42.4 ml is the volume in milliliters of the lead ball if a lead ball is added to a graduated cylinder containing 50.6 ml of water.
<h3>What is a graduated cylinder?</h3>
A tall narrow container with a volume scale is used especially for measuring liquids.
The graduated cylinder contains water
mL is a volume unit.
Water volume = 50.6 ml
The lead ball caused an increase in volume from 50.6 ml to 93.0 mL.
The new volume is the lead ball volume plus the original water volume :
Final volume = Vlead ball+ Water original volume
Hence, 42.4 ml is the volume in milliliters of the lead ball.
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Answer:
#1 Exposition
#2 Background information
#3 Complication
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Answer:
The final volume is 3.07L
Explanation:
The general gas law will be used:
P1V1 /T1 = P2V2 /T2
V2 =P1 V1 T2 / P2 T1
Give the variables to the standard unit:
P1 = 345 torr = 345 /760 atm = 0.4539atm
T1 = -15°C = -15 + 273 = 258K
V1 = 3.48L
T2 = 36°C = 36+ 273 = 309K
P2 = 468 torr = 468 * 1/ 760 atm = 0.6158atm
V2 = ?
Equate the values into the gas equation, you have:
V2 = 0.4539 * 3.48 * 309 / 0.6158 * 258
V2 = 488.0877 /158.8764
V2 = 3.07
The final volume is 3.07L
Answer:
1 year- 1 mole
time in general- amount of matter
1 second- 1 atom/ particle
