Answer:
0.89
Explanation:
You are mixing an acid and a base, so there will be a neutralization reaction.
One of the reactants will be in excess, so we must determine its concentration and then calculate the pH.
<em>Moles of NaOH
</em>
Moles of NaOH = 25 × 0.173
Moles of NaOH = 4.32 mmol
===============
<em>Moles of HCl
</em>
Moles of HCl = 35 × 0.342
Moles of HCl = 12.0 mmol
===============
<em>Amount of excess reactant
</em>
NaOH + HCl ⟶ NaCl +H₂O
<em>n</em>/mmol: 4.32 12.0
The 4.32 mmol of NaOH reacts completely with 4.32 mmol of HCl.
Excess HCl = 12.0 – 4.32
Excess HCl = 7.6 mmol
===============
<em>Concentration of the excess HCl
</em>
Total volume = 25 + 35
Total volume = 60 mL
<em>c </em>= millimoles HCl/millilitres HCl
<em>c</em> = 7.6/60
<em>c</em> = 0.13 mol/L
===============
<em>Calculate the pH
</em>
The HCl dissociates completely to hydronium ions, so
[H₃O⁺] = 0.13 mol·L⁻¹
pH = -log[H₃O⁺]
pH = -log0.13
pH = 0.89
The question does not provide the equation
Answer:-
72.89 grams
Explanation:-
The balanced chemical equation for this reaction is
CuSO4 + Fe --> FeSO4 + Cu
Molecular weight of CuSO4 = 63.55 x 1 + 32 x 1 + 16 x 4
= 159.55 gram
Atomic weight of Cu = 63.55 gram.
According to the balanced chemical equation
1 CuSO4 gives 1 Cu
∴159.55 gram of CuSO4 would give 63.55 gram of Cu.
183 gram of CuSO4 would give 63.55 x 183 / 159.55
= 72.89 grams of Cu
Answer:
An atom is composed of a positively-charged nucleus, with a cloud of negatively-charged electrons surrounding it, bound together by electrostatic force. ... Protons and neutrons are bound together to form a nucleus by the nuclear force.
If an object<span> has a net </span>force<span> acting on it, it </span>will<span> accelerate. The </span>object will speed<span> up, slow down or </span>change direction<span>. An </span>unbalanced force<span> (net </span>force) acting on anobject<span> changes its </span>speed<span> and/or </span>direction<span> of motion. An </span>unbalanced force<span> is an unopposed </span>force<span> that causes a </span>change<span> in motion.</span>
