Answer:
72.66%
Explanation:
NH₄Cl reacts in presence of NaOH producing ammonia, NH₃, as follows:
NH₄Cl + NaOH → NaCl + NH₃ + H₂O
The residual NaOH reacts with H₂SO₄ as follows:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.
The equivalent-gram of H₂SO₄ are:
16mL * 0.1N * 1.06 = 1.696mEq.
As the complete residual solution is 100mL but the neutralization was made only with 10mL, the mEq you need to neutralize the residual NaOH is:
1.696mEq * (100mL / 10mL) = 16.96mEq.
The mEq of NaOH you add in the first are:
25mL * 0.95mEq = 23.75mEq
That means the NaOH that reacts = moles of NH₄Cl is:
23.75mEq - 16.96mEq = 6.79mEq = 6.79mmoles NH₄Cl =
6.79x10⁻³ moles NH₄Cl
In grams (Using molar mass NH₄Cl = 53.5g/mol):
6.79x10⁻³ moles NH₄Cl * (53.5g / mol) =
0.3633g of NH₄Cl are in the original mixture.
% is:
0.3633g/ 0.5g * 100 = 72.66%
Based on the information provided, it appears that you will need to calculate the amount of heat absorbed by the water from the peanut that was burned. We are given the following information:
specific heat capacity, c = 1.0 cal/g°C
mass of water = 76 g
Ti = 22°C
Tf = 46°C
change in temperature, ΔT = 24°C
We can use the formula q = mcΔT to measure the amount of energy absorbed by the water to increase in tempature:
q = (76 g)(1.0 cal/g°C)(24°C)
q = 1824 cal
Therefore, the water absorbed 1824 calories from the peanut that was burned.
d. Fe(s) and Al(s)
<h3>Further explanation</h3>
In the redox reaction, it is also known
Reducing agents are substances that experience oxidation
Oxidizing agents are substances that experience reduction
The metal activity series is expressed in voltaic series
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent
The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent
So that the metal located on the left can push the metal on the right in the redox reaction
The electrodes which are easier to reduce than hydrogen (H), have E cells = +
The electrodes which are easier to oxidize than hydrogen have a sign E cell = -
So the above metals or metal ions will reduce Pb²⁺ (aq) will be located to the left of the Pb in the voltaic series or which have a more negative E cell value (greater reduction power)
The metal : d. Fe(s) and Al(s)
Answer:
A I think
Explanation:
im not sure so do with that what you will
