Lar mass of Ca<span> = 40.08 </span>grams/mole 77.4 g Ca<span> * ( 1 </span>mole Ca<span>/ 40.08 ... n = m / M 1mol </span>Ca<span>weights 40 gmol-1 n = 77,4 / 40 = 1.93 </span>mol<span>.</span>
Answer:
1 mol
Explanation:
Using the general gas law equation as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the provided information in the question;
V = 22.4L
T = 273K
P = 1 atm
R = 0.0821 Latm/molK
n = ?
Using PV = nRT
n = PV/RT
n = (1 × 22.4) ÷ (0.0821 × 273)
n = 22.4 ÷ 22.4
n = 1mol
Answer:
The empirical formula is Ag2O.
The empirical formula is Ag2O.Explanation:
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.
The empirical formula is Ag2O.Explanation:The empirical formula is the simplest whole-number ratio of atoms in a compound.The ratio of atoms is the same as the ratio of moles. So our job is to calculate the molar ratio of Ag to 2O.
do the steps ...
To get this into an integer ratio, we divide both numbers by the smaller value.
From this point on, I like to summarize the calculations in a table.
ElementAgMass/gXMolesXllRatiomllIntegers
—————————————————−———mAgXXXm7.96Xm0.07377Xll2.00mmm2
mlOXXXXl0.59mm0.0369Xml1mmmml1
There are 2 mol of Ag for 1 mol of O.
doesnt salt desolve ice? so wouldn't the salt dissolve in the ice water?
