Answer:
Explanation:
An oxidizing accepts an electron and becomes reduced while a reducing agent loses an electron and become oxidized.
Chemical equation:
1) 2 N₂H₄ + N₂O₄ → 3 N₂ + 4 H₂O
2) Hydrazine ( N₂H₄) is being oxidized
Dinitrogen tetroxide N₂O₄ is being reduced
3) The reducing agent is Hydrazine ( N₂H₄) and the oxidizing agent is dinitrogen tetroxide (N₂O₄)
Answer:
The correct answer is Option-D (6 Hydrogen atoms).
Explanation:
Carbon atom has a unique property of linking to it self and making a chain of carbon called as Catenation. In a molecule having two carbon atoms there must be a bond between the two carbon atoms. The bond can be saturated (single bond) or unsaturated (double or triple bond). But as the statement states maximum number of hydrogen atoms so, we will asume the bond between two carbon atoms to be single.
As carbon has four valence electrons so, it has the ability to make four single bonds. In given molecule each carbon has already made a single bond with another carbon therefore, each carbon is left with three more unpaired electrons which will make covalent bond with three hydrogen atoms each as shown in attached structure.
Answer:
1.17 grams
Explanation:
Let's consider the balanced equation for the combustion of ethylene.
C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l)
We can establish the following relations:
- 1411 kJ are released (-1411 kJ) when 1 mole of C₂H₄ burns.
- The molar mass of C₂H₄ is 28.05 g/mol.
The grams of C₂H₄ burned to give 59.0 kJ of heat (q = -59.0 kJ) is:

First, we determine the number of moles of barium phosphate. This is done using:
Moles = mass / Mr
Moles = 1 / 602
Moles = 0.002
Now, we see from the formula of the compound that each mole of barium phosphate has 2 moles of phosphorus (P). Therefore, the moles of phosphorus are:
0.002 * 2 = 0.004
The number of particles in one mole of substance is 6.02 x 10²³. The number of phosphorus atoms will be:
0.004 * 6.02 x 10²³
2.41 x 10²¹ atoms of phosphorus are present
Answer is: molality of urea is 5.84 m.
If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.