Number of moles of CO2 =
Mass /Ar
= 50.2 / (12 + 32)
1.14 mols
For every 1 mol of gas, there will be
24000 cm^3 of gas
Vol. = 1.14 x 24 dm^3
= 27.36 dm^3
I'm not exactly sure which one but I do know that an acid and a base react in a aqueous solution to form water, so i would probably eliminate the ones that aren't aqueous solutions.
Answer:
1.8 moles of NaCl must be produced.
Explanation:
Based on the reaction:
HCl + NaOH → NaCl + H2O
<em>1 mol of HCl reacts with 1 mol of NaOH to produce 1mol of NaCl</em>
<em />
To solve this question we must find, as first, the <em>limiting reactant:</em>
<em />
1.8 moles of HCl will need 1.8 moles of NaOH for a complete reaction (Ratio of reaction 1:1). As there are 3.3 moles of NaOH,
<em>HCl is limiting reactant</em>
<em />
When the 1.8 moles of HCl react completely,
1.8 moles of NaCl must be produced because 1 mole of HCl produce 1 mole of NaCl
TNT has the molecular formula: C7H5N3O6. And hence, when reacted in oxygen gas, you get what is known as <span>combustion</span> reaction. the reaction is: <span><span>C7</span><span>H5</span><span>N3</span><span>O6</span>+<span>O2</span>→C<span>O2</span>+<span>N2</span>+<span>H2</span><span>O</span></span>
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
![HN_3 + OH- ---> N^-_{3} + H_2O](https://tex.z-dn.net/?f=HN_3%20%2B%20OH-%20---%3E%20N%5E-_%7B3%7D%20%2B%20H_2O)
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of
=
= 0.0458 M
Concentration of
=
= 0.0521 M
GIven that :
Ka = ![1.9 x 10^{-5}](https://tex.z-dn.net/?f=1.9%20x%2010%5E%7B-5%7D)
Thus; it's pKa = 4.72
![pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})](https://tex.z-dn.net/?f=pH%20%3D4.72%20%2B%20%20log%28%5Cdfrac%7B%20%5C%200.0521%7D%7B0.0458%7D%29)
![pH =4.72 + log(1.1376)](https://tex.z-dn.net/?f=pH%20%3D4.72%20%2B%20log%281.1376%29)
![pH =4.72 + 0.05598](https://tex.z-dn.net/?f=pH%20%3D4.72%20%2B%200.05598)
![pH =4.77598](https://tex.z-dn.net/?f=pH%20%3D4.77598)
pH ≅ 4.80