Ozone which is present in the stratospheric region of atmosphere is helpful for preventing harmful UV rays from reaching the surface of earth. Due to human activity, several compounds (specifically chlorofluorocarbons) are released in atmosphere. Due to inherent chemical stability of these compounds, the remain stable in lower region of atmosphere and slowly diffuse into stratosphere. On reaching the stratosphere, these compounds reacts with ozone and thereby depletes the effective concentration of ozone present in atmosphere. Hence, <span>the Montreal Protocol was signed in 1987 by major countries of the world. This aim of this protocol was to protect the stratospheric ozone layer by phasing out the production and consumption of ozone-depleting substances.</span>
In water particles a much closer together and they can quickly transmit the vibration energy from one particle to the next. This means the sound wave travels over four times faster than it would in air but a takes a lot of energy to start the vibration.
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Answer:
817.66 g/mol
Explanation:
The <em>freezing point change</em> can be calculated by the formula:
Where m is the molality of the solute in NO, Kf is the cryoscopic constant (in K·kg/mol), and i is 1 in this case (because the substance in non-dissociating).
We can <u>calculate Kf of NO</u> with the information given by the problem and the formula:
Where R is the universal gas constant (8.314 J·mol⁻¹·K⁻¹), M is the molar mass of NO (30 g/mol, or 0.03 kg/mol), T f is the freezing point of NO (in K), and ΔH is the heat of fusion.
- Kf = 8.314 J·mol⁻¹·K⁻¹ * 0.03 kg/mol * (109.16 K)² ÷ 2300 J/mol = 1.292 K·kg/mol
Now we calculate the molality of the solute in NO:
<em>Molality is equal to the moles of solute per kilogram of solvent</em>:
- 0.1370 m = moles solute/ 1 kg NO
- moles solute = 0.1370 moles
With the given mass of the solute we <u>can calculate the molar mass</u>:
- 112 g / 0.1370 moles = 817.66 g/mol
Answer:
HgSO₄
Explanation:
% => g => moles => ratio => reduce => empirical ratio
%Hg = 67.6% => 67.6g/201g/mol = 0.34mol
%S = 10.8% => 10.8g/32g/mol = 0.34mol
%O = 21.6% => 21.6g/16g/mol = 1.35mol
Hg:S:O => 0.34:0.34:1.35
Reduce to whole number ratio by dividing by the smaller mole value...
Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4
∴ Empirical Formula is HgSO₄