The answer would be the <span>Chemical formula for ammonia by
hydrogen and nitrogen gas: 1 N2 + 3 H2 --> 2 NH3 t<span>hen 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.
<span>3.44 mol H2 * (2 mol NH3 / 3 mol H2) = 2.29 mol NH3</span></span></span>
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.
<em>Octopus and squids breathe</em> <em>like </em><em>fishes </em><em>they </em><em>breathe </em><em>from </em><em>gills </em>
<em>so </em><em>even </em><em>octopus</em><em> and</em><em> squids</em><em> </em><em>breathe </em><em>through </em><em>gills </em><em>too.</em>
<em><u>maybe </u></em><em><u>this </u></em><em><u>answer</u></em><em><u> </u></em><em><u>would</u></em><em><u> </u></em><em><u>help</u></em><em><u> </u></em><em><u>u</u></em>
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 = 
= 6.84 x 10⁻¹⁹ joule
Find the wavelength of a photon we use E = hν
⇒ E = 
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 = 
= 2.9 x 10⁻⁷ m
= 290 nm
∵ 1 nm = 10⁻⁹ m
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