The normal pH of surface sea water is _____. 5.0 6.0 7.0 9.0
2 answers:
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To answer this question, you need to know the concept of half-life, which is how a radioactive material decreases in mass over time.
The half life of U-235 is 703.8 million years. The first part of this problem is to find the scale factor. To do this, divide the time that has past by the half life, like this:
Now, take this scale factor and multiply it by the current mass, like this:
This number is what you add to the current mass to get the original mass. That is because the scale factor showed us that it was just over one half life. Since after one half life, the mass is cut in half, and this is over one half life, when we add to the original it will be a little over double. This equation illustrates the final addition:
I hope this helped you. Fell free to ask any further questions.
The half life of U-235 is 703.8 million years. The first part of this problem is to find the scale factor. To do this, divide the time that has past by the half life, like this:
Now, take this scale factor and multiply it by the current mass, like this:
This number is what you add to the current mass to get the original mass. That is because the scale factor showed us that it was just over one half life. Since after one half life, the mass is cut in half, and this is over one half life, when we add to the original it will be a little over double. This equation illustrates the final addition:
I hope this helped you. Fell free to ask any further questions.
Answer:
Explanation:
Given that:
Pressure = 791 mmHg
Temperature = 20.0°C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (20 + 273.15) K = 293.15 K
T = 293.15 K
Volume = 100 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 62.3637 L.mmHg/K.mol
Applying the equation as:
791 mmHg × 1.14 L = n × 62.3637 L.mmHg/K.mol × 293.15 K
⇒n of produced = 0.0493 moles
According to the reaction:-
1 mole of carbon dioxide is produced 1 mole of calcium carbonate reacts
0.0493 mole of carbon dioxide is produced 0.0493 mole of calcium carbonate reacts
Moles of calcium carbonate reacted = 0.0493 moles
Molar mass of = 100.0869 g/mol
The formula for the calculation of moles is shown below:
Thus,
Impure sample mass = 5.28 g
Percent mass is percentage by the mass of the compound present in the sample.