A picometer (pm) is equal to 1 x 10^-12 m. The O-H bond lengths in water are 95.8 pm,while S-H bond lengths in dihydrogen sulfide are 135 pm. Why are S-H bonds longer than O-H bonds?
Is it because Sulfurs mass is greater than Oxygens mass?
Please correct me if I'm wrong.
Oxygen = 15.99
Sulfur = 32.06
Sulfur is greater than oxygen because the atomic number of sulfur is 16 and oxygen is 8 so sulfur has a greater atomic radius than oxygen. Sulfur has three electron layers but oxygen only has two electron layers. One electron layer expands the atomic radius a lot and causes the last electron to stick closer to the nucleus. Another important reason for this is tha oxygen atom is more electronegative than sulfur that is it attracts more the boding electrons so the bond length in OH is longer than that in SH
wave lengths in your eye grab light and make it the colour you see
Why do you think that arrows of varying lengths are used to represent forces?
X-rays are shorter waves than ultraviolet ones.
The issue is not frequency and wavelength, a relationship is the problem AM Wave length is longer, than FM Wave length. Shorter wave lengths have a tendency to be shorter in the pm. AM Wave lengths were used before FM wave lengths.
it has to do with distance temp. and wave lengths
The number used to multiply the lengths of a figure to stretch or shrink it to a similar image. If we use a scale factor of 3, all the corresponding lengths in the original side lengths will be multiplied by three.
The scale factor is the number that the side lengths of one figure can be multiplied by to give the corresponding side lengths of the other figure.
The scale factor is the number that the side lengths of one figure can be multiplied by to give the corresponding side lengths of the other figure.
The ratio of areas is the square of the ratio of lengths. Ratio lengths = 1 : 2 → ratio areas = 1² : 2² = 1 : 4 → if the lengths are doubled, the areas of quadrupled (multiplied by 4).
negative reciprocal slopes
The sides of a triangle are its lengths are cannot be negative. However, you could place a triangle on coordinate system and some points where the vertices are could be negative numbers.
If the lengths of the two parallel sides are a and b, and the height of the trapezium (the distance between them) is h, then area_of_trapezium = 1/2 (a + b) h that is half of the sum of the lengths of the two parallel sides multiplied by the distance between them.
Three lengths multiplied yield a volume not an area.
Adding the lengths of the straight sides of any polygon will give its perimeter. In the case of regular polygons it will be the length of the side multiplied by the number of sides. But in the case of a non-regular polygon all the individual sides' lengths will be added to give you the perimeter.
No, Dijkstra's algorithm can not be used when there are negative arc lengths. In Dijkstra's, the vertex that can be reached from the current set of labeled vertices and that of having the minimum weight among the alternatives is permanently labeled in that iteration. Since a negative arc weight would result in changing the label of a pre-permanently-labeled vertex, the algo collapses. Bellman's algorithm is used with negative arc lengths.
You need to find the perimeter of one by adding together the lengths of all its sides. The perimeter of the similar shape is the answer multiplied by the similarity ratio.
rhombus