Answer:
The molarity of the HCl solution should be 4.04 M
Explanation:
<u>Step 1:</u> Data given
volume of HCl solution = 10.00 mL = 0.01 L
volume of a 1.6 M NaOH solution = 25.24 mL = 0.02524 L
<u>Step 2:</u> The balanced equation
HCl + NaOH → NaCL + H2O
Step 3: Calculate molarity of HCl
n1*C1*V1 = n2*C2*V2
Since the mole ratio for HCl and NaOH is 1:1 we can just write:
C1*V1 =C2*V2
⇒ with C1 : the molarity of HCl = TO BE DETERMINED
⇒ with V1 = the volume og HCl = 10 mL = 0.01 L
⇒ with C2 = The molarity of NaOH = 1.6 M
⇒ with V2 = volume of NaOH = 25.24 mL = 0.02524 L
C1 * 0.01 = 1.6 * 0.02524
C1 = (1.6*0.02524)/0.01
C1 = 4.04M
The molarity of the HCl solution should be 4.04 M
12gHe/1 × 1molHe/4.0026g × 6.02x10^23atomHe/1mol = 1.8 atoms
False because the sugar is just changing it’s form
Make sure have same amounts of species on both sides
Cu (s) + 2 AgNO3 (aq) -> Cu(NO3)2 (aq) + 2 Ag (s)
Number of moles of FeCl2 used = mass/ molar mass
Number of moles = 507/126.751 = 4.
If one mole of Fe reacts with two moles of sodium
Then 4 moles of Fe produces 8 moles of sodium.
Number of moles of sodium = mass/molar mass
Molar mass of sodium chloride = 23 +35.5 = 58.5 g/mol
Hence mass = 8 * 58.5 = 468 g. Hence Option A.
