Explanation:
Many meteorites have iron or nickel in them, so they are heavier (and denser) than Earth rocks. Some meteorites have pits (regmaglypts) on the outside, which look like deep thumbprints. Meteorites are not bubbly, and do not have holes. Meteorite are usually not round.
An individual is hospitalized and the initial blood work indicates high levels of
in the blood and a pH of 7. 47. This would indicate the individual probably has compensated respiratory acidosis.
A chronic illness usually leads to compensated respiratory acidosis because the kidneys have time to adjust to the delayed onset. Even if the
is elevated in a compensated respiratory acidosis, the pH is within the usual range.
The kidneys counteract a respiratory acidosis by increasing the amount of
that tubular cells reabsorb from the tubular fluid, the amount of
that collecting duct cells secrete while also producing
, and the amount of
buffer that is formed through ammoniagenesis.
Respiratory acidosis is frequently brought on by hypoventilation as a result of: breathing depression , paralysis of the respiratory muscles, diseases of the chest wall , abnormalities of the lung parenchyma and abdominal squeezing.
Learn more about Respiratory acidosis here;
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<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
All of the questions here are pertaining to the colligative properties of a solution and the preparation of solutions. Maybe, it would be best if you understand the equations to be used in order to answer these questions.<span>
Freezing point depression or Boiling point elevation:
</span><span>ΔT = -K (m) (i)
</span>ΔT is the change in the freezing point or the boiling point not the freezing point/boiling point. Therefore, it should be added to the original value of the property of the solvent.
<span>
K is a constant called the molal freezing point depression constant and for the boiling point is the boiling point elevation constant. It is a property of the solvent.
</span><span>
m is the concentration of the solute in the solvent in terms of molality or kg solute/kg solvent.
</span><span>
i is the vant hoff factor which will represent the number of ions which the solute dissociates when in solution.</span>
4.6to the twelth power plus ten or something like that it has been a while since I have done the scientific notation but where ever you put in the number of zeros. is twelve