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BlackZzzverrR [31]

2 years ago

12

Defferent between crysttiline solid and amerphus solid

2 answers:

Alexxx [7]2 years ago

5 0

Answer:

Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points. In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures.

Explanation:

Hope this helped!

Olegator [25]2 years ago

4 0

Answer:

Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points. In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures.

Explanation:

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What is the total number of hydrogen atoms in a molecule of heptyne?

Step2247 [10]

Answer:

12

Explanation:

alkyne formula is CnH2n-2

heptyne = 7

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3 years ago

What happens to the electrons in an ionic bond?

Anton [14]

Answer:

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Explanation:

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3 years ago

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I NEED ANSWER ASAP Which of the following statements is true for a scientific theory? It is a well-tested hypothesis. It is veri

Digiron [165]

Answer:

A.) It is a well-tested hypothesis.

7 0

3 years ago

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Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after

Alenkinab [10]

<u>Answer:</u>

<u>For a:</u> The number of moles of air present in the RV is 0.047 moles

<u>For b:</u> The number of molecules of gas is $2.83\times 10^{22}$

<u>Explanation:</u>

<u>For a:</u>

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the air = $37^oC=[37+273]K=310K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of air = ?

Putting values in above equation, we get:

$1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol$

Hence, the number of moles of air present in the RV is 0.047 moles

<u>For b:</u>

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.047 moles of air will contain $(0.047\times 6.022\times 10^{23})=2.83\times 10^{22}$ number of gas molecules.

Hence, the number of molecules of gas is $2.83\times 10^{22}$

7 0

2 years ago

For water at 30C and 1 atm: a= 3.04x10^-4 k^-1 , k =4.52x10^-5 atm ^-1 = 4.46 x10^-10 m^2/N, cpm= 75.3j/9molk), Vm =18.1cm^3/mol

Lapatulllka [165]

Answer:

$C_{vm}$ of water at 30C and 1 atm is 256.834 J/mol·K.

Explanation:

To solve the question, we note the Maxwell relation such as

$C_{pm}-C_{vm}=\frac{9T\alpha ^2 V }{K}$

Where:

$C_{pm}$ = Specific heat of gas at constant pressure = 75.3 J/mol·K

$C_{vm}$ = Specific heat of gas at constant volume = Required

T = Temperature = 30 °C = 303.15 K

α = Linear expansion coefficient = 3.04 × 10⁻⁴ K⁻¹

K = Volume comprehensibility = 4.52 × 10⁻⁵ atm⁻¹

Therefore,

75.3 - $C_v$ = $\frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81 \times 10^{-5} }{4.52 \times 10^{-5} }$

$C_{vm}$ = $\frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81 \times 10^{-5} }{4.52 \times 10^{-5} }$ - 75.3 = 256.834 J/mol·K.

8 0

3 years ago