The materials which allow heat to pass through them easily are conductors of heat. For examples,aluminum, iron, and copper.
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The enthalpies of formation of each of the compound involved in the chemical reaction presented above are given below:
CO2: -393.5 kJ/mol
CO: -99 kJ/mol
O2: 0 kJ/mol
As observed O2 will not have enthalpy of formation as it is a pure substance.
To calculate for the enthalpy of reaction,
enthalpy of formation of products - enthalpy of formation of reactants
= (-99 kJ/mol) - (-393.5 kJ/mol)
= 294.5 kJ/mol
ANSWER: 294.5 kJ/mol
Answer:
2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.
Explanation:
<em>∵ pH = - log[H₃O⁺]</em>
∴ 4.6 = - log[H₃O⁺].
∴ log[H₃O⁺] = - 4.6.
∴ [H₃O⁺] = 2.51 x 10⁻⁵.
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
[H₃O⁺] = 2.51 x 10⁻⁵ M.
∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.51 x 10⁻⁵ M) = 3.98 × 10⁻¹⁰ M ≅ 4.0 × 10⁻¹⁰ M.
<em>So, the right choice is: 2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.</em>
First convert the kg to g ----- 0.03kg = 30g
Then divide the mass by the volume ----- 30g ÷ 25mL = 1.2
The density is 1.2g/mL<span />
