The mass of calcium hydroxide that is formed when 10.0 g of CaO reacts with 10.0 g of water is 13.024 grams
calculation
from the equation
CaO + H2O → Ca(OH)2,
1 moles of CaO reacted with 1 moles of H2O to form 1 moles of Ca(OH)2
find the moles of each reactant
moles=mass/molar mass
moles of CaO= 10 g/56 g/mol=0.179 moles
moles of H2O = 10 g/18 g/mol 0.556 moles
CaO is the limiting reagent therefore by use of mole ratio of CaO:Ca(OH)2 which is 1:1 moles of Ca(OH)2 is = 0.179 moles
mass= moles x molar mass
= 0.176 moles x 74 g/mol = 13.024 grams
Answer:
Explanation:
The acidity of a solution is measured by its pH, which is the logarithm of the inverse of the molar concentration of hydronium (H₃O⁺) ions:
- pH = log 1 / [H₃O⁺] = - log [H₃O⁺]
When you know the pH value you can find hydronium concentration using the antilogaritm function:
![pH=-log[H_3O^{+}]\\ \\ {[H_3O^+]}=10^{-pH}\\ \\ {[H_3O^+]}=10^{-2.50}\\ \\ {[H_3O^+]}=0.0032](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%7B%2B%7D%5D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D10%5E%7B-pH%7D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D10%5E%7B-2.50%7D%5C%5C%20%5C%5C%20%7B%5BH_3O%5E%2B%5D%7D%3D0.0032)
The unit of molar concentration is M.
To prove your answer you can take the logarithm of 0.0316:
The balanced chemical reaction is written as:
Sb2S3 + 6HCl = 6SbCl<span>3 + 3H2S
We are given the amount of </span><span>antimony(III) sulfide to be used in the reaction. This is amount will be used for the calculations. We do as follows:
2.85 g Sb2S3 ( 1 mol / </span><span>339.715 g ) ( 6 mol SbCl3 / 1 mol Sb2S3 ) (</span> 228.13 g / mol ) = 11.48 g SbCl3
Answer:
=16.49 L
Explanation:
Using the equation
P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273
P1V1/T1= P2V2/T2
0.6×30/298= 1×V2/273
V2=16.49L
Answer:
B.) 1.3 atm
Explanation:
To find the new pressure, you need to use Gay-Lussac's Law:
P₁ / T₁ = P₂ / T₂
In this equation, "P₁" and "T₁" represent the initial pressure and temperature. "P₂" and "T₂" represent the final pressure and temperature. After converting the temperatures from Celsius to Kelvin, you can plug the given values into the equation and simplify to find P₂.
P₁ = 1.2 atm P₂ = ? atm
T₁ = 20 °C + 273 = 293 K T₂ = 35 °C + 273 = 308 K
P₁ / T₁ = P₂ / T₂ <----- Gay-Lussac's Law
(1.2 atm) / (293 K) = P₂ / (308 K) <----- Insert values
0.0041 = P₂ / (308 K) <----- Simplify left side
1.3 = P₂ <----- Multiply both sides by 308
