Answer: 0.5 moles
Explanation:
Cr2(SO3)2 is the chemical formula for chromium sulphate.
Given that,
Amount of moles of Cr2(SO3)2 (n) = ?
Mass of Cr2(SO3)2 in grams = 128.9g
For molar mass of Cr2(SO3)2, use the atomic masses:
Chromium, Cr = 52g;
Sulphur, S = 32g;
Oxygen, O = 16g
Cr2(SO3)2 =
(52g x 2) + [(32g + 16g x 3) x 2]
= 104g + [(32g + 48g) x 2]
= 104g + [80g x 2]
= 104g + 160g
= 264g/mol
Since, n = mass in grams / molar mass
n = 128.9g / 264g/mol
n = 0.488 mole [Round the value of n to the nearest tenth which is 0.5
Thus, there are 0.5 moles in 128.9 grams of Cr2(SO3)2
I think d is wrong because Everything else is correct for sure. also, minerals vary in the position they hold and have different chemical compounds.
Answer:
3.47 × 10²³ C atoms
Solution:
Data Given:
Mass of Diamond = 48 mg = 0.048 g
M.Mass of Diamond = 12.01 g.mol⁻¹ (as it is purely Carbon)
Step 1: Calculate Moles of Diamond as,
Moles = Mass ÷ M.Mass
Putting values,
Moles = 0.048 g ÷ 12.01 g.mol⁻¹
Moles = 0.576 mol
Step 2: Calculate number of Carbon atoms,
As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Carbon atoms can be written as,
Moles = Number of C Atoms ÷ 6.022 × 10²³ Atoms.mol⁻¹
Solving for Number of C atoms,
Number of C atoms = Moles × 6.022 × 10²³ Atoms.mol⁻¹
Putting value of moles,
Number of C atoms = 0.576 mol × 6.022 × 10²³ Atoms.mol⁻¹
Number of C atoms = 3.47 × 10²³ C atoms
The wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
<em>"Your question is not complete, it seems to be missing the diagram of the emission spectrum"</em>
the diagram of the emission spectrum has been added.
<em>From the given</em><em> chart;</em>
The wavelength of the atomic emission corresponding to the orange line is 610 nm = 610 x 10⁻⁹ m
The frequency of this emission is calculated as follows;
c = fλ
where;
- <em>c is the speed of light = 3 x 10⁸ m/s</em>
- <em>f is the frequency of the wave</em>
- <em>λ is the wavelength</em>

The energy of the emitted photon corresponding to the orange line is calculated as follows;
E = hf
where;
- <em>h is Planck's constant = 6.626 x 10⁻³⁴ Js</em>
<em />
E = (6.626 x 10⁻³⁴) x (4.92 x 10¹⁴)
E = 3.26 x 10⁻¹⁹ J.
Thus, the wavelength of the orange line is 610 nm, the frequency of this emission is 4.92 x 10¹⁴ Hz and the energy of the emitted photon corresponding to this <em>orange line</em> is 3.26 x 10⁻¹⁹ J.
Learn more here:brainly.com/question/15962928