HCl and NaOH react based on the following equation:
HCl + NaOH ..............> NaCl + H2O
This reaction is known as a double replacement reaction. In this types of reactions, the electrolyte (and the solvent) is water and the driving force is mainly the formation of a non-electrolyte produce.
Based on this, the driving force in the above neutralization reaction is the formation if the non-electrolyte salt (NaCl)
Answer:
<em>o</em>-bromotoluene, <em>m</em>-bromotoluene and <em>p</em>-bromotoluene.
Explanation:
Hello,
In this case, on the attached picture you will find the reaction which yields <em>o</em>-bromotoluene as the first product, <em>m</em>-bromotoluene as the second product and <em>p</em>-bromotoluene as the last one since the substitution could be done at the second (ortho), third (meta) or fourth (para) carbons on the toluene.
Regards.
Answer:
0.02405 g/L is the solubility of argon in water at 25 °C.
Explanation:
Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.
To calculate the molar solubility, we use the equation given by Henry's law, which is:

where,
= Henry's constant = 
= partial pressure of carbonated drink = 0.51atm
Putting values in above equation, we get:

Molar mass of argon = 39.95 g/mol
Solubility of the argon gas :

0.02405 g/L is the solubility of argon in water at 25 °C.
Answer:
21.6 g
Explanation:
The reaction that takes place is:
First we<u> convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄
- 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂
0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.
Now we <u>calculate how many moles of water are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.6 mol CH₄ *
= 1.2 mol H₂O
Finally we<u> convert 1.2 moles of water into grams</u>, using its <em>molar mass</em>:
- 1.2 mol * 18 g/mol = 21.6 g
C + O2= CO2



CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %