Answer:
Separate the reaction into half equations
Explanation:
When we want to balance redox equations, we must take cognizance of the fact that a specie was oxidized and another specie was reduced.
Hence we must identify the specie that was oxidized and the one that was reduced and then break up the whole redox reaction into oxidation and reduction half equations.
Liquids have more kinetic energy in their particles compared to solids. this allows the particles to move more freely, hence why they are fluids
Liquids diffuse from a region of high concentration to a region of low concentration, until equilibrium is reached
When heat is applied the particles gain more kinetic energy so they now have enough energy to overcome the bonds holding them in the liquid. this means they can evaporate off
Answer:
E) are electrically attracted to each other
Explanation:
Water molecule is polar because there is a difference in electronegativity values between hydrogen and oxygen. The hydrogen side of the molecule has a slight positive charge and the oxygen side is slightly negatively.
Positively and negatively charged ends cause water molecules to attract one another and for this reason water shows the properties mentioned in the question: cohesion, high specific heat, and high heat of vaporization.
Answer:

Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
M_r: 32 60
CH₃OH + CO ⟶ CH₃COOH
m/g: 160
(a) Moles of CH₃OH

(b) Moles of CH₃COOH

(c) Mass of CH₃COOH

Answer:
The number ratio is 4:7
Explanation:
Step 1: Data given
Compound 1 has 50.48 % oxygen
Compound 2 has 36.81 % oxygen
Molar mass oxygen = 16 g/mol
Molar mass manganese = 54.94 g/mol
Step 2: Calculate % manganes
Compound 1: 100 - 50.48 = 49.52 %
Compound 2: 100 - 36.81 = 63.19 %
Step 3: Calculate mass
Suppose mass of compounds = 100 grams
Compound 1:
50.48 % O = 50.48 grams
49.52 % Mn = 49.52 grams
Compound 2:
36.81 % O = 36.81 grams
63.19 % Mn = 63.19 grams
Step 4: Calculate moles
Compound 1
Moles O = 50.48 grams / 16.0 g/mol = 3.155 moles
Moles Mn = 49.52 grams / 54.94 g/mol = 0.9013 moles
Compound 2
Moles O = 36.81 grams / 16.0 g/mol = 2.301 moles
Moles Mn = 63.19 grams / 54.94 g/mol = 1.150 moles
Step 5: calculate mol ratio
We will divide by the smallest amount of moles
Compound 1
O: 3.155/0.9013 = 3.5
Mn: 0.9013 / 0.9013 = 1
Mn2O7
Compound 2
O: 2.301 / 1.150 = 2
Mn: 1.150 / 1.150 = 1
MnO2
The number ratio is 2:3.5 or 4:7