Mg + 1/2 O2 → MgO
1 mol = 24 g of Mg
X mol = 12 g of Mg
x = 0.5 moles of Mg
Mg :MgO = 1:1 (coefficient from equations using mole ratio)
So
0.5 moles of MgO
1 mol MgO = (24+16) g = 40 g
0.5 moles of MgO = 0.5 × 40
= 20 g of MgO produced
I am not sure but this is what I think
This question uses the formula connecting mass, density and volume
Which is Density= Mass/Volume
Convert the mass in g
92.5 kg = 92,500g
7.87g/ml = 92,500/ Volume
Volume= 92,500/7.87
= 11,753.5 ml
Now since we have to give the answer in liters we can just divide by 1000 and get
11.75 litres
Considering that CCL3F gas behave like an ideal gas then we can use the Ideal Gas Law
<span>PV = nRT, however is an approximation and not the only way to resolve this problem with the given data..So,at the end of the solution I am posting some sources for further understanding and a expanded point of view. </span>
<span>Data: P= 856torr, T = 300K, V= 1.1L, R = 62.36 L Torr / KMol </span>
<span>Solving and substituting in the Gas equation for n = PV / RT = (856)(1.1L) /( 62.36)(300) = 0.05 Mol. This RESULT is of any gas. To tie it up to our gas we need to look for its molecular weight:MW of CCL3F = 137.7 gm/mol. </span>
<span>Then : 0.05x 137.5 = 6.88gm of vapor </span>
<span>If we sustract the vapor weight from the TOTAL weight of liquid we have: 11.5gm - 6.88gm = 4.62 gm of liquid.d</span>
<u>Answer:</u> The concentration of solute is 0.503 mol/L
<u>Explanation:</u>
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
where,
= osmotic pressure of the solution = 24 atm
i = Van't hoff factor = 2 (for NaCl)
c = concentration of solute = ?
R = Gas constant = 
T = temperature of the solution = ![25^oC=[273+25]=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5D%3D298K)
Putting values in above equation, we get:
Hence, the concentration of solute is 0.503 mol/L
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