Answer:True
Explanation: An anion has a larger radius than a neutral atom because it gains valence electrons. There are added electron/electron repulsions in the valence shell that expand the size of the electron cloud, which results in a larger radius for the anion.
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Answer:
When the volume of product increases.
When the weight of the product decreases.
Option (a) and (d) are correct.
Explanation:
The overall density of the product can be decreased:
a. Increase the volume of the product (and keep the material same)
d. Decrease the weight of the product ( and keep the same material ).
Density is calculated as the ratio of mass to the volume.
Density is inversely related to volume and directly related to mass.
So, when the mass of the product is increased then the density will increase keeping the material same. Density will also increase when the volume of the product decreased.
Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = <span>h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /</span>λ
value for c is 3 x 10⁸ m/s
E = (6.626 x 10 ⁻³⁴ J)(3 x 10⁸ m/s) / 434 x 10⁻⁹m
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules
For example, ionic compounds, which are very polar, are often soluble in the polar solvent water. Nonpolar substances are likely to dissolve in nonpolar solvents. For example, nonpolar molecular substances are likely to dissolve in hexane, a common nonpolar solvent.
Answer:
2.53 L is the volume of H₂ needed
Explanation:
The reaction is: C₁₈H₃₀O₂ + 3H₂ → C₁₈H₃₆O₂
By the way we can say, that 1 mol of linolenic acid reacts with 3 moles of oxygen in order to produce, 1 mol of stearic acid.
By stoichiometry, ratio is 1:3
Let's convert the mass of the linolenic acid to moles:
10.5 g . 1 mol / 278.42 g = 0.0377 moles
We apply a rule of three:
1 mol of linolenic acid needs 3 moles of H₂ to react
Then, 0.0377 moles will react with (0.0377 . 3 )/1 = 0.113 moles of hydrogen
We apply the Ideal Gases Law to find out the volume (condition of measure are STP) → P . V = n . R . T → V = ( n . R .T ) / P
V = (0.113 mol . 0.082 L.atm/mol.K . 273.15K) 1 atm = 2.53 L
