Answer:
The correct answers are:
a) 180 g
b) 93.7 cm³
Explanation:
The density of a substance is the mass of the substance per unit of volume. So, it is calculated as follows:
density= mass/volume
From the data provided in the problem:
density = 0.8 g/cm³
a) Given: volume= 225 cm³
mass= density x volume = 0.8 g/cm³ x 225 cm³ = 180 g
b) Given: mass= 75.0 g
volume = mass/density = 75.0 g/(0.8 g/cm³)= 93.75 cm³≅ 93.7 cm³
Answer:
Cl₂O₇
Explanation:
For the reaction:
ClₓOₙ + H₂ → HCl + H₂O
Moles of HCl and moles of H₂O are:
HCl: 0.233g HCl ₓ (1mol / 36.46g) = 6.39x10⁻³ mol HCl
H₂O: 0.403g H₂O ₓ (1mol / 18.02g) = 2.236x10⁻² mol H₂O
As you can see, moles of HCl are equivalent to moles of Cl in the compound and moles of H₂O are equivalent to moles of O in the compound, that means:
6.39x10⁻³ mol Cl
2.236x10⁻² mol O
Empirical formula is the simplest ratio of atoms presents in a molecule. If Cl is <em>1</em>, Oxygen will be:
2.236x10⁻² mol / 6.39x10⁻³ = <em>3.5</em>
As empirical formula must be given in natural numbers, the empirical formula is:
<em>Cl₂O₇</em>
<em></em>
Answer:
HS-PS1-3
Explanation:
Plan and carry out an investigation to gather evidence to compare the structure of substances at the bulk scale to infer the strength of electrical forces between particles. Examples of bulk properties of substances could include the melting point and boiling point, vapor pressure, and surface tension.
Answer:
C
Explanation:
Looking at the periodic table, we can see that sodium is in group 1, so a sodium ion would be Na⁺, with a charge of +1. Oxygen is in group 16, so an oxygen ion would be O²⁻, with a charge of -2.
A compound formed only by a single sodium ion and a single oxygen ion would thus have a charge of -1, and in order to have a stable ionic compound its charge must be zero.
Answer is: <span>Mutations sometimes improve the chances of survival for a plant.
</span>Mutations are very important because they change <span>variability in populations and in that way enable evolutionary change.
</span>There are three types of mutations:
1) good or advantageous mutations - <span> improve the chances of survival for a plant.
2) </span>bad or deleterious - decrease the chances of survival for a plant.
3) neutral - not affect he chances of survival for a plant.