Answer:
N and P
Explanation:
Anion:
When an atom gain the electrons anion is formed. The negative sign shows that atom gain electron because number of electron are greater than protons or we can say that negative charge becomes greater than positive charge.
Cation:
When atom lose electron cation is formed. The atom thus have positive charge because number of positive charge i.e protons are increased are greater than negative charge or electron.
In given problem N and phosphorus both can gain three electrons which means negative charge becomes greater that's why the extra electron gained by atoms are written as -3 and both form anion with charge -3.
while Al form cation with charge +3 Mg form cation with charge +2 and iodine and bromine both form anion with charge of -1.
It is a covalent bond. Whenever a compound uses such suffixes like mono, di, tri, tetra, and so on, it is a covalent compound- thus having covalent bonds.
Answer:
hard and brittle
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Answer:
6.88 mg
Explanation:
Step 1: Calculate the mass of ³²P in 175 mg of Na₃³²PO₄
The mass ratio of Na₃³²PO₄ to ³²P is 148.91:31.97.
175 mg g Na₃³²PO₄ × 31.97 g ³²P/148.91 g Na₃³²PO₄ = 37.6 mg ³²P
Step 2: Calculate the rate constant for the decay of ³²P
The half-life (t1/2) is 14.3 days. We can calculate k using the following expression.
k = ln2/ t1/2 = ln2 / 14.3 d = 0.0485 d⁻¹
Step 3: Calculate the amount of P, given the initial amount (P₀) is 37.6 mg and the time elapsed (t) is 35.0 days
For first-order kinetics, we will use the following expression.
ln P = ln P₀ - k × t
ln P = ln 37.6 mg - 0.0485 d⁻¹ × 35.0 d
P = 6.88 mg
Answer:
a. Cyclohexanone
Explanation:
The principle of IR technique is based on the <u>vibration of the bonds</u> by using the energy that is in this region of the electromagnetic spectrum. For each bond, there is <em>a specific energy that generates a specific vibration</em>. In this case, you want to study the vibration that is given in the carbonyl group C=O. Which is located around 1700 cm-1.
Now, we must remember that the <u>lower the wavenumber we will have less energy</u>. So, what we should look for in these molecules, is a carbonyl group in which less energy is needed to vibrate since we look for the molecule with a smaller wavenumber.
If we look at the structure of all the molecules we will find that in the last three we have <u>heteroatoms</u> (atoms different to carbon I hydrogen) on the right side of the carbonyl group. These atoms allow the production of <u>resonance structures</u> which makes the molecule more stable. If the molecule is more stable we will need more energy to make it vibrate and therefore greater wavenumbers.
The molecule that fulfills this condition is the <u>cyclohexanone.</u>
See figure 1
I hope it helps!