How should the following reaction be classified?

Ammonia (nh3) is widely used as a fertilizer and in many household cleaners. how much ammonia is produced when 6.44 mol of hydro

attashe74 [19]

The answer would be the Chemical formula for ammonia by hydrogen and nitrogen gas: 1 N2 + 3 H2 --> 2 NH3 then use stoichiometry. You know you have enough nitrogen gas to react, so you can just straight convert mol H2 to mol NH3, where 3 mol H2 = 2 mol NH3.
3.44 mol H2 * (2 mol NH3 / 3 mol H2) = 2.29 mol NH3

8 0

3 years ago

8.7 Two products are formed in the following reaction in a 50:50 mixture. Would the resulting solution be optically active

k0ka [10]

Answer:

Yes. The solution would be optically active.

Explanation:

Diastereomer are defined as the image that is non mirror and non -identical. It is made up of two stereoisomers. They are formed when the two stereoisomers or more than two stereoisomers of the compound have the same configuration at the equivalent stereocenters.

In the given context, as the product given is a diastereomeric mixture, the product would have an optical activity in total.

So the answer is Yes.

4 0

2 years ago

Octopus and squids breathe through

svet-max [94.6K]

Octopus and squids breathe like fishes they breathe from gills 

so even octopus and squids breathe through gills too.

maybe this answer would help u

4 0

2 years ago

It takes 412. KJ/mol to break a carbon-hydrogen single bond. Calculate the maximum wavelength of light for which a carbon-hydrog

olga_2 [115]

Answer:

The maximum wavelength of light for which a carbon-hydrogen single bond could be broken by absorbing a single photon = 290 nm

Explanation:

So to break a single C - H bond require = $\frac{412}{6.023 X (10)^{23} }$

= 6.84 x 10⁻¹⁹ joule

Find the wavelength of a photon we use E = hν

⇒ E = $\frac{hc}{Wavelength}$

Where h = Planck's constant = 6.626 x 10⁻³⁴ J.K⁻¹.Mole⁻¹

c = speed of light = 3 x 10⁸ m/sec

Wavelength = $\frac{6.626 X 10^{-34} X 3 X 10^{8} }{6.84 X10^{-19} }$

= 2.9 x 10⁻⁷ m

= 290 nm

∵ 1 nm = 10⁻⁹ m

7 0

3 years ago

What is the empirical formula for a compound that contains 79.86 % iodine and 20.14 % oxygen by mass?

serg [7]

Answer:

IO₂

Explanation:

We have been given the mass percentages of the elements that makes up the compound:

Mass percentage given are:

Iodine = 79.86%

Oxygen = 20.14%

To calculate the empirical formula which is the simplest formula of the compound, we follow these steps:

> Express the mass percentages as the mass of the elements of the compound.

> Find the number of moles by dividing through by the atomic masses

> Divide by the smallest and either approximate to nearest whole number or multiply through by a factor.

> The ratio is the empirical formula of the compound.

Solution:

I O

% of elements 79.86 20.14

Mass (in g) 79.86 20.14

Moles(divide by

Atomic mass) 79.86/127 20.14/16

Moles 0.634 1.259

Dividing by

Smallest 0.634/0.634 1.259/0.634

1 2

The empirical formula is IO₂

7 0

3 years ago