Answer:
1. AT
2. NT
3. AT
4. ST
5. NT
6. AT
7.ST.
Explanation:
1. AT
This is so because a mole of a pure substance contains that number of atoms at any point in time. The only criterion needed there is if the compound is pure or not.
2. NT.
3.AT
IN CCl4, we have one atom of carbon and 4 atoms of chlorine.
4. ST
This is only true in some instances and not at all times. The relation is only true at s.t.p. It does not hold at other conditions which are not s.t.p
5. NT
Methane contains 75% carbon. It is only right that 100gm methane would contain more than what the hydrogen would have as it has a higher percentage by mass and hence the 25gm is wrong.
6. AT
YES. The molecular formula of methane is also it's empirical formula
7. ST
This is not always true as we have more than one way in which the outside multiple n can multiply the inner numbers to yield the molecular formula. In the case quoted above, n = 3. Analysis can also show that n = 6 in which case the empirical formula would be CH2O.
Coastal ecosystems have more sunlight, more nutrients, and higher levels of productivity than open ocean ecosystems. These areas are shallow (lots of light) and close to river mouths (bringing high amounts of nutrients)
An object will be floating in water if the density is lower and will be sink if the density is higher. Water density would be 1000g/liter or 1g/ml. To determine whether the object float or sink you need to calculate both object density.
The density of object A would be: 2g/ 0.5liter= 4g/liter
The density of object B would be: 3g/ 5liter= 0.6g/liter
It is clear that the object is much lighter than water so both objects would float.
Answer:
6.61 × 10∧-29 m³
Explanation:
Given data:
Atomic radius= 143 pm = 143 × 10∧-12 m
volume = ?
Formula:
r = a/2√2
143 × 10∧-12 m = a/ 2√2
a= 143 × 10∧-12 m × 2√2
a= 404.4 × 10∧-12 m
where a is edge length, so we can calculate the volume by using following formula:
volume= a³
V= (404.4 × 10∧-12 m)³
v= 6.61 × 10∧-29 m³
