Answer: A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.
Explanation:
Given:
= 0.20 M,
= 15.0 mL
= 0.10 M,
= ?
Formula used is as follows.

Substitute the values into above formula s follows.
![M_{1}V_{1} = M_{2}V_{2}\\0.20 M ]times 15.0 mL = 0.10 M ]times V_{2}\\V_{2} = 30 mL](https://tex.z-dn.net/?f=M_%7B1%7DV_%7B1%7D%20%3D%20M_%7B2%7DV_%7B2%7D%5C%5C0.20%20M%20%5Dtimes%2015.0%20mL%20%3D%200.10%20M%20%5Dtimes%20V_%7B2%7D%5C%5CV_%7B2%7D%20%3D%2030%20mL)
Thus, we can conclude that a 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.
132 g of C , 22 g of H , 176 g of O
132 + 22 + 176 => 330 g <span>of the substance
</span>Now convert the masses in <span>moles :
</span>
C = 12.0 u H = 1.0 u O = 16.0 u
C = 132 / 12.0 => 11 moles
H = 22 / 1.0 => 22 moles
O = 176 / 16.0 => 11 moles
Using the values obtained the lowest proportion in mols of elements present, simply divide the values found for the least of them<span>:
</span>
C = 11 / 11 => 1
H = 22 / 11 => 2
O = 11 / 11 => 1
formula empirically <span>is : CH</span>₂O
hope this helps!
Answer:
n = 0.573mol
Explanation:
PV = nRT => n = PV/RT
P = 1.5atm
V = 8.56L
R = 0.08206Latm/molK
T = 0°C = 273K
n = (1.5atm)(8.56L)/(0.08206Latm/molK)(273K) = 0.573mol
Given:
<span>CS2 + 3O2 → CO2 + 2SO2
</span><span>114 grams of CS2 are burned in an excess of O2
</span>
moles CS2 = 114 g/76.143 g/mol → 114g * mol/76.143 g = 1.497 mol
<span>the ratio between CS2 and SO2 is 1 : 2 </span>
moles SO2 formed = 1.497 x 2 = 2.994 moles → 2nd option