Answer:
The range of [H⁺] is from 2.51 x 10⁻⁶ M to 6.31 x 10⁻⁶ M,
Explanation:
To answer this problem we need to keep in mind the <u>definition of pH</u>:
So now we <u>calculate [H⁺] using a pH value of 5.2 and of 5.6</u>:
-5.2 = log [H⁺]
= [H⁺]
6.31 x 10⁻⁶ M = [H⁺]
-5.6 = log [H⁺]
= [H⁺]
2.51 x 10⁻⁶ M = [H⁺]
Answer:
237.2 mL.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
(XMV) acid = (XMV) base.
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HCl = (XMV) NaOH.</em>
<em></em>
For HCl; X = 1, M = 0.5 M, V = ??? mL.
For NaOH, X = 1, M = 1.54 M, V = 77.0 mL.
<em>∴ V of HCl = (XMV) NaOH / (XV) HCl = (</em>1)(1.54 M)(77.0 mL) / (1)(0.5 M) = <em>237.2 mL.</em>
Answer:
1. CaCO3 + 2HCl → CaCl2 + H2O + CO2
2. C6H12O2 + 8O2 → 6CO2 + 6H2O
Explanation:
The mass of NaCl formed is 8.307 grams
<u><em> calculation</em></u>
step 1: write the equation for reaction
Na₂CO₃ + 2HCl → 2 NaCl +CO₂ +H₂O
Step 2: find the moles of Na₂CO₃
moles = mass/molar mass
The molar mass of Na₂CO₃ is = (23 x2) + 12 + ( 16 x3) = 106 g/mol
moles = 7.5 g/106 g/mol =0.071 moles
Step 3: use the mole ratio to determine the mole of NaCl
Na₂CO₃:NaCl is 1:2 therefore the moles of NaCl =0.07 x2 =0.142 moles
Step 4: calculate mass of NaCl
mass= moles x molar mass
the molar mass of NaCl= 23 +35.5 =58.5 g/mol
mass = 0.142 moles x 58.5 g/mol =8.307 grams
