A 3.5 kg iron shovel is left outside through the winter. The shovel, now orange with rust, is rediscovered in the spring. Its ma

1) To find the molar mass of C6H8O6, you must refer to the atomic weights of C, H, and O from the periodic table: C is 12; H is 1; O is 16 <span> (12x6)+(1x8)+(16x6)= 176g/mol


</span> <span> 176 g = 1 mol
0.5 g = x mol = 500 mg = 0.5 grams

Molar mass = mass ÷ moles
176 = 0.5 ÷ x
x = 2.84 x 10⁻³ mol
2) To calculate the total number of molecules in those </span> 2.84 x 10⁻³ mol, multiply the moles by <span> Avogadro's constant.
Number of molecules = Avogadro's constant x number of moles
Number of molecules = 6.022 x 10²³ x 2.84 x 10⁻³ </span> = 1.71 x 10²¹ molecules of vitamin C. <span>
</span>

Answer:

B, C

Explanation:

The Lewis dot structures that are valid are for B and C.

Regarding A;

The Lewis representation of carbon is correct. Each of the four dots symbolizes the four valence electrons.

However, the nitrogen with a single dot above, to the left, and below and a -3 charge isincorrect. For it to have a -3 charge, it must possess 8 Lewis dots (two on each side).

The nitrogen showing four dots (on top, right, bottom and left) is incorrect.

In B;

An oxygen having two dots positioned above and below with one dot on both sides is correct , as the 6 dots indicate oxygen's valence electrons.

In C;

A carbon shows two dots at each position and has a charge of +4. This is correct because the charge reflects that it has gained four electrons giving it 8 valence electrons.

In D;

An oxygen shows two dots above, to the left and below, with a charge of -2. This is incorrect as not all Lewis dots are present. Two are missing.

Answer:

Explanation:

In KCl, the two elements that combine to create KCl are potassium (K) and chlorine (Cl).

Potassium, as a Group 1 element, possesses one valence electron in its outermost shell which it readily donates during bonding. Every element aims to achieve a stable electron configuration, typically with 2 or 8 electrons in its outer shell. Potassium is characterized by its lower electronegativity and higher ionization energy, making it more likely to donate its electron than to accept one. On the other hand, chlorine belongs to Group 17 and has 7 electrons in its outer shell, requiring just one additional electron to complete its octet. Chlorine’s higher electronegativity and lower ionization energy facilitate its tendency to accept an electron rather than donate it.

The bond between potassium and chlorine that results in KCl is termed an electrovalent bond.

Reaction equation:

K + Cl → KCl

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

10 mL of this solution is further diluted to 250 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

10 mL of this solution is further diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

5 mL of this solution is diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

10 mL of this solution is further diluted to 500 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

10 mL of this solution is further diluted to 1000 mL

Convert μM:

Thus, this dilution scheme will yield a 200μM solution.