 1.019 atm

6 0

3 years ago

A polymer P is made up of two monodisperse fractions; fraction A with molecular weight of 1000 g/mole and fraction B with a mole

vesna_86 [32]

Answer:

a)Number average molecular weight is 50, 500 g/mol

b) Weight average molecular weight is 99, 019.8 g/mol

Explanation:

We have a polymer P made up of two monodisperse fractions.

A with molecular weight of MA = 1000 g/mol and B with MB =100000 g/mol.

The batch contains an equal mole fraction of each component A and B.

Let's suppose a total number (Nt) of mols 2 moles. Equal fraction means XA = 0.5 and XB =0.5

Nt = 2 mol

Na = 2*0.5 = 1 mol

Nb = 2*0.5 = 1 mol.

So, we have 1 mol of A, 1 mol of B and 2 moles in total.

a) The number average molecular weight (NAM) is calculate using the mole numbers of each component. In this case, we will multiple each component molecular weight by the number of moles of each one. After that we will sum them and finally to divide by the total number of moles.

NAM = (Na*MA + Nb*MB)/(Nt)

NAM = (1 mol *1000 g/mol + 1*100000 g/mol ) /(2 mol)

NAM = 50500 g/mol

The number average molecular weight for the polymer P is 50,500 g/mol

b) Weight average molecular weight (WAM) is calculated using the mass quantities of each component. Weight mass of A (WA), weight mass of B (WB) are calculate using the moles of A, B and their molecular weights respectively. Total Weight (WT)

WA = Na*MA = 1 mol *1000 g/mol = 1000 g A

WB = Nb*MB = 1mol * 100000 g/mol = 100 000 gB

WT = WA + WB = 101 000 g

Now we will calculate average molecular using weights, we will multiple each component molecular weight by the mass of each one. After that we will sum them and finally to divide by the total mass.

WAM = (WA*MA + WB*MB)/(WT)

WAM = (1000 g *1000 g/mol + 100000 g*100000 g/mol )/(101 000 g)

WAM = 99 019.8 g/mol

The weight average molecular weight for polymer P is 99, 019.8 g/mol

6 0

4 years ago