Answer:
0.0187 M
Explanation:
Step 1: Write the balanced neutralization reaction
NaOH + HCl ⇒ NaCl + H₂O
Step 2: Calculate the reacting moles of HCl
18.7 mL of 0.01500 M HCl react.
0.0187 L × 0.01500 mol/L = 2.81 × 10⁻⁴ mol
Step 3: Calculate the reacting moles of NaOH
The molar ratio of HCl to NaOH is 1:1. The reacting moles of NaOH are 1/1 × 2.81 × 10⁻⁴ mol = 2.81 × 10⁻⁴ mol.
Step 4: Calculate the molarity of NaOH
2.81 × 10⁻⁴ moles are in 15.00 mL of NaOH.
[NaOH] = 2.81 × 10⁻⁴ mol/0.01500 L = 0.0187 M
Use stoichiometry 3.75moles k* 39.10g k\1 mole k= 146.63g k
The balanced chemical reaction would be :
S + O2 = SO2
2SO2 + O2 = 2SO3
We are given the amount of oxygen to be used in the reaction. This will be the starting value in the calculations. We do as follows:
2.56 L O2 ( 1 mol / 22.4 L) ( 1 mol SO2 / 1 mol O2 ) ( 2 mol SO3 / 2 mol SO2 ) 22.4 L / 1 mol = 2.56 mol SO3
Answer:
Explanation:
282000 j/mol *1 mol/6*10^(23) = 4.7*10^(-19) J for one electron
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mass = density x volume
Substitute your values in
mass = 2.7 x 21
Solve:
mass = 56.7g
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