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BlackZzzverrR [31]
3 years ago
12

Defferent between crysttiline solid and amerphus solid​

Chemistry
2 answers:
Alexxx [7]3 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]3 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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The wavelength of a particular color of violet light is 433 nm. The energy of this wavelength of light is kJ/photon. (109 nm = 1
mash [69]

Answer:

4.59 × 10⁻³⁶ kJ/photon

Explanation:

Step 1: Given and required data

  • Wavelength of the violet light (λ): 433 nm
  • Planck's constant (h): 6.63 × 10⁻³⁴ J.s
  • Speed of light (c): 3.00 × 10⁸ m/s

Step 2: Convert "λ" to meters

We will use the conversion factor 1 m = 10⁹ nm.

433 nm × 1 m/10⁹ nm = 4.33 × 10⁷ m

Step 3: Calculate the energy (E) of the photon

We will use the Planck-Einstein's relation.

E = h × c/λ

E = 6.63 × 10⁻³⁴ J.s × (3.00 × 10⁸ m/s)/4.33 × 10⁷ m

E = 4.59 × 10⁻³³ J = 4.59 × 10⁻³⁶ kJ

5 0
3 years ago
Q5 what's the answer to A?
Snezhnost [94]
The answer is CH3CH2CH2CHO
8 0
3 years ago
Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs
saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of Ag^+ when CuBr just begins to precipitate is, 1.34\times 10^{-6}M

Percent of Ag^+ remains is, 0.0018 %

Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

CuBr\rightleftharpoons Cu^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][Br^-]

4.2\times 10^{-8}=0.073\times [Br^-]

[Br^-]=5.75\times 10^{-7}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][Br^-]

7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M

[Ag^+]=1.34\times 10^{-6}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

Percent of Ag^+ remains = 0.0018 %

3 0
3 years ago
During a titration the following data were collected. A 50.0 mL portion of an HCl solution was titrated with 0.500 M NaOH; 200.
Travka [436]

<u>Answer:</u> The mass of HCl present in 500 mL of acid solution is 36.5 grams

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=200.mL

Putting values in above equation, we get:

1\times M_1\times 50.00=1\times 0.500\times 200\\\\M_1=\frac{1\times 0.500\times 200}{1\times 50.00}=2M

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Molar mass of HCl = 36.5 g/mol

Molarity of solution = 2 M

Volume of solution = 500 mL

Putting values in above equation, we get:

2mol/L=\frac{\text{Mass of solute}\times 1000}{36.5g/mol\times 500}\\\\\text{Mass of solute}=\frac{2\times 36.5\times 500}{1000}=36.5g

Hence, the mass of HCl present in 500 mL of acid solution is 36.5 grams

4 0
3 years ago
A 10.-liter flask at a given temperature and pressure contains 6.0 × 1023 molecules of hydrogen gas. Under the same conditions o
valentina_108 [34]

Answer:

The same number of molecules, 6.0 × 10²³ molecules.

Explanation:

The amount of any given gas that can be stored in a container depends on the <u>temperature, pressure and volume </u>of the container. It does not depend on the nature (or identity) of the gas.

So if a 10-liter flask contains 6.0 × 10²³ molecules of hydrogen gas, it will contain the same amount of molecules of any other gas when temperature and pressure remain constant.

6 0
3 years ago
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