All categories

3 years ago

What is the theoretical yield of ammonia that can be obtained from the reaction of 17.1 g of H2 and excess N2?

1 answer:

6 0

The theoretical yield is the yield after a certain reaction occurs. So if you are given 17.1 g of H2 and an excess of N2, you should know the balanced equation.

N2 + 3H2 → 2NH3

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17 g/mol

(17.1 g of H2)(1 mol H2/2.02g H2)(2 mol NH3/3mol H2)(17g NH3/1mol NH3) = 95.94g NH3

You might be interested in

A boy jumps into a cold swimming pool and his body temperature goes down. His muscles, blood vessels, and nervous system work to

mixas84 [53]

Answer:

balance

Explanation:

4 0

2 years ago

Calculate the density of CO2 in g/cm3 at room temperature(25 degrees Celsuis) and pressure(1 atm) assuming it acts as an ideal g

Readme [11.4K]

Answer:

$density=1.8x10^{-3}g/mL$

Explanation:

Hello,

Considering the ideal equation of state:

$PV=nRT$

The moles are defined in terms of mass as follows:

$n=\frac{m}{M}$

Whereas $M$ the gas' molar mass, thus:

$PV=\frac{mRT}{M}$

Now, since the density is defined as the quotient between the mass and the volume, we get:

$P=\frac{m}{V} \frac{RT}{M}$

Solving for $m/V$ :

$density= m/V=\frac{PM}{RT}$

Thus, the result is given by:

$density=\frac{(1atm)(44g/mol)}{[0.082atm*L/(mol*K)]*298.15K} \\density=1.8g/L=1.8x10^{-3}g/mL$

Best regards.

8 0

3 years ago

51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0

3 years ago

What is another name for a condensation reaction? what is another name for a condensation reaction? water formation catabolism d

gregori [183]

Condensation is a chemical reaction in which two molecules are combined together and as a product formed a large molecule. There is also loss of small molecules in the condensation reaction. And mostly the functional groups combined in the reaction.
The other name for a condensation reaction is dehydration. As there is loss of molecule in condensation reaction, when that molecule is water it is said to be dehydration.

3 0

3 years ago

Read 2 more answers

In a two-step synthesis, C6H11Br is converted into C6H12O. From the structure of the product, molecular formula of the starting

tamaranim1 [39]

Answer:

See explanation below

Explanation:

The question is incomplete. However in picture 1, you have the starting materials and the structure of the product, which you miss in this part.

Now, in picture 2, you have the starting reactant and the product, and the mechanism that is taking place here.

First, all what we have here is an acid base reaction. In the first step, we are using the acid medium to convert the reactant into an alcohol. The bromine there, is not leaving the molecule yet, because it's neccesary for the next step. The starting reactant is an alkene, in that way, we can convert the reactant in the first step into a secondary alcohol. In other words, the first reaction is a alkene hydration.

In the second step, we use a strong base. You may say this is a strong nucleophile and will do a Sn2 reaction to form another alcohol there, but it's not the case, because, before any kind of reaction happens, the priority here is always the acid base, so the base will react with the acidic hydrogen. In this case, it will substract an hydrogen from the OH. When this happens, the lone pair will do an auto condensation here, and attacks the bromine in the molecule. In this way, the molecule will become a cyclomolecule, and that way it form the final product.

See picture 2, for mechanism

8 0

3 years ago