Answer:
marine biology
coastal ecology
astronomy
meteorology
Explanation:
As a college student, to study oceanography one will have to take classes in the field of marine biology, coastal ecology, astronomy and meteorology.
An oceanographer is someone or a professional that studies the ocean in order have more scientific knowledge about them.
Oceanography is a merger of geology, biology, chemistry, physics as it pertains to the ocean.
- There is no need to study human anatomy to study oceanography.
- Marine biology and coastal ecology deals with study of life forms in their marine environment.
- Astronomy and meteorology helps to gain insight about the formation of the ocean and how weather relates to the ocean.
Answer:
Height = 1.9493 cm
Width = 1.9493 cm
Depth = 1.9493 cm
Solution:
Data Given:
Mass = 20 g
Density = 2.7 g/mL
Step 1: Calculate the Volume,
As,
Density = Mass ÷ Volume
Or,
Volume = Mass ÷ Density
Putting values,
Volume = 20 g ÷ 2.7 g/mL
Volume = 7.407 mL or 7.407 cm³
Step 2: Calculate Dimensions of the Cube:
As we know,
Volume = length × width × depth
So, we will take the cube root of 7.407 cm³ which is 1.9493 cm.
Hence,
Volume = 1.9493 cm × 1.9493 cm × 1.9493 cm
Volume = 7.407 cm³
Answer:
The Bohr model suggested that electrons orbited the nucleus in circular paths where as the modern model views the atom to consist of positively charged nucleus surrounded by electrons.
Explanation:
In the modern model, the nucleus contains two sub-atomic particles, the protons which are positively charged and the neutrons which are not charged.According to Bohr's model,the electron in a hydrogen atom travel around the nucleus in a circular orbit. In the modern model, electrons do not move around nucleus around circular obits.
4.22 grams.
1. First find out how much AgNO3 weighs with one mole (107.87 g Ag + 14.007 g N + 48 g O = 169.89 grams)
2. Find the percent of Ag you have. So, (107.87 g/mol Ag)/(169.89 g/mol AgNO3)= 0.63 * 100 = 63%.
3. If you have 6.7 grams total, you know 63% of it is going to be silver, so just multiply 6.7 grams by .63 and you get 4.22 g Ag
