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Ymorist
29 days ago
11

The highest energy occupied molecular orbital in the li−li bond of the li2 molecule is _____.

Chemistry
2 answers:
KiRa [2.7K]29 days ago
6 0

Response:

2s

Reasoning:

Hello,

To find the designated orbital, we first need to construct the electron configuration for Li:

1s^22s^1

Thus, the 2s orbital represents the highest energy level that is occupied, indicating it is in the most stable state.

Best regards.

castortr0y [2.7K]29 days ago
4 0
The answer is likely to be toluene.
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Write the chemical formula for iridium(III) nitride?
lorasvet [2515]

Answer:

Ir(NO2)3

Explanation:

The molar mass is 330.2335, in case that's also required.

3 0
1 month ago
Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calcul
Anarel [2600]

Answer:

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Explanation:

Assuming all calculations occur at standard pressure and a temperature of -1.72°C :

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Where

n is the number of moles of hydrogen

n is the mass of hydrogen

\rho is the density of hydrogen

6 0
1 month ago
Read 2 more answers
5 grams of NaHCO3 was added to 50 cm3 of 1.5 mol dm-3 hydrochloric acid, causing the temperature to decrease by 6.8 K. What is t
Anarel [2600]

The balanced chemical equation for the neutralization of HCl with NaHCO_{3} is:

HCl (aq) + NaHCO_{3}(aq)--> NaCl (aq) + H_{2}O(l)+CO_{2} (g)

Given weight of NaHCO_{3} = 5g

Moles of NaHCO_{3} = 5 g *\frac{1 mol}{84 g} = 0.05952 mol NaHCO_{3}

Volume of HCl solution = 50 cm^{3} * \frac{1 mL}{1cm^{3}} = 50 mL

Assuming the density of the solution is 1.0 g/mL

Mass of HCl solution = 50 g

Overall mass of the solution = 50 g + 5 g = 55 g

To find the heat of neutralization, we calculate:

Q = m C ΔT

where m equals the mass of the solution = 55 g

C represents the specific heat capacity of the solution = 4.184\frac{J}{g. ^{0}C}

ΔT signifies the temperature change = 6.8 K = (6.8 - 273) C = -266.2^{0}C

Q = 55 g * 4.184 \frac{J}{g. K}(6.8K) = 1565 J

The enthalpy of neutralization per mole of NaHCO_{3}

= \frac{1565J}{0.05952 mol} = 26294 \frac{J}{mol}*\frac{1 kJ}{1000J} =26.294kJ/mol

3 0
1 month ago
Read 2 more answers
In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was form
KiRa [2711]

I predict that there will be an increase in the seconds recorded in the time column. This is because, as more water is mixed with sodium thiosulfate, its concentration diminishes in each flask. Additionally, a lower concentration results in a slower reaction rate since fewer molecules of sodium thiosulfate means there are less frequent collisions with sulfuric acid. With fewer collisions occurring in the reaction, it takes a longer time for the reaction to complete, leading to increased time when sodium thiosulfate is diluted.

Explanation:

I can confirm that this explanation is accurate.

6 0
1 month ago
21.7 mL of gas at 98.8 kPa is allowed to expand at constant temperature into a 52.7 mL container. What is the new pressure of th
alisha [2704]
The new pressure of the gas is calculated to be 40.7 kPa. Using the principle that P1 • V1 = P2 • V2, we can set 98.8 kPa (P1) multiplied by 21.7 mL (V1) equal to P2 (unknown pressure) multiplied by 52.7 mL (V2). To isolate P2, we rearrange the equation to P2 = (98.8 kPa • 21.7 mL) / 52.7 mL, resulting in P2 equal to 40.7 kPa.
8 0
22 days ago
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