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Answer:
Grams of mercury= 0.06 g of Hg
Note: The question is incomplete. The complete question is as follows:
A compact fluorescent light bulb contains 4 mg of mercury. How many grams of mercury would be contained in 15 compact fluorescent light bulbs?
Explanation:
Since one fluorescent light bulb contains 4 mg of mercury,
15 such bulbs will contain 15 * 4 mg of mercury = 60 mg
1 mg = 0.001 g
Therefore, 60 mg = 0.001 g * 60 = 0.06 g of mercury.
Compact fluorescent lightbulbs (CFLs) are tubes containing mercury and noble gases. When electricity is passed through the bulb, electron-streams flow from a tungsten-coated coil. They collide with mercury atoms, exciting their electrons and creating flashes of ultraviolet light. A phosphor coating on the inside of the tube absorbs this UV light flashes and re-emits it as visible light. The amount of mercury in a fluorescent lamp varies from 3 to 46 mg, depending on lamp size and age.
Answer:
The final temperature is 348.024°C.
Explanation:
Given data:
Specific heat of copper = 0.385 j/g.°C
Energy absorbed = 7.67 Kj (7.67×1000 = 7670 j)
Mass of copper = 62.0 g
Initial temperature T1 = 26.7°C
Final temperature T2 = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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 = T2 - T1
Q = m.c. ΔT
7670 J = 62.0 g × 0.385 j/g °C ×( T2- 26.7 °C
)
7670 J = 23.87 j.°C ×( T2- 26.7 °C
)
7670 J / 23.87 j/°C = T2- 26.7 °C
T2- 26.7 °C = 321.324°C
T2 = 321.324°C + 26.7 °C
T2 = 348.024°C
The final temperature is 348.024°C.
Answer:
Its basically a series of events that take place in a cell. A cell spends most of its time in what is called inter phase, and during this time it grows, replicates its chromosomes, and prepares for cell division. Then after that The cell then leaves interphase, undergoes mitosis, and completes its division.
Explanation:
-Hailey: )
Answer: 4.15234 m
512 g H2O * 
= 0.512 kg H2O
Nitric Acid: HNO3 = 1.008 + 14.007 + 3(15.999) = 63.012 g/mol
H = 1.008 g/mol
N = 14.007 g/mol
O3 = 3*15.999
134 g HNO₃ * 
= 2.126 mol
m = 
= 4.15234 m
