According to Hund's rule, if three electrons are available to fill a set of p orbitals, how will the electrons be arranged? *

4. Which state of matter has a definite volume and takes the shape of<br> gas<br> liquid<br> solid

DedPeter [7]

Answer:

Solido

Explanation:

Es mas definido

4 0

3 years ago

300 mL of 0.200 M Pb(NO3)2 is added to 200 mL of 0.0500 M NaCl(aq) at 25°C. Will a precipitate of PbCl2 form at 25 °C? Tip: You

Bas_tet [7]

Explanation:

It is given that volume of $Pb(NO_{3})_{2}$ is 300 mL and molarity is 0.200 M.

Volume of NaCl is 200 mL and molarity is 0.050 M.

The chemical reaction will be as follows.

$PbCl_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^{-}(aq)$

$K_{sp}$ for $PbCl_{2}$ is given as $1.7 \times 10^{-5}$ .

As, molarity is number of moles present in liter of solution.

Hence, moles of $Pb^{2+}$ (aq) will be calculated as follows.

moles of $Pb^{2+}$ (aq) = $0.300 L \times 0.200 M$

= 0.06 mol

$[Pb^{2+}]$ = $\frac{0.06 mol}{0.5 L}$

= 0.120 M

Mole of $Cl^{-}(aq)$ = $0.2 L \times 0.05 M$

= 0.010 M

Now, Q = $[Pb^{2+}][Cl^{-}]^{2}$

= $0.120 \times (0.010)^{2}$

= $1.2 \times 10^{-5}$

As, Q < $K_{sp}$ hence, there will be no formation of $PbCl_{2}$ precipitate.

3 0

3 years ago

Hemoglobin, a protein in red blood cells, carries o2 from the lungs to the body's cells. iron (as ferrous ion, fe2+) makes up 0.

jenyasd209 [6]

In order to determine the number of Fe ions, we first determine the mass mass of iron present. This is determined using:

Mass of Fe = 0.33% of mass of hemoglobin

Mass of hemoglobin = 6.8 x 10⁴ g/mol

Mass of Fe = 0.0033 * 6.8 x 10⁴
Mass of Fe = 224.4 g/mol

Now, we know that the atomic mass of Fe is 55.85 g/mol

We may determine the mass of iron by the mass of each molecule of ion to determine the number present, which is:

224/55.85 = 4 moles of Fe ions present per mole of hemoglobin.

6 0

3 years ago

What happens to the number of waves when you decrease the wavelength?

Keith_Richards [23]

C

Decreasing the wavelength increases the frequency of a wave and vice versa, as long as the energy of the wave remains constant.

Explanation:

The energy of a wave is given by the formula;

E= hf

Whereby;

E = energy

h = Plank’s constant

f = frequency

Remember that f can be also represented as;

¹/ λ

Whereby;

λ = wavelength

Wavelength and frequency are therefore inversely proportional.

Decreasing the wavelength increases the frequency of a wave and vice versa, as long as the energy of the wave remains constant.

Learn More:

For more on frequency and wavelength of waves check out;

brainly.com/question/13451801

brainly.com/question/9501542

brainly.com/question/14068647

brainly.com/question/12924624

brainly.com/question/10955773

#LearnWithBrainly

6 0

4 years ago

A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an

erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0

3 years ago