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Grace [21]
2 years ago
12

A hydroxide group (OH) is a component of most...

Chemistry
1 answer:
qaws [65]2 years ago
5 0

Answer: bases

Explanation:

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Potassium (K) and calcium fluoride (CaF2) can both be classified as<br> (Please help)
nasty-shy [4]
Potassium and calcium fluoride are both metals
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Complexes containing metals with d10 electron configurations are typically colorless because ________. Complexes containing meta
algol [13]

Answer:

there is no d electron that can be promoted via the absorption of visible light

Explanation:

One of the properties of transition elements is the possession of incompletely filled d orbitals. This property accounts for their unique colours.

The colours of transition metal compounds stem from d-d transition of electrons due to the presence of vacant d orbitals of appropriate energy to which electrons could be promoted.

For elements whose atoms have a d10 configuration, such vacant orbitals does not exist hence their compounds are not colored.

Sometimes, the colour of transition metal compounds stem from ligand to metal charge transfer(LMCT) for instance in KMnO4.

8 0
2 years ago
What is the temperature shown on the thermometer below?
olchik [2.2K]
The temperature is -8 F

Answer D
7 0
2 years ago
Read 2 more answers
Aspirin synthesis involves the addition of an acetyl group to salicylic acid in a condensation reaction with an alcohol. The ace
BARSIC [14]

Answer:

The correct appropriate will be Option 1 (Acid anhydrides are less stable than esters so the equilibrium favors the ester product.)

Explanation:

  • Acid anhydride, instead of just a carboxyl group, is typically favored for esterification. The predominant theory would be that Anhydride acid is somewhat more volatile than acid. This is favored equilibrium changes more toward the right of the whole ester structure.
  • Extremely responsive than carboxylic acid become acid anhydride as well as acyl chloride. Thus, for esterification, individuals were most favored.

The other options offered are not relevant to something like the scenario presented. So, the solution here is just the right one.

3 0
3 years ago
When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular
Luda [366]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is C_2H_2O_4 and C_6H_4O_2

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=1.03g

Mass of H_2O=0.14g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, \frac{12}{44}\times 1.03=0.28g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, \frac{2}{18}\times 0.14=0.016g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = \frac{0.023}{0.00775}=2.96\approx 3

For Hydrogen  = \frac{0.016}{0.00775}=2.06\approx 2

For Oxygen  = \frac{0.00775}{0.00775}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is C_3H_{2}O_1=C_3H_2O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

n=\frac{108.10g/mol}{54g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2

Thus, the empirical and molecular formula for the given organic compound is C_3H_2O and C_6H_4O_2

6 0
3 years ago
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