Answer:
The isotope notation for an ion of silver-109 with a charge of positive 1 is 
.
Explanation:
The elements with same atomic number with different mass numbers are called isotopes.
The isotopes of silver is as follows.
From the above two isotopes have same atomic number and different mass number.
From the given,the isotope notation for an ion of silver-109 with a charge of positive 1
It can be represented is as follows.
General representation of isotope is as in attachment.
Answer:
33.8 m
Explanation:
(a) We want to know the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO₃ by mass).
Step 1: Determine the mass of HNO₃ and water in 100 grams of solution.
Step 2: Convert the mass of HNO₃ to moles.
Step 3: Convert the mass of water to kilograms.
Step 4: Calculate the molality.
(b)
Step 1
In 100 g of solution, there are 68.0 g of HNO₃ and 100 g - 68.0 g = 32.0 g of water.
Step 2
The molar mass of HNO₃ is 63.01 g/mol. The moles corresponding to 68.0 g are:
68.0 g × (1 mol/63.01 g) = 1.08 mol
Step 3
The mass of water is 32.0 g = 0.0320 kg
Step 4
The molality of HNO₃ is:
m = moles of solute / kilograms of solvent
m = 1.08 mol / 0.0320 kg
m = 33.8 m
Answer:
wind
Explanation:
it would be naturally caused by the pressure
The balanced chemical reaction is:
Na2SO4 + Ba(OH)2 = 2NaOH +BaSO4
We are given the amount of the reactants to be used. These values will be the starting point of our calculations.
0.0820 mol/L Ba(OH)2 (2.27 L solution) = 0.18614 mol Ba(OH)2
0.0664 mol/L Na2SO4 (3.06 L solution) = 0.20318 mol Na2SO4
The limiting reactant is the Ba(OH)2. The amount for this compound will be used.
0.18614 mol Ba(OH)2 (1 mol BaSO4 / 1 mol Ba(OH)2 ) (233.43 g BaSO4 / 1 mol BaSO4)= 43.4507 g BaSO4
