Answer:
a) quadruple covalent bond
Explanation:
Quadruple covalent bonds are the strongest covalent bonds. The more the number of bonding pairs, the stronger the bond becomes.
- A covalent bond is formed by sharing of electrons between two species.
- They are usually between compound with similar electronegativities.
- The number of bonds between the atom determines the bond strength.
Single < double < triple < quadruple covalent bonds.
They are changes in bonding energy between the molecules. ... Immediately after the molecular bonds in the ice are broken the molecules are moving (vibrating) at the same average speed as before, so their average kinetic energy remains the same, and, thus, their Kelvin temperature remains the same."
Answer:
A. NO₃⁻ ⇒ +5
B. NO₄⁺ ⇒ +9
C. N₂ ⇒ 0
D. NH₂OH ⇒ -1
E. NO₂⁻ ⇒ +3
Explanation:
To calculate the oxidation number of an element in a compound, we have to know the oxidation number of the other elements. Then, we have to consider that the sum of the oxidation number of each atom multiplied by the subscripts is equal to the net charge of the compound.
A. NO₃⁻
This is the ion nitrate. Oxygen atoms (O) has an oxidation number of -2 because it derives from an oxide. In this case, the net charge of the ion is -1. Thus, we calculate the oxidation number of N as follows:
N + (3 x (-2)) = -1
N - 6 = -1 ⇒ N= -1 + 6 = +5
B. NO₄⁺
In this case, the sum of the oxidation numbers of O and N multiplied by the subscripts is equal to +1:
N + (4 x (-2)) = +1
N - 8 = +1 ⇒ N = +1 +8 = +9
C. N₂
The oxidation number of N is 0 because N₂ is an elemental substance.
D. NH₂OH
The oxidation number of H is +1 and -2 for O. The net charge of the molecule is 0.
N + (H x 3) + (O) = 0
N + (+1 x 3) + (-2) = 0
N + 3 -2 = 0 ⇒ N = 2 - 3 = -1
E. NO₂⁻
The net charge of the ion nitrite is -1.
N + (2 x O) = -1
N + (2 x (-2)) = -1 ⇒ N = -1 + 4 = +3
Therefore:
Highest oxidation number = B. NO₄⁺ (+9)
Lowest oxidation number = D. NH₂OH (-1)
Answer:
The answer to your question is V2 = 825.5 ml
Explanation:
Data
Volume 1 = 750 ml
Temperature 1 = 25°C
Volume 2= ?
Temperature 2 = 55°C
Process
Use the Charles' law to solve this problem
V1/T1 = V2/T2
-Solve for V2
V2 = V1T2 / T1
-Convert temperature to °K
T1 = 25 + 273 = 298°K
T2 = 55 + 273 = 328°K
-Substitution
V2 = (750 x 328) / 298
-Simplification
V2 = 246000 / 298
-Result
V2 = 825.5 ml
Since chlorine is one of the 7 diatomic elements we know that chlorine appears as Cl₂ gas naturally. That means that the molar mass of a chlorine gas is 70.9g/mol. That being said, first you need to find the number of moles of chlorine gas that are present in a 35.5g sample. To do this divide 35.5g by the molar mass of chlorine gas (70.9g/mol) to get 0.501mol of chlorine. Then you have to multiply 0.501mol by 6.02×10²³ to get the number of chlorine gas molecules. Therefore 3.01×10²³ molecules of chlorine gas are present in a 35.5g sample.
I hope that helps. Let me know in the comments if anything is unclear.
