CHEMISTRY HIGH SCHOOL

Amines are derivatives of the ______ molecule.aldehyde
ammonia
acetone
nitric acid

Answers

Answer 1
Answer: Amines are derivatives of the ammonia molecule. Ammonia or azane is a compound of nitrogen and hydrogen, it is a colorless gas with a characteristic pungent smell. 
Answer 2
Answer:

Answer: ammonia

Explanation:- Functional groups are specific group of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules.

1. Aldydes have functional group -O=C-H.

Example: Ethanal with molecular formula CH_3CHO

2. Amines have functional group -NH_2.

Example: Methylamine with molecular formula CH_3NH_2. they are formed by substituting one hydrogen of ammonia by a R substituent.

3. Ketones have functional group -C=O.

Example: Acetone with molecular formula CH_3COCH_3

4. Nitric acid is a inorganic acid with molecular formula HNO_3.

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Which atomic model states that it is impossible to know the exact location of electrons around the nucleus?

Answers

The correct answer is electron-cloud model. This model states that electrons don't go around in an eliptical way but rather completely erratically and uncontrolably and this is why it can't be known where they are or how they will behave. They form sort of like a cloud of many electrons that move erratically so you have to check it to see where it is.

Answer:

Electron -cloud model

Explanation:

Edge 2021

The temperature of a sample of matter is ameasure of the
(1) average potential energy of the particles of
the sample
(2) average kinetic energy of the particles of the
sample
(3) total nuclear energy of the sample
(4) total thermal energy of the sample

Answers

The temperature of a sample of matter is ameasure of the... average kinetic energy of the particles of thesample.

Balance this equation please.

Answers

2K + 2H2O = 2KOH + H2
2K + 2H2O = 2KOH + H2  Actually Really what That other guy said

The empirical formula of a compound is found to be P2O5. Experiments show that the molar mass of the compound is 426g/mol. What is the molecular formula of the compound?Answer: P4O10

Answers

Answer:

Molecular formula = P₄O₁₀

Explanation:

Given data:

Empirical formula of compound = P₂O₅

Molar mass of compound = 426 g/mol

Molecular formula = ?

Solution:

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass=  P₂O₅  = 283.89 g/mol

n = 426 g/mol / 283.89 g/mol

n = 2

Molecular formula = n (empirical formula)

Molecular formula = 2 (P₂O₅ )

Molecular formula = P₄O₁₀

Calculate the number of kilojoules to warm 125 g of iron from 23.5 °C to 78.0 °C.

Answers

Final answer:

Using the formula q = mcΔT, and substituting the values for mass, specific heat capacity of iron, and temperature change, it is calculated that it takes approximately 3.058 KJ to warm 125 g of iron from 23.5 °C to 78.0 °C.

Explanation:

To calculate the amount of heatneeded to warm 125 g of iron from 23.5 °C to 78.0 °C, we use the formula q = mcΔT, where 'm' is the mass in kilograms, 'c' is the specific heat capacity, and 'ΔT' is the temperature change. In this case, the mass 'm' is 0.125 kg (since 1 g = 10^-3 kg), the specific heat capacity 'c' of iron is 0.449 J/g°C (or 449 J/kg°C), and 'ΔT' is 78.0 °C - 23.5 °C = 54.5 °C.

Substituting these values into the formula, we get q = (0.125 kg) * (449 J/kg°C) * (54.5 °C), which gives a result of approximately 3.058 KJ.

Therefore, it would take approximately 3,058 KJ to warm 125 g of iron from 23.5 °C to 78.0 °C.

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Final answer:

To warm 125 g of iron from 23.5 °C to 78.0 °C, it requires approximately 3.93 kilojoules of energy.

Explanation:

To calculate the number of kilojoules required to warm 125 g of iron from 23.5 °C to 78.0 °C, we can use the formula:

q = m * c * ΔT

Where:

  • q is the energy absorbed or released
  • m is the mass of the substance
  • c is the specific heat capacity of the substance
  • ΔT is the change in temperature

Using the given values:

  • m = 125 g
  • c = 0.450 J/g°C (specific heat capacity of iron)
  • ΔT = 78.0 °C - 23.5 °C

Substituting the values into the formula:

q = 125 g * 0.450 J/g°C * (78.0 °C - 23.5 °C)

Simplifying the equation:

q = 125 * 0.450 * (78.0 - 23.5)

q ≈ 3933.75 J ≈ 3.93 kJ

Therefore, it requires approximately 3.93 kilojoules of energy to warm 125 grams of iron from 23.5 °C to 78.0 °C.

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What is the correct molar mass for the compound CaBr3

Answers

CaBr₃ = 40 + 80 * 3 = 280 g/mol

hope this helps!
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