Covalent bonds are between two non-metals. Hydrogen and oxygen are non-metals. Hence is is, "yes, hydrogen and oxygen are nonmetals."
For the future, you might want to look at some videos to help you! :) (Tyler DeWitt is a really good chem you-tube teacher!)
Answer:
Q = 246 kJ
Explanation:
It is given that,
Mass of water, m = 200 g
Let initial temperature, 
Final temperature of water, 
We know that the specific heat capacity of water, 
So, the heat energy needed to raise the temperature is given by :
or
Q = 246 kJ
So, the heat energy of 246 kJ is needed.
Answer:
3.0x10⁻²M
Explanation:
Silver sulfate, Ag₂SO₄, has a product constant solubility equilbrium of:
Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻
When an excess of silver sulfate is added, some Ag₂SO₄ will react producing Ag⁺ and SO₄²⁻ until reach the equilbrium determined for the formula:
ksp = 1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
Assuming the Ag₂SO₄ that react until reach equilibrium is X, we can replace in Ksp expression:
1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
1.4x10⁻⁵ = [2X]² [X]
1.4x10⁻⁵ = 4X³
3.5x10⁻⁶ = X³
0.015 = X
As [Ag⁺] is 2X:
[Ag⁺] = 0.030 = 3.0x10⁻²M
The answer is:
<h3>3.0x10⁻²M</h3>
Answer:
28 g/mol, N2
Explanation:
Given data:
Volume of gas = 5.0 L
Mass of gas = 6.3 g
Pressure = 1 atm
Temperature = 273 K
Molar mass of gas = ?
Solution:
We will calculate the density first.
d = mass/ volume
d = 6.3 g/ 5.0 L
d = 1.26 g/L
Molar mass:
d = PM/RT
M = dRT/P
M = 1.26 g/L× 0.0821 atm.L/mol.K × 273 K/ 1 atm
M = 28 g/mol
Molar mass of N₂ is 28 g/mol thus given gas is N₂.
<h3>
Answer:</h3>
A. 1.4 V
<h3>
Explanation:</h3>
We are given the half reactions;
Ni²⁺(aq) + 2e → Ni(s)
Al(s) → Al³⁺(aq) + 3e
We are required to determine the cell potential of an electrochemical cell with the above half-reactions.
E°cell = E(red) - E(ox)
From the above reaction;
Ni²⁺ underwent reduction(gain of electrons) to form Ni
Al on the other hand underwent oxidation (loss of electrons) to form Al³⁺
The E.m.f of Ni/Ni²⁺ is -0.25 V and that of Al/Al³⁺ is -1.66 V
Therefore;
E°cell = -0.25 V - (-1.66 V)
= -0.250V + 1.66 V
= + 1.41 V
= + 1.4 V
Therefore, the cell potential will be +1.4 V
