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3 years ago

The fundamental unit used to measure mass is??

Chemistry

2 answers:

Soloha48 [4]3 years ago

6 0

The fundamental unit used to measure mass is Kg (kilograms).

yawa3891 [41]3 years ago

4 0

The unit used to measure mass is Kilograms (Kg)

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The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.

Marina86 [1]

Answer:

Therefore it will take 7.66 hours for 80% of the lead decay.

Explanation:

The differential equation for decay is

$\frac{dA}{dt}= kA$

$\Rightarrow \frac{dA}{A}=kdt$

Integrating both sides

ln A= kt+c₁

$\Rightarrow A= e^{kt+c_1}$

$\Rightarrow A=Ce^{kt}$ [ $e^{c_1}=C$ ]

The initial condition is A(0)= A₀,

$\therefore A_0=Ce^{0.k}$

$\Rightarrow C=A_0$

$\therefore A=A_0e^{kt}$ .........(1)

Given that the $Pb_{209}$ has half life of 3.3 hours.

For half life $A=\frac12 A_0$ putting this in equation (1)

$\frac12A_0=A_0e^{k\times3.3}$

$\Rightarrow ln(\frac12)= 3.3k$ [taking ln both sides, $ln \ e^a=a$ ]

$\Rightarrow k=\frac{ln \frac12}{3.3}$

⇒k= - 0.21

Now A₀= 1 gram, 80%=0.8

and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram

Now putting the value of k,A and A₀in the equation (1)

$\therefore 0.2=1e^{(-0.21)\times t}$

$\Rightarrow e^{-0.21t}=0.2$

$\Rightarrow -0.21t= ln(0. 2)$

$\Rightarrow t= \frac{ln (0.2)}{-0.21}$

⇒ t≈7.66

Therefore it will take 7.66 hours for 80% of the lead decay.

3 0

3 years ago

PLS HELP

vivado [14]

Answer:

D. All of these.

Explanation:

5 0

3 years ago

While heating up a 25 gram sample of concrete (specific heat = 0.210-cal/g°C), your initial tempărature is room temperature (25°

Lana71 [14]

Answer:

Final temperature = 83.1 °C

Explanation:

Given data:

Mass of concrete = 25 g

Specific heat capacity = 0.210 cal/g. °C

Initial temperature = 25°C

Calories gain = 305 cal

Final temperature = ?

Solution:

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

305 cal = 25 g ×0.210 cal/g.°C × T2 - 25°C

305 cal = 5.25cal/°C × T2 - 25°C

305 cal / 5.25cal/°C = T2 - 25°C

58.1 °C = T2 - 25°C

T2 = 58.1 °C + 25°C

T2 = 83.1 °C

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The answer is statement #3.

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