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ICE Princess25 [194]

3 years ago

12

Imagine the following scenario involving marbles as "radioactive isotopes".

1 answer:

sveta [45]3 years ago

8 0

Answer:

1 million years

Explanation:

The half life of a radioactive isotope refers to the time it takes for the radioactive isotope to decay to half of its original amount.

The half life of a radioactive isotope is independent of the amount of starting material. Hence, whether the amount of starting material is small or large, the half life of the substance remains the same.

Hence the half life of 2 green marbles and 500,000 green marbles is 1 million years.

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What is one observation you can make aboit venus

PilotLPTM [1.2K]

Answer:

The temperature is very hot.

Explanation:

The gas around the planet traps the suns heat and makes it the hottest planet in our solar system.

8 0

3 years ago

1. A gas has a volume of 45 L at a pressure of 140 kPa. What is the volume when

pishuonlain [190]

Answer:

The new volume of the gas is 21 L.

Explanation:

Volume of a gas is inversely proportional to its pressure at constant temperature such that,

$V\propto \dfrac{1}{P}$

or

$P_1V_1=P_2V_2$

We have,

$V_1=45\ L\\\\P_1=140\ kPa \\\\P_2=300\ kPa$

It is required to find V₂. Using above law or Boyle's law such that :

$V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{140\times 45}{300}\\\\V_2=21\ L$

So, the new volume of the gas is 21 L.

7 0

3 years ago

The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

artcher [175]

Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0

3 years ago

A reproducible measurement is an accurate one. A. True B. False

4vir4ik [10]

The answer is
B. False

5 0

3 years ago

Read 2 more answers

Draw the Lewis structure for carbonate

Nadusha1986 [10]

There you go try that out.

8 0

3 years ago