Answer:
The calorimeter constant is = 447 J/°C
Explanation:
The heat absorbed or released (Q) by water can be calculated with the following expression:
Q = c × m × ΔT
where,
c is the specific heat
m is the mass
ΔT is the change in temperature
The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.
The heat absorbed by the calorimeter (Q) can be calculated with the following expression:
Q = C × ΔT
where,
C is the calorimeter constant
The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).
Qabs + Qrel = 0
Qabs = - Qrel
Qcal + Qw₁ = - Qw₂
Qcal = - (Qw₂ + Qw₁)
Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)
Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]
Ccal = 447 J/°C
A weak acid has a low concentration of H+ Ions and a dilute acid is a solution where acid is dissolved in a more volume of water than that of acid.
Answer:
1) d = 2.4 g/cm³
2) m = 25 g
3) v = 126.7 cm³
Explanation:
Given data:
Mass of material = 24 g
Volume of material = 10 cm³
Density of material = ?
Solution:
Formula:
d = m/v
by putting value,
d = 24 g / 10 cm³
d = 2.4 g/cm³
2) Given data:
Density of material = 5 g/cm³
Volume of material = 5 cm³
Mass of material = ?
Solution:
Formula:
d = m/v
5 g/cm³ = m / 5 cm³
m = 5 g/cm³×5 cm³
m = 25 g
3)Given data:
Density of material = 3 g/cm³
Mass of material = 380 g
Volume of material = ?
Solution:
Formula:
d = m/v
3 g/cm³ = 380 g / v
v = 380 g /3 g/cm³
v = 126.7 cm³
Answer:
112 mL
Explanation:
The formula for percent by volume is

If you have 250 mL of a solution that is 44.8 % v/v,

Answer:

Explanation:
Hello there!
In this case, it is possible to comprehend these mass-particles problems by means of the concept of mole, molar mass and the Avogadro's number because one mole of any substance has 6.022x10²³ particles and have a mass equal to the molar mass.
In such a way, for C₆H₁₂O₆, whose molar mass is about 180.16 g/mol, the referred mass would be:

Best regards!