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zheka24 [161]
4 years ago
7

Which of these statements about electron orbitals is not true?

Chemistry
1 answer:
dusya [7]4 years ago
4 0
One point that is not true is an electron orbital has the same amount of energy in all orbital.
hope this helps!
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Based on the kinetic theory, which statement is true? (5 points)
Hoochie [10]

Answer:

the particles of matter are arranged in different ways for the different states

Explanation:

because solid liquid and gas all three matters have different states for example the particles in a solid are closely packEd and form of movement is vibration

7 0
2 years ago
3. Light acts like (1 point)
alexandr402 [8]
Both a particle and a wave?
6 0
3 years ago
In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t
solmaris [256]

Answer:

4.81 moles

Explanation:

The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.

Pressure at which gauge reads zero = 14.7 psi

Pressure read by the gauge = 988 psi

Total pressure = 14.7 + 988 psi = 1002.7 psi

Also, P (psi) = P (atm) / 14.696

Pressure = 1002.7 / 14.696  = 68.2297 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 4.81 moles

4 0
3 years ago
Question
madam [21]

Answer:

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

Given,

For Metal sample,

mass = 13 grams

T = 73°C

For Water sample,

mass = 60 grams

T = 22°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the  addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that , water sample temperature changed from  22°C to 27°C and metal sample temperature changed from 73°C to 27°C.

Since, Specific heat of water = 4.184 J/g°C

Let Cp be the specific heat of the metal.

Substituting values,

(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)

By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.

5 0
3 years ago
What is the volume of 1.56 kg of a compound whose molar mass is 81.86 g/mole and whose density is 41.2 g/ml?
hjlf

Answer:

v = 37.9 ml

Explanation:

Given data:

Mass of compound = 1.56 kg

Density = 41.2 g/ml

Volume of compound = ?

Solution:

First of all we will convert the mass into g.

1.56 ×1000 = 1560 g

Formula:

D=m/v

D= density

m=mass

V=volume

v = m/d

v =  1560 g / 41.2 g/ml

v = 37.9 ml

7 0
3 years ago
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