Question:
The information given is:
Trial # Tiwater T f ΔT Masswater (m)
#1: 21.2 10.8 10.8 24.990
#2: 20.8 9.50 9.5 25.000
#3: 20.9 9.20 9.2 25.010
Answer:
The heat of the reaction is -5985 J
Explanation:
The heat absorbed by the water is given by
ΔQ = m·c·ΔT
From which
∑ (ΔT·m)/3 = 278.34 kg·°C
ΔQ = c×∑ (ΔT·m)/3 = 4.184 J/g·°C×278.34 kg·°C = 1164.565 J
ΔQ Calorimter = Specific heat capacity of calorimeter, 
× ΔT
Where the = 443 J/°C for example, we have
ΔQ Calorimter = 443×11.133 = 4820.733 J
From which the heat of reaction is then
Using 4 digits, we get
.
Answer:
1 gram
Explanation:
1 gram= weight of 1 cubic centimeter of pure water at temperature 4°C = 1000 milligrams = 0.001 kilogram
PO4 anion has a 3- charge.A sodium cation , has a 1+ charge.Now , you have to think how many Na ions you need to fully neutralize the PO4 's 3- charge. Answer 3. That's because a molecule must have a neutral charge (a 0 charge). 3+(-3)=0.So , sodium phosphate has the formula Na3PO4.
Answer:
(3) 5.36
Explanation:
Since this is a titration of a weak acid before reaching equivalence point, we will have effectively a buffer solution. Then we can use the Henderson-Hasselbalch equation to answer this question.
The reaction is:
HAc + NaOH ⇒ NaAc + H₂O
V NaOH = 40 mL x 1 L/1000 mL = 0.040 L
mol NaOH reacted with HAc = 0.040 L x 0.05 mol/L = 0.002 mol
mol HAC originally present = 0.050 L x 0.05 mol/L = 0.0025 mol
mol HAc left after reaction = 0.0025 - 0.002 = 0.0005
Now that we have calculated the quantities of the weak acid and its conjugate base in the buffer, we just plug the values into the equation
pH = pKa + log ((Ac⁻)/(HAc))
(Notice we do not have to calculate the molarities of Ac⁻ and HAc because the volumes cancel in the quotient)
pH = -log (1.75 x 10⁻⁵) + log (0.002/0.0005) = 5.36
THe answer is 5.36
