False, as oceans can act as carbon sinks along with forests.
Answer:
30.17 × 10²³ atoms
Explanation:
Given data:
Number of moles of lead = 5.01 mol
Number of atoms = ?
Solution:
Avogadro number:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
In given question:
1 mole = 6.022 × 10²³ atoms
5.01 mol × 6.022 × 10²³ atoms / 1 mol
30.17 × 10²³ atoms
As the question tells you, you need to use the formula
% mass= mass of solute/ mass of solution x 100
mass solute= 30.0 g
mass of solution= 30.0 + 270.0= 300.0 g
% mass= 30.0/ 300.0 x 100= 10%
answer is B
The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)
Answer:
He realized he needs to have the upper body and lower body held in place and needed the buckle as far down beside the person's hip so it could hold the body properly
Explanation: ''I realized both the upper and lower body must be held securely in place with one strap across the chest and one across the hips,'' Mr. Bohlin once said. ''The belt also needed an immovable anchorage point for the buckle as far down beside the occupant's hip, so it could hold the body properly during a collision.
