Answer:
= konstanta produk kelarutan
= kation dalam larutan berair
= anion dalam larutan berair
= konsentrasi relatif a dan b
DARI WEB
Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g . Solubility of NaNO3=21.9g or NaNO3 x 100 g/ 25 g =87.6. Calculate the molar mass of the dissolved compound as the sum of mass of all atoms in the molecule.
Answer:
Plants and animals are pH sensitive. The growth of plants is dependent on the nature of the soil. If the pH of the soil is greater than 7 and is alkaline, then the plants cannot grow in the soil.
Explanation:
Answer:
72.6 grams
Explanation:
I got this answer through stoichiometry. For every 1 mole of Mg, 2 moles of CuBr are consumed. Because of this, multiply the moles of Mg by 2. Then, convert moles to grams.
The most reasonable way to measure absolute zero would have been to extrapolate the ideal gas law.
<h3>What is Absolute zero?</h3>
This is referred to the temperature at which a thermodynamic system has the lowest form of energy.
Guillaume Amontons used gas equation to prove that absence of heat was theoretically possible which would have involved only extrapolating the ideal gas law.
Read more about Absolute zero here brainly.com/question/18560146
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Answer:- The natural abundance of 
is 0.478 or 47.8% and 
is 0.522 or 52.2% .
Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:
Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)
We have been given with atomic masses for and as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.
Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of as n then the abundance of would be 1-n .
Let's plug in the values in the formula:
151.964=150.919860n+152.921243-152.921243n
on keeping similar terms on same side:
negative sign is on both sides so it is canceled:
The abundance of is 0.478 which is 47.8%.
The abundance of is = 
= 0.522 which is 52.2%
Hence, the natural abundance of is 0.478 or 47.8% and is 0.522 or 52.2% .
