The mass percent of sulfurous acid in the new solution : 38.9%
<h3>Further explanation</h3>
<em>In a container you have 800 g of a 35% by mass solution of sulfurous acid, from which 80 ml of water evaporates. What is the mass percent of sulfurous acid in the new solution? data: density of water is 1g / ml.</em>
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solution 1
composition :
solution 2(new solution)
composition :
- Total mass of new solution after water evaporated
- %mass of acid in a new solution
Answer:
Q = -3980.9 j
Explanation:
Given data:
Mass of sample = 30 g
Initial temperature = 56.7 °C
Final temperature = 25 °C
Specific heat of water = 4.186 j/g.°C
Amount of heat released = ?
Formula:
Q = m.c.ΔT
Q = heat released
m = mass of sample
c = specific heat of given sample
ΔT = change in temperature
Solution:
ΔT = T2 -T1
ΔT = 25 °C - 56.7 °C = - 31.7°C
Q = m.c.ΔT
Q = 30 g × 4.186 j/g.°C × - 31.7°C
Q = -3980.9 j
The best way to display this data for analysis would be a line graph
because line graphs are used to track changes over short periods of
time. when smaller changes exist. Line graphs can also be used to
compare changes over the same period of time for more than one group.
Kinetic energy=Ek
Ek=(1/2)mv²
Ek=480 J
v=8 m/s
mass=?
Ek=(1/2)mv²
480 J=(1/2)m(8 m/s)²
480 J=(32 m²/s²) m
m=(480 J)/(32 m²/s²)=15 kg
answer: the mass of the object is 15 kilograms.
