Answer:
The specific heat of gold is 0.129 J/g°C
Explanation:
Step 1: Data given
Mass of gold = 15.3 grams
Heat absorbed = 87.2 J
Initial temperature = 35.0 °C
Final temperature = 79.2 °C
Step 2:
Q = m*c*ΔT
⇒ Q =the heat absorbed = 87.2 J
⇒ m = the mass of gold = 15.3 grams
⇒ c = the specific heat of gold = TO BE DETERMINED
⇒ ΔT = The change in temperature = T2 - T1 = 79.2 - 35.0 = 44.2 °C
87.2 J = 15.3g * c * 44.2°C
c = 87.2 / (15.3 * 44.2)
c = 0.129 J/g°C
The specific heat of gold is 0.129 J/g°C
Answer:
Kc = 12.58
Explanation:
Kc = [0.229]^2*[0.687]^6/[0.221]^4*[0.5685]^3
Kc = (0.052441)(0.10513)/(0.002385)(0.18373)
Kc = 0.0005513/0.000438
Kc = 12.58
Hope that helps!!
Answer:
In aqueous solution the pH scale varies from 0 to 14, which indicates this concentration of hydrogen. Solutions with pH less than 7 are acidic (the value of the exponent of the concentration is higher, because there are more ions in the solution) and alkaline (basic) those with a pH higher than 7. If the solvent is pure water, the pH = 7 indicates neutrality of the solution
Explanation:
PH is a measure of how acidic or basic a liquid is. Specifically, from a dissolution. The acidity of a solution is essentially due to the concentration of hydrogen ions dissolved in it. In reality, the ions are not found alone, but are in the form of hydronium ions consisting of one oxygen molecule and three positively charged hydrogen. PH precisely measures this concentration. And to do it, we can use simple and very visual methods.
Answer:
The solution becomes diluted.
Explanation:
When you add water to a solution, the number of moles of the solvent stays the same while the volume increases. Therefore, the molarity decreases.
Hope this helps!
Ans: 15.1 grams
Given reaction:
Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3
Mass of Na2CO3 = 20.0 g
Molar mass of Na2CO3 = 105.985 g/mol
# moles of Na2CO3 = 20/105.985 = 0.1887 moles
Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH
# moles of NaOH produced = 0.1887*2 = 0.3774 moles
Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol
Mass of NaOH produced = 0.3774*39.996 = 15.09 grams
