For equal moles of gas, temperature can be calculated from ideal gas equation as follows:
P×V=n×R×T ...... (1)
Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.
2.4 atm ×3.25 L=n×R×297.5 K
Rearranging,
n\times R=0.0262 atm L/K
Similarly at final pressure and volume from equation (1),
1.5 atm ×4.25 L=n×R×T
Putting the value of n×R in above equation,
1.5 atm ×4.25 L=0.0262 (atm L/K)×T
Thus, T=243.32 K
<u>Answer:</u> The solubility of oxygen at 682 torr is ![4.58\times 10^{-3}M](https://tex.z-dn.net/?f=4.58%5Ctimes%2010%5E%7B-3%7DM)
<u>Explanation:</u>
To calculate the molar solubility, we use the equation given by Henry's law, which is:
![C_{A}=K_H\times p_{A}](https://tex.z-dn.net/?f=C_%7BA%7D%3DK_H%5Ctimes%20p_%7BA%7D)
Or,
![\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}](https://tex.z-dn.net/?f=%5Cfrac%7BC_%7B1%7D%7D%7BC_%7B2%7D%7D%3D%5Cfrac%7Bp_%7B1%7D%7D%7Bp_2%7D)
where,
are the initial concentration and partial pressure of oxygen gas
are the final concentration and partial pressure of oxygen gas
We are given:
Conversion factor used: 1 atm = 760 torr
![C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm](https://tex.z-dn.net/?f=C_1%3D1.38%5Ctimes%2010%5E%7B-3%7DM%5C%5Cp_1%3D0.27atm%5C%5CC_2%3D%3F%5C%5Cp_2%3D682torr%3D0.897atm)
Putting values in above equation, we get:
![\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M](https://tex.z-dn.net/?f=%5Cfrac%7B1.38%5Ctimes%2010%5E%7B-3%7D%7D%7BC_2%7D%3D%5Cfrac%7B0.27atm%7D%7B0.897atm%7D%5C%5C%5C%5CC_2%3D%5Cfrac%7B1.38%5Ctimes%2010%5E%7B-3%7D%5Ctimes%200.897atm%7D%7B0.27atm%7D%3D4.58%5Ctimes%2010%5E%7B-3%7DM)
Hence, the solubility of oxygen gas at 628 torr is ![4.58\times 10^{-3}M](https://tex.z-dn.net/?f=4.58%5Ctimes%2010%5E%7B-3%7DM)
<span>1962.252(a)
whenever the material is dropped from 20feet to any poiny lying outside the exterior wall of building, an enclosed chute of wood, or equivalent material. For the purpose of paragraph, an enclose chute is a slide, closed in all sides, through which material is placed from high plac eto lower one.
1962.252(b)
when the disposal is dropped through holes in the floor without the use of chute, the area onto which the material is dropped shall be completely enclosed with barricades not less than 42 inch high and not less than 6 feet back from the projected edge of of the opening valve. signs warning of hazards of falling material shall be posted at each level.
1962.252(c)
All scrap lumper, waste material, and rubbish shall be removed from the immediate work area as the work progresses.
1962.252(d)
Disposal of waste material or debris by burning shall comply with local fire regulation
1962.252(e)
All sovent waste, oily rags, and flammable liquid should be kept in fire resistant covered container until removed from work site.</span>
From the reaction between Cu and HNO₃, the formed gas is NO₂ instead of NO₃. Hence the correct balanced equation would be,
Cu(s) + 4HNO₃(aq) → Cu(NO₃)₂(aq) + 2NO₂(g) + 2H₂O<span>(l)
Here, Cu goes to </span>Cu(NO₃)₂ by changing its oxidation number from 0 to +2 while NO₃⁻ goes to NO₂ by reducing its oxidation state from +5 to +4 . Hence Cu is oxidized by HNO₃ in the reaction.