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DochEvi
10 days ago
8

Citric acid is a naturally occurring compound. what orbitals are used to form each indicated bond? be sure to answer all parts.2

xsafari

Chemistry
1 answer:
alisha [964]10 days ago
7 0

Response: Below are the orbitals responsible for each bond identified in citric acid per the attachment.

Response 1) σ Bond a: Carbon uses SP^{2} and Oxygen employs SP^{2}.

Clarification: The sigma bonds are formed through the hybrid orbitals of carbon and oxygen. This occurs at the 'a' location in the citric acid structure.

Response 2) π Bond a: Both Carbon and Oxygen have π orbitals.

Clarification: The π-bond at position 'a' consists of interactions between the π orbitals of carbon and oxygen.

Response 3) Bond b: Oxygen SP^{3} and Hydrogen solely utilizes the S orbital.

Clarification: The bonding at position 'b' includes oxygen and hydrogen atoms, with hydrogen utilizing its S orbital.

Response 4) Bond c: Carbon is SP^{3} and Oxygen is also SP^{3}.

Clarification: The bonding process at position 'c' involves both carbon and oxygen atoms with their respective hybrid orbitals.

Response 5) Bond d: Carbon atom has SP^{3} and the second carbon has SP^{3}.

Clarification: In position 'd', the bond formed between carbon atoms is SP^{3}, utilizing orbitals that underwent SP^{3} hybridization which are SP^{3}.

Response 6) Bond e: C1 has O SP^{2}.

C2 has SP^{3}.

Clarification: The carbon that contains oxygen and a double bond utilizes SP^{2} hybridized orbitals; conversely, carbon at C2 employs SP^{3} hybridized orbitals in this bonding at position 'e'.

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Consider butter (density= 0.860 g/mL) and sand (density= 2.28 g/mL). If 1.00 mL of butter were mixed with 1.00 mL of sand and mi
Anarel [852]

The mixture’s density is 1.57 g/cm³.


Step 1: Determine the mass of the butter.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{0.680 g} }{\text{1 cm}^{3 }} = \text{0.860 g}\\

Step 2: Determine the mass of the sand.


\text{Mass} = \text{1.00 cm}^{3 } \times \frac{\text{2.28 g} }{\text{1 cm}^{3 }} = \text{2.28 g}\\

Step 3: Determine the density of the mixture.

Total mass = 0.860 g + 2.28 g = 3.14 g.

Total volume = 1 cm³ + 1 cm³ = 2 cm³

\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{3.14 g} }{\text{2 cm}^{3 }} = \textbf{1.57 g/cm}{^{3}\\

6 0
1 day ago
At what temperature would the volume of a gas be 0.550 L if it had a volume of 0.432 L at –20.0 o C?
castortr0y [927]
To find the temperature at which the volume of the gas would be 0.550 L, given that it is 0.432 L at -20.0 °C, apply Charles’s Law.

The formula is v1/T1 = v2/T2
Known values:
V1 = 0.550 L
T1 = ?
T2 = -20°C + 273 = 253 K
V2 = 0.432 L

Rearranging for T1:
T1 = (V1 × T2) / V2

Calculating:
T1 = (0.55 L × 253) / 0.432 L = 322.11 K or 49.11°C
8 0
13 days ago
Read 2 more answers
Two hypothetical ionic compounds are discovered with the chemical formulas XCl2 and YCl2, where X and Y represent symbols of the
Tems11 [854]

Answer:

THE MOLAR MASS OF XCL2 IS 400 g/mol

THE MOLAR MASS OF YCL2 IS 250 g/mol.

Explanation:

We derive the molar mass of XCl2 and YCl2 by recalling the molar mass formula when both mass and the number of moles are known.

Number of moles = mass / molar mass

Molar mass = mass / number of moles.

For XCl2,

mass = 100 g

number of moles = 0.25 mol

Thus, molar mass = mass / number of moles

Molar mass = 100 g / 0.25 mol

Molar mass = 400 g/mol.

For YCl2,

mass = 125 g

number of moles = 0.50 mol

Molar mass = 125 g / 0.50 mol

Molar mass = 250 g/mol.

Accordingly, the molar masses for XCl2 and YCl2 are 400 g/mol and 250 g/mol, respectively.

3 0
5 days ago
Write the chemical formula for iridium(III) nitride?
lorasvet [960]

Answer:

Ir(NO2)3

Explanation:

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

3 0
9 days ago
Given the connection between Aw and K (Aw=2k) could you use the ideal gas law and derive the Boltzmann constant. Water freezes a
lions [1003]

Answer:

Explanation:

The relationship between the new temperature scale and the absolute temperature scale is defined as follows

Aw = 2 K

for K = 273.15 (the freezing point of water on the absolute scale)

Aw = 2 x 273.15 = 546.3 K

Each division of the new scale is equivalent to half that of each division on the absolute scale

each division of the new scale is minimal.

The value of R = 8.314 J per mole per K

Here, per K corresponds to 2Aw

Hence, the value of R in the new scale = 8.314/2 J per mole per Aw

= 4.157 J per mole per Aw

k = R / N

= 4.157 / 6.02 x 10²³

= .69 x 10⁻²³

= 6.9 x 10⁻²⁴ J per molecule per Aw .

7 0
15 days ago
Read 2 more answers
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