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Oksana_A [137]
2 years ago
13

Consider each possible structure of carbon dioxide, c o 2. Determine whether the structure is correct and justify your decision.

Chemistry
1 answer:
Mrrafil [7]2 years ago
6 0

This structure is correct due to the total number of bonds and electrons

We employ the octet rule, which states that all atoms in a compound are expected to follow, to check the accuracy of any chemical structure. The octet rule is precisely satisfied for each atom in the depicted structure of carbon IV oxide. The valence shell of each atom in the molecule is surrounded by eight electrons. We may thus infer that this structure is accurate given the total number of bonds and electrons since CO2 has sixteen valence electrons.

It possesses a negative charge of 1.602176634 coulombs, the basic unit of electric charge, which is comparable to its negative charge. The rest mass of an electron is 9.1093837015 10^-31 kg, or just 1/1,836 the mass of a proton. An electron is therefore considered to have practically negligible mass in comparison to a proton or neutron, and its matter is not taken into consideration when calculating an atom's mass number.

The question is incomplete, the complete question is found in the image attached to this answer

To know more about   bonds visit : brainly.com/question/13190131

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A 475 ml sample of a gas was collected at room temperature of 23.5 °C and a pressure of
Molodets [167]

Answer:

The volume when the conditions were altered is 0.5109 L or 510.9 mL

Explanation:

Using the general gas equation,

P1 V1 / T1 = P2 V2 / T2

where;

P1 = 756 mmHg

V1 = 475 ml = 0.475 L

T1 = 23.5°C = 23.5 + 273K = 275.5 K

P2 = 722 mm Hg

T2 = 10°C = 10 + 273 K = 283 K

V2 = ?

Rearranging to make V2 the subject of the formula, we obtain:

V2 = P1 V1 T2 / P2 T1

V2 = 756 * 0.475 * 283 / 722 * 275.5

V2 = 101, 625.3 / 198911

V2 = 0.5109 L or  510.9 mL

3 0
4 years ago
Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al (s) + 3O2 (g) → 2Al2O3 (s) In
jek_recluse [69]

Answer: The percent yield of the reaction is 74 %

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol

For oxygen gas:

\text{Moles of oxygen gas}=\frac{2.5g}{32g/mol}=0.078mol

The chemical equation for the reaction of titanium and chlorine gas follows:

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

By Stoichiometry of the reaction:

4 moles of aluminium reacts with 3 moles of oxygen.

So, 0.092 moles of aluminium reacts with = \frac{3}{4}\times 0.092=0.069mol of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

4 moles of aluminium produce = 2 moles of Al_2O_3

So, 0.092 moles of aluminium will produce = \frac{2}{4}\times 0.092=0.046moles of Al_2O_3

Now, calculating the mass of aluminium oxide:

\text{Mass of aluminium oxide}=moles\times {\text {molar mas}}=0.046mol\times 102g/mol=4.7g

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield  = 3.5 g

Theoretical yield = 4.7 g

Putting values in above equation, we get:

\%\text{ yield of reaction}=\frac{3.5g}{4.7g}\times 100\\\\\% \text{yield of reaction}=74\%

Hence, the percent yield of the reaction is 74 %

6 0
3 years ago
One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was 20.826 ℃, and the temperature af
gtnhenbr [62]

Answer:fH = - 3,255.7 kJ/mol

Explanation:Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.

Qsystem + Qcombustion = 0

Qsystem = heat capacity*ΔT

10000*(25.000 - 20.826) + Qc = 0

Qcombustion = - 41,740 J = - 41.74 kJ

So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:

n = mass/molar mass = 1/ 78

n = 0.01282 mol

fH = -41.74/0.01282

fH = - 3,255.7 kJ/mol

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Under normal conditions, describe how increasing the temperatures effects the solubility of a typical salt
AnnZ [28]
The higher the temperature the higher the solubility
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