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r-ruslan [8.4K]
3 years ago
5

HELPPP How can we use x ray wavelengths for?

Chemistry
1 answer:
Ket [755]3 years ago
7 0

Answer:

May you post a picture for us to understand?

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Smh why do people always have to use fake profile pictures? Its not like anyone is gonna judge you it isnt facebook its a websit
agasfer [191]
Seriously though
people love to do that stuff .-.

7 0
3 years ago
A mixture of hydrogen (2.02 g) and chlorine (35.90 g) in a container at 300 K has a total gas pressure of 748 mm Hg. What is the
Llana [10]

The partial atmospheric pressure (atm) of hydrogen in the mixture is 0.59 atm.

<h3>How do we calculate the partial pressure of gas?</h3>

Partial pressure of particular gas will be calculated as:

p = nP, where

  • P = total pressure = 748 mmHg
  • n is the mole fraction which can be calculated as:
  • n = moles of gas / total moles of gas

Moles will be calculated as:

  • n = W/M, where
  • W = given mass
  • M = molar mass

Moles of Hydrogen gas = 2.02g / 2.014g/mol = 1 mole

Moles of Chlorine gas = 35.90g / 70.9g/mol = 0.5 mole

Mole fraction of hydrogen = 1 / (1+0.5) = 0.6

Partial pressure of hydrogen = (0.6)(748) = 448.8 mmHg = 0.59 atm

Hence, required partial atmospheric pressure of hydrogen is 0.59 atm.

To know more about partial pressure, visit the below link:
brainly.com/question/15302032

#SPJ1

3 0
2 years ago
What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

\Delta H_{fus} = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{freezing}}{T_m}

\Delta S=\frac{-3170J/mol}{175.4K}

\Delta S=-18.07J/mol.K

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0
3 years ago
A student placed equal volumes of honey and of water in two, identical, open dishes, and left them at room temperature for 8 hou
Ahat [919]

Answer: Honey has a much lower vapor pressure than pure water has. So, pure water evaporates at a much higher rate.

Explanation:

8 0
3 years ago
Read 2 more answers
What is the most effective way to differentiate between potassium feldspar and plagioclase feldspar?
Tresset [83]
<span>plagioclase feldspars have striations and potassium feldspars don't have striations</span>
3 0
4 years ago
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