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2 years ago

What happens when iodoform is heated with silver powder.

2 answers:

7 0

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When iodoform is heated with silver it forms acetylene (ethyne) and silver iodide

$\sf CHI_3 + 6Ag +CHI_3 \xrightarrow{ \Delta}CH≡CH + 6AgI$

wel2 years ago

4 0

When iodoform is heated with silver powder then it will form acetylene.

<h3>What is iodoform?</h3>

CHI3 is the chemical formula for iodoform, an organo-iodine compound. A pale yellow, crystalline, volatile chemical, it has a deep and unique smell and, akin to chloroform, sweetish taste. It's even used as a cleanser.

The reaction of iodoform is written as.

When iodoform will react with silver powder then it will form acetylene and silver nitrate.

$CHI_{3} +6Ag+ CHI_{3}$ → $C_{2} H_{2} + 6Ag I$ .

Therefore, when iodoform is heated with silver powder then it will form acetylene.

To know more about iodoform click here.

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H2SO4 +HI → __ H2S+12 +H2O balance the equation

Citrus2011 [14]

Answer:

H2SO4 + 8HI → H2S + 4I2 + 4H2O

3 0

2 years ago

Boron has an average mass of 10.81. One isotope of boron has a mass of 10.012938 and a relative abundance of 19.80 percent. The

Andrej [43]

The average mass of an atom is calculated with the formula:

average mass = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2) + ... an so on

For the boron we have two isotopes, so the formula will become:

average mass of boron = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2)

We plug in the values:

10.81 = 0.1980 × 10.012938 + 0.8020 × mass of isotope (2)

10.81 = 1.98 + 0.8020 × mass of isotope (2)

10.81 - 1.98 = 0.8020 × mass of isotope (2)

8.83 = 0.8020 × mass of isotope (2)

mass of isotope (2) = 8.83 / 0.8020

mass of isotope (2) = 11.009975

mass of isotope (1) = 10.012938 (given by the question)

5 0

3 years ago

Helpppppp helpppppp helppppp

soldier1979 [14.2K]

where is the diagram?

without the diagram i can't help

7 0

3 years ago

Select all statements that are correct:____

topjm [15]

Answer:

A, C and D are correct.

Explanation:

Hello.

In this case, since the relationship between the vapor pressure of a solution is directly proportional to the mole fraction of the solvent and the vapor pressure of the pure solvent as stated by the Raoult's law:

$P_{vap}^{solution}=x_{solvent}P_{solvent}$

Since the solute is not volatile, the mole fraction of the solute is not taken into account for vapor pressure of the solution, therefore A is correct whereas B is incorrect.

Moreover, since the higher the vapor pressure, the weaker the intermolecular forces due to the fact that less more molecules are like to change from liquid to vapor and therefore more energy is required for such change, we can evidence that both C and D are correct.

Best regards.

4 0

3 years ago

Cl2(aq) + H2O H+(aq) + Cl–(aq) + HOCl(aq)

dimulka [17.4K]

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6 0

3 years ago

What happens when iodoform is heated with silver powder.​

What happens when iodoform is heated with silver powder.