Answer:
Q = -1045 J
Explanation:
Given data:
Mass of water = 5.00 g
Initial temperature = 348.0 K
Final temperature = 298.0 K
Heat given off = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water is 4.18 J/g.K
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 298.0 K - 348.0 K
ΔT = - 50 K
Q = 5.0 g ×4.18 J/g.K× - 50 K
Q = -1045 J
Answer:
48.03grams
Explanation:
Density of a substance is calculated as follows:
Density (g/cm³) = mass (g) / volume (cm³)
Based on the information given in this particular question:
Density of lead (Pb) = 11.3 g/cm³
Volume of lead (Pb) = 4.25 cm³
Density = m/V
11.3 = mass/4.25
Mass = 48.025
= 48.03grams.
Answer: Phosphate is colorless , soft, waxy solid that glows in the dark when at room temperature.
Explanation:
Answer:
B
B
A
C
Explanation:
3
B. Vanillin as a phenoxide ion (conjugate base)
4
B. Protonation of acetic anhydride
5
A. 1.97 mmol of vanillin and 8.45 mmol acetic anhydride
6
C. 1 mmol of vanillin and 10.6 mmol acetic anhydride