Answer:
0.21 g
Explanation:
The equation of the reaction is;
NaCl(aq) + AgNO3(aq) -----> NaNO3(aq) + AgCl(s)
Number of moles of NaCl= 0.0860 g /58.5 g/mol = 0.00147 moles
Number of moles of AgNO3 = 30/1000 L × 0.050 M = 0.0015 moles
Since the reaction is 1:1, NaCl is the limiting reactant.
1 mole of NaCl yields 1 mole of AgCl
0.00147 moles of NaCl yields 0.00147 moles of AgCl
Mass of precipitate formed = 0.00147 moles of AgCl × 143.32 g/mol
= 0.21 g
The molarity is count by dividing the mole of the solute within 1 liter of solvent. In this case, the KNO3 is 16.8g with 101.11 g/mol molar mass. Then we need to find the mol first. The calculation would be: 16.8g / (101.11g/mol)= 0.0166 mol.
Then the molarity would be: 0.0166mol/ 0.3l= 0.0498= 0.0553 M
that's your work bro do by your own
Answer:
1.99 x 10⁻¹⁸J
Explanation:
Given parameters:
Frequency of the wave = 3 x 10¹⁵Hz
Unknown:
Energy of the photon = ?
Solution:
To solve this problem, we use the expression below;
E = hf
Where E is the energy, h is the Planck's constant and f is the frequency
Now insert the parameters and solve for E;
E = 6.63 x 10⁻³⁴ x 3 x 10¹⁵ = 19.9 x 10⁻¹⁹J or 1.99 x 10⁻¹⁸J
<span>2 BAO2 = 2 BAO + O<span>2 is the balanced equation</span></span>
