Answer:
3.62 g/cm³
Explanation:
density = mass ÷ volume
Therefore, do 12.69 divided by 3.5
Answer:
Your question is complex, because I think you wrote it wrong.
Although in front of this what I can help you is that the carbons are associated between a single, double or triple union.
This depends on whether they are attached to more or less carbons or hydrogens, the carbons have the possibility of joining 4 radicals, both other carbons and hydrogens.
Simple junctions talks about compound organisms called ALKANS.
The double unions, in organic these compounds are called as ALQUENOS.
And as for the tertiary unions, the organic chemistry names them as ALQUINOS.
These compounds that we write, a simple union, the less energy, the less this union, that is why the triple bond is the one that contains the most energy when breaking or destroying it in a reaction.
Explanation:
In a chemical compound the change of these unions if we modified them we would generate changes even in the classifications naming them as well as different compounds and not only that until they change their properties
The subscriot 2 means that in the formula there are two parts of K, and the subscript 1 (implicit) for S, indicates that there is one part of S.
This is, the formula gives the ratio of the elements K and S in the compound, which is:
2 atoms of K : 1 atom of S.
Answer: there are 2 atoms of K and 1 atom of S in a molecule of K2S.
Answer:
The atomic mass of element is 65.5 amu.
Explanation:
Given data:
Abundance of X-63 = 50.000%
Atomic mass of X-63 = 63.00 amu
Atomic mass of X-68 = 68.00 amu
Atomic mass of element = ?
Solution:
Abundance of X-68 = 100-50 = 50%
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (50×63)+(50×68) /100
Average atomic mass = 3150 + 3400 / 100
Average atomic mass = 6550 / 100
Average atomic mass = 65.5 amu.
The atomic mass of element is 65.5 amu.
Answer: 6.1 g
Explanation:
between Mg and MgO theres a 1;1 MOLE RATIO
here's the balanced equation
2Mg + O2 ==> 2MgO
24g of magnesium is approximately 1 mole of magnesium so it produces 40 g of mgo which is also 1 mole of mgo thus 10/40 =0.25 moles of MgO so 0.25 moles of magnesium would be needed which is approximately 6.1 g
