Answer:
34.15% is the mass percentage of calcium in the limestone.
Explanation:
Mass of precipitate that is calcium oxalate = 140.2 mg = 0.1402 g
1 mg = 0.001 g
Moles of calcium oxalate = 
1 mole of calcium oxalate have 1 mole of calcium atom.
Then 0.001095 moles of calcium oxalate will have 0.001095 moles of calcium atom.
Mass of 0.001095 moles of calcium :
0.001095 mol × 40 g/mol = 0.04381 g
Mass of sample of limestone = 128.3 mg = 0.1283 g
Percentage of calcium in limestone:
34.15% is the mass percentage of calcium in the limestone.
The answer is 79.9 g.
Copper takes 92.0% of aluminum bronze and it is a limiting factor. We have aluminum in excess, so we need to make a proportion.
If 73.5 g of copper are 92.0%, how many g of aluminum bronze will be 100%:
73.5 g : 92.0% = x : 100%.
x = 73.5 g : 100% * 92.0%
x = 79.9 g
Therefore, from 73.5 g of copper and 6.4 g of aluminum (since 79.9 g - 73.5 g = 6.4 g), maximum 79.9 g of aluminum bronze can be prepared.
Answer : 
Explanation : In the molecule of the hydrogen bonding forces tend to resist any phase transition form liquid state to gaseous phase, and these hydrogen bonding forces are greater usually when the hydrogen is bound to an electronegative atom like chlorine, fluorine,oxygen,etc. This then lowers the viscosity of the compound.
Correct answer: Option D, <span>
K = 5.04 × 10^52</span>
Reason:
We know that,
Ecell =
,
where n = number of electrons = 2 (in present case)
K = equilibrium constant.
Also, Ecell = <span>+1.56 v
Therefore, 1.56 = </span>
Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52
Explanation:
Expresa los gramos de soluto por cada 100 gramos de disolución. Porcentaje masa = masa de soluto___ x 100 masa de la disolución Cuando trabajamos con la masa, podemos sumar el soluto y el disolvente para obtener la disolución.
