Answer:
Elements are classified into- metals – Nonmetals- Metalloids – Noble gases. State which of A, B, C, D is a:
1) Metallic element
2) Non-metallic element
3) Metalloid
4) Noble gas.
A) Is non-malleable, non-ductile and a poor conductor of electricity
B) Has lustre, is malleable and ductile and a good conductor of electricity
C) Is unreactive and inert and present in traces in air
D) Shows properties of both metals and nonmetals
Explanation:
Here we have to explain and predict the product of the reaction between hept-3-yne and bromine (Br₂)
The reaction between hept-3-yne and bromine produces <em>trans</em>-3,4-dibromo-3-heptene.
The reaction between the unsaturated hydrocarbon with halogen remove the unsaturation in the molecule and produces di halo compound. The reaction proceed through the formation of bromonium ion.
In this reaction as the unsaturation present in the middle of the alkyl chain i.e. hept-3-yne. The reaction will stop after the formation of di-halo compound with one double bond as shown in the figure.
The product will be exclusively trans product as the bromine is a bulky group and the formation of bromonium ion (intermediate) will be one side of the unsaturated compound.
Answer:
B. Wall applies equal reaction force on hand
Explanation:
This is newton's third law. You push something, that something pushes you back with same force. The direction is opposite, so if I push a wall, the wall pushes me back with same force just in opposite direction (so wall towards me). This is why option A is incorrect b/c it says force in same direction as wall. C and D talk about unequal force which is incorrect.
Answer:
Mass percent N₂ = 89%
Mass percent H₂ = 11%
Explanation:
First we <u>use PV=nRT to calculate n</u>, which is the total number of moles of nitrogen and hydrogen:
- 1.03 atm * 7.45 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305 K
So now we know that
- MolH₂ + MolN₂ = 0.307 mol
and
- MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49 g
So we have a <u>system of two equations and two unknowns</u>. We use algebra to solve it:
Express MolH₂ in terms of MolN₂:
- MolH₂ + MolN₂ = 0.307 mol
Replace that value in the second equation:
- MolH₂ * 2 g/mol + MolN₂ * 28 g/mol = 3.49
- (0.307-MolN₂) * 2 + MolN₂ * 28 = 3.49
- 0.614 - 2MolN₂ + 28molN₂ = 3.49
Now we calculate MolH₂:
- MolH₂ + MolN₂ = 0.307 mol
Finally, we convert each of those mol numbers to mass, to <u>calculate the mass percent</u>:
- N₂ ⇒ 0.111 mol * 28 g/mol = 3.108 g N₂
- H₂ ⇒ 0.196 mol * 2 g/mol = 0.392 g H₂
Mass % N₂ = 3.108/3.49 * 100% = 89.05% ≅ 89%
Mass % H₂ = 0.392/3.49 * 100% = 11.15% ≅ 11%
