The answers are a.) 0.03 mol KOH requires 0.03 mol HCl, b.) 2 mol NH3 requires 2 mol HCl and c.) 0.1 mol Ca(OH)2 requires 0.2 mol HCl.
Solution:
We need to write the balanced equations for each reactions to find out the stoichiometry for each reactants.
a.) HCl (aq) + KOH (aq) → KCl (aq) + H2O(ℓ)
From the balanced equations, we can see that 1 HCl reacts with 1 KOH, therefore if 0.03 mol KOH is reacted then 0.03 mol HCl must also be present.
b.) HCl(aq) + NH3(aq) ) → NH4Cl(aq)
If 2 moles of NH3 are reacted then 2 moles of HCl must also be present since 1 HCl reacts with 1 NH3 from the balanced reaction.
c.) 2HCl(aq) + Ca(OH)2(s) → CaCl2(aq) + 2H2O(ℓ)
We can see that 2 HCl react with 1 Ca(OH)2, hence if 0.1 mol of Ca(OH)2 is reacted then 0.2 mol HCl must also be present.
Answer:
8.1107 g
Explanation:
The given reaction:
Given that:
Mass of silver sulfadiazine = 25.0 g
Molar mass of silver sulfadiazine = 357.14 g/mol
The formula for the calculation of moles is shown below:
Thus,
From the reaction,
2 moles of silver sulfadiazine are formed from 1 mole of silver oxide
So,
1 mole of silver sulfadiazine are formed from 1/2 mole of silver oxide
0.07 mole of silver sulfadiazine are formed from 1/2*0.07 mole of silver oxide
Moles of silver oxide = 0.035 moles
Molar mass of silver oxide = 231.735 g/mol
Mass = Moles * Molar mass = 0.035 moles * 231.735 g/mol = 8.1107 g
Answer:
wave length=speed/frequency
=365/23
wave length=speed/frequency
=365/35000
Atomic number is 57, and the atomic mass is 138.90547 u + 0.00007 u
Answer: The number of N atoms in 137.0 g of N2O3 21.67 x 10∧23 atoms.
Explanation:
- We must obtain the number of moles of the compound: (n = mass/molar mass), mass = 137.0 g and molar mass of N2O3 = 76.01 g/mol.
- n = (137.0 g)/ (76.01 g/mol) = 1.80 mol.
- It is necessary to determine the number of molecules of this sample.
- Every mole contains Avagadro's number (6.02 x 10^23) of molecules.
- The number of molecules = (6.02 x 10^23)(1.80) = 10.84 x 10∧23 molecules.
- Every molecule of N2O3 contain 2 atoms of N.
- The number of N atoms in 137.0 g of N2O3 = (10.84 x 10∧23 molecule) (2 atoms) = 21.67 x 10∧23 atoms.
