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2 years ago

Write the chemical formula for iridium(III) nitride?

Chemistry

1 answer:

lubasha [3.4K]2 years ago

3 0

Answer:

Ir(NO2)3

Explanation:

it's Molar Mass is 330.2335 if you need that too

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A student obtained an unknown metal sample that weighed 65.3 g and at a temperature of 99.8oC, he placed it in a calorimeter con

lawyer [7]

Answer:

0.377 J/gºC

Explanation:

From the question given above, the following data were obtained:

Mass of metal (Mₘ) = 65.3 g

Initial temperature of metal (Tₘ) = 99.8 °C

Mass of water (Mᵥᵥ) = 43.7 g

Initial temperature of water (Tᵥᵥ) = 25.7 °C

Equilibrium temperature (Tₑ) = 34.5 °C

Specific heat capacity of water (Cᵥᵥ) = 4.18 J/gºC

Specific heat capacity of metal (Cₘ) =?

The specific heat capacity of metal can be obtained as illustrated below:

Heat lost by metal = heat gained by water.

MₘCₘ(Tₘ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Cᵥᵥ)

65.3 × Cₘ (99.8 – 34.5) = 43.7 × 4.18 (34.5 – 25.7)

65.3Cₘ × 65.3 = 182.666 × 8.8

4264.09Cₘ = 1607.4608

Divide both side by 4264.09

Cₘ = 1607.4608 / 4264.09

Cₘ = 0.377 J/gºC

Therefore the specific heat capacity of the metal is 0.377 J/gºC

3 0

2 years ago

Which statement describes the transfer of heat energy that occurs when an ice cube is added to an insulated container with 100 m

nadya68 [22]

Number 3 is the most likely answer

4 0

2 years ago

In two or more complete sentences describe all of the van der Waals forces that exist between molecules of sulfur

zavuch27 [327]

Answer:

Dipole-Dipole attraction

Explanation:

Dipole-dipole attraction is a type of vander waals forces found in the molecules of sulfur dioxide.

Vander waals forces are weak attractions joining non-polar and polar molecules together. They are of two types:

London dispersion forces which are weak attractions found between non-polar molecules.
Dipole-Dipole attraction are the forces of attraction which exists between polar molecules. Such molecules have permanent dipoles. This implies that the positive pole of one molecule attracts the negative pole of another. This is what happens between the oxygen and sulfur molecules.

3 0

2 years ago

Read 2 more answers

Using the mass of the proton 1.0073 amu and assuming its diameter is 1.0×10−15m, calculate the density of a proton in g/cm3.

icang [17]

Answer : $3.2 X 10^{15} g/cm^{3}$

Explanation : To convert amu i.e. atomic mass unit in grams we have the conversion factor as 1 amu = $1.66054 X 10^{-24} g$

we know the mass of the proton is 1.0073 amu

So converting it into grams we have to multiply;

1.0073 amu X $1.66054 X 10^{-24} g/amu$ = $1.673 X 10^{-24} g$

Now, Volume = 1/6πd³ as diameter is given as $1.0 X 10^{-15} m$ converting it to cm will require to multiply with 100

∴ Volume = 1/6π $(1.0 X 10^{-15}mX 100 cm / 1 m)^{3}$

Hence, volume = $5.236 X 10^{-40} cm^{3}$

Therefore, Density = mass / volume

∴ Density = $1.673 X 10^{-24} g / 5.236 X 10^{-40} cm^{3}$

Therefore, Density will be $3.2 X 10^{15} g/cm^{3}$ .

6 0

2 years ago

Read 2 more answers

The diagram shows the movement of particles from one end of the container to the opposite end of the container. A cylindrical co

trapecia [35]

Answer:

C. effusion because there is a movement of a gas through a small opening into a larger volume

Explanation:

Effusion makes fluid/gas molecules move to the container with less pressure or larger volume. In diffusion, the movement should work two ways even though one side might receive more. But in effusion, the movement is rather one way.

This case shows how effusion work because its not the concentration that makes the balls moving to the bottom part of the container. No ball moving from bottom container to top either.

3 0

2 years ago

Read 2 more answers