Can you help me answer the question ?

What are the possible values of ml for l=2<br>

Marrrta [24]

Answer:

-2, -1, 0, 1, 2

Explanation:

There are four types of quantum numbers;

1) Principal quantum number (n)

2) Azimuthal quantum number (l)

3) magnetic quantum number (ml)

4) Spin quantum number (s)

The azimuthal quantum number (l) describes the orbital angular momentum and shape of an orbital while the magnetic quantum number shows the projections of the orbital angular momentum along a specified axis. This implies that the magnetic quantum number shows the orientation of various orbitals along the Cartesian axes. The values of the magnetic quantum number ranges from -l to + l

For l= 2, the possible values of the magnetic quantum number are; -2, -1, 0, 1, 2

5 0

2 years ago

PLS ANSWER FAST THANK UU

andriy [413]

Answer
I think it might be B
Explanation

4 0

3 years ago

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago

Balance the following chemical equations

jok3333 [9.3K]

Answer:

2N2 + O2 ⇒ 2N2O

Explanation:

Since I cannot see the product's of the second chemical equation I will only solve the first one.

In order to balance a chemical equation you need to make sure that the atoms on both sides are equal.

In this case we have...

N2 + O2 = N2O

N = 2

O = 2

N = 2

O = 2

Change the coefficient:

N2O = 2N20

2 × 2 = 4

1 × 2 = 2

2N2 + O2 ⇒ 2N2O

Hope this helps.

4 0

2 years ago

If I have 3.5 moles of C, and excess Fe2O3 , how many moles of Fe can I produce?

Wittaler [7]

Answer:

3.5 moles Fe

Explanation:

From the equation, Reaction of 2 moles of Fe₂O₃ with 1 mole of C produces 1 mole of Fe. When excess Fe₂O₃ is used, the only liming factor is C.

The ratio of amount of C used to the amount of Fe produced is 1:1

Therefore, if 3.5 moles of C are used, 3.5 moles of Fe are also produced.

6 0

2 years ago