Answer:
- <em>The coefficients in a chemical equation represent the </em><u>relative number of moles of each reactant and product that interven in the chemical reaction.</u>
Explanation:
The <em>coefficients</em> are the numbers that you put in front of each chemical formula that represents the reactants and products in the <em>chemical equation</em>. They indicate the mole ratio in which the elements or compounds react to form the products, as per the chemical equation.
See an example:
- Word equation: hydrogen and oxygen produce water
- Chemical (skeleton) equation: H₂ (g) + O₂(g) → H₂O (g)
This equation is not balanced: the number of atoms of oxygenin the reactant side is 2 while the number of atoms of oxygen isn the product side is 1. In order to balance the equation you need to add some coefficients.
When no coefficients are shown it is understood that the coefficient is 1.
- Balanced chemical equation: 2H₂ (g) + O₂(g) → 2H₂O (g)
The coefficients 2 in front of H₂ and 1 (understood) in front of O₂, in the reactant side, and 2 in front of H₂O, in the product side, balance the equation.
Those coefficients mean that the 2 molecules (or mole of molecules) of H₂ react with 1 molecule (or mole of molecules) of O₂ to form 2 molecules (or moles) of H₂O (product side).
That is the mole ratio: 2 H₂ : 1 O₂ : 2 H₂O.
Notice that, in spite of the aboslute numbers may change, the mole ratio is unique for any chemical reaction. For example 4 : 2 : 4 is the same ratio that 2 : 1 : 2, or 8 : 4 : 8, but the most common practice is to use the most simple form of the ratio, i.e. 2: 1: 2.
Forests and prairies are examples of ecosystems on land. An ecosystem is a community of living things. Members survive by interacting with each other and with their environment. At first glance, the ocean seems like one big ecosystem.
Answer:
The empirical formule is CO
Explanation:
Step 1: Data given
Suppose the mass of a compound is 100 grams
Suppose the compound contains:
42.88 % C = 42.88 grams C
57.12 % O = 57.12 grams O
Molar mass C = 12.01 g/mol
Molar mass O = 16.0 g/mol
Step 2: Calculate moles
Moles = mass / molar mass
Moles C = mass C / molar mass C
Moles C = 42.88 grams / 12.01 g/mol
Moles C = 3.57 moles
Moles O = 57.12 grams / 16.0 g/mol
Moles O = 3.57 moles
Step 3: Calculate the mol ratio
We divide by the smallest amount of moles
C: 3.57 moles / 3.57 moles = 1
O: 3.57 moles / 3.57 moles = 1
The empirical formule is CO
3.07g H2
27.4/26.98/2x3x1.01x2=3.07
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.
1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.
2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.
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3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.<span>
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