Energy diagrams are use to depict the energy changes that occur during a chemical reaction. There are two types of reaction based on the energy change, these are exothermic and endothermic reactions. In endothermic reactions energy are gained while in exothermic reactions energy are lost to the environment. To identify an exothermic reaction on a potential energy diagram, one has to compare the potential energy of the reactants and the products. If the potential energy of the product is less than that of the reactants, the reaction is exothermic.
The shells further away from the nucleus are LARGER and can hold MORE electrons
Answer:
3.09kg
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
2C8H18 + 25O2 —> 16CO2 + 18H2O
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 from the balanced equation = 2 x 114 = 228g
Converting 228g of C8H18 to kg, we obtained:
228/1000 = 0.228kg
Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol
Mass of CO2 from the balanced equation = 16 x 44 = 704g
Converting 704g of CO2 to kg, we obtained:
704/1000 = 0.704kg
From the equation,
0.228kg of C8H18 produced 0.704kg of CO2.
Therefore, 1kg of C8H18 will produce = 0.704/0.228 = 3.09kg of CO2
Answer:
7.16x10⁻⁸M = [Ag+]
Explanation:
Using the equation:
E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)
<em>Where E</em>⁰<em>= 0.4249V</em>
<em>E(Cell) = -(-0.0019V) -Measured value-</em>
<em>[Cu2+] = 1M</em>
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Replacing:
0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)
-0.423V = - 0.0296 • log (1M/[Ag+]²)
14.29 = log (1M/[Ag+]²)
1.95x10¹⁴ = 1M / [Ag+]²
[Ag+]² = 5.12x10⁻¹⁵M
7.16x10⁻⁸M = [Ag+]
Answer:
45.3°C
Explanation:
Step 1:
Data obtained from the question.
Initial pressure (P1) = 82KPa
Initial temperature (T1) = 26°C
Final pressure (P2) = 87.3KPa.
Final temperature (T2) =.?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T(°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K.
Step 3:
Determination of the new temperature of the gas. This can be obtained as follow:
P1/T1 = P2/T2
82/299 = 87.3/T2
Cross multiply to express in linear form
82 x T2 = 299 x 87.3
Divide both side by 82
T2 = (299 x 87.3) /82
T2 = 318.3K
Step 4:
Conversion of 318.3K to celsius temperature. This is illustrated below:
T(°C) = T(K) – 273
T(K) = 318.3K
T(°C) = 318.3 – 273
T(°C) = 45.3°C.
Therefore, the new temperature of the gas in th tire is 45.3°C
