Answer:
About 547 grams.
Explanation:
We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.
To do so, we can use the initial value and convert it to grams using the molar mass.
Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:
Dimensional Analysis:
In conclusion, about 547 grams of copper (II) bicarbonate is produced.
Answer:
1.25 gram of cesium-137 will remain.
Explanation:
Given data:
Half life of cesium-137 = 30 year
Mass of cesium-137 = 5.0 g
Mass remain after 60 years = ?
Solution:
Number of half lives passed = Time elapsed / half life
Number of half lives passed = 60 year / 30 year
Number of half lives passed = 2
At time zero = 5.0 g
At first half life = 5.0 g/2 = 2.5 g
At 2nd half life = 2.5 g/ 2 = 1.25 g
Thus. 1.25 gram of cesium-137 will remain.
1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.
Concentration in chemistry is calculated by dividing a constituent's abundance by the mixture's total volume.
It is calculated in mg/ml.
The unit of measurement frequently used for electrolytes is the milliequivalent (mEq). This value compares an element's chemical activity, or combining power, to that of 1 mg of hydrogen.
Formula for calculating concentration in mg/ml is
Conc. (mg/ml) = M(eq) /ml × Molecular weight / Valency
Given
M(eq) NaCl/ ml = 23.5
Molecular weight pf NaCl = 58.5 g/mol
Valency = 1
Putting the values into the formula
Conc. (mg/ml) = 23.5 ×58.5/1
= 1374.75 mg/ml
Hence, 1374.75 is the concentration in milligrams per ml of a solution containing 23.5 meq sodium chloride per milliliter.
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You can automatically rule out CH₄ since it has no lone pairs at all around the central atom. Water has 2. Ammonia is the only Lewis structure that contains one lone pair.
Answer:
There are 1.4754246675000002e+24 atoms of Hydrogen within the measurement of 2.45 moles of hydrogen!
Explanation:
