Answer:
Explanation:
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One important measure of the rate at which a radioactive substance decays is called half-life, or t1/2. Half-life is the amount of time needed for one half of a given quantity of a substance to decay. Half-lives as short as 10–6 second and as long as 109 years are common.
1) D = 13.6 g / mL
2)ethyl alcohol weighs 158g
3)ρ
_copper = 8.9 g 
Explanation:
1)
D = m / V
=306.0 g / 22.5 mL
D= 13.6 g / mL
2)
density = mass / volume
mass = density × volume
=0.789g /ml × 200.0 ml
M=158g
Ethyl alcohol weighs 158g
3)
ρ (density) = Mass / Volume
ρ
_copper = 1896 g / 8.4cm × 5.5cm × 4.6cm
= 1896g / 212.5
ρ
_copper=8.9 g
There are several information's already given in the question. Based on those information's, the answer can be easily deduced.
Amount of gasoline required by Harry's car to travel 25 miles = 1 gallon
Then
amount of gasoline required
by Harry's car to travel 15000 miles = 15000/25
= 600 gallons
So
Amount of CO2 released by burning 1 gallon of gasoline = 20 pounds
Then
Amount of CO2 released
by burning 600 gallon of gasoline = 600 * 20
= 12000 pounds
From the above deduction, it can be concluded that the amount of CO2 that will be added by Harry's car to the atmosphere is 12000 pounds.
The absorbance reported by the defective instrument was 0.3933.
Absorbance A = - log₁₀ T
Tm = transmittance measured by spectrophotometer
Tm = 0.44
Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654
True absorbance can be calculated by true transmittance, Tm = T+S(α-T)
S = fraction of stray light = 6%= 6/100 = 0.06
α= 1, ideal case
T = true transmittance of the sample
Tm = T+S(α-T)
now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233
therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)
i.e; 0.3933
To know more about transmittance click here:
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Answer:
Volume
Explanation:
Volume is the amount of space taken up or occupied by an object.
