Answer:
Cr₂S₃
Explanation:
From the question given above, the following data were obtained:
Mass of chromium (Cr) = 0.67 g
Mass of chromium sulfide = 1.2888 g
Empirical formula =?
Next, we shall determine the mass of sulphur (S) in the compound. This can be obtained as follow:
Mass of chromium (Cr) = 0.67 g
Mass of chromium sulfide = 1.2888 g
Mass of sulphur (S) =?
Mass of S = (Mass of chromium sulfide) – (Mass of Cr)
Mass of S = 1.2888 – 0.67
Mass of S = 0.6188 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
Mass of Cr = 0.67 g
Mass of S = 0.6188 g
Divide by their molar mass
Cr = 0.67 / 52 = 0.013
S = 0.6188 / 32 = 0.019
Divide by the smallest
Cr = 0.013 / 0.013 = 1
S = 0.019 / 0.013 = 1.46
Multiply by 2 to express in whole number
Cr = 1 × 2 = 2
S = 1.46 × 2 = 3
Therefore, the empirical formula of the compound is Cr₂S₃
Answer:
<h3>Hlo there !! </h3>
<u>One mole of any substance contains 6.022*1023 structural units (atoms, molecules, ions, etc.). This number is known as the Avogadro constant.</u>
<u>One mole of any substance contains 6.022*1023 structural units (atoms, molecules, ions, etc.). This number is known as the Avogadro constant.So 1.04*107 mol of Al contains 1.40*107 * 6.022*1023 = 8.43*1030 structural units (in case of Al – atoms).</u>
<h3><u>8.43*1030 particles Al.</u></h3>
Explanation:
<h3>Hope this helps !!</h3>
The answer you are looking for would be Colloid, because in colloid mixtures, the particles are big enough to reflect them in a way. Colloids can be distinguished from solutions using the Tyndall effect. Light passing through a colloidal dispersion, such as smoke or foggy air, will be reflected by the larger particles and the light beam will be visible. <span />
Here, we should use combined gas law which can be derived from combined gas law, “PV=nRT”. Rearranging, we can get PV/T=nR. Then we can set the two states in the problem together to get
P1V1/T1 = P2V2/T2
Then just plug in and solve algebraically.
Hope this helps
