GEORGEOCTOBER 18, 2012Floor Wax is Harmful to the Environment
Floor wax is applied on floor surfaces to make it scuff-resistant, water-resistant, slip resistant and glossy. It provides a thin, protective and hard surface layer when applied to flooring. Conventional floor wax has five main ingredients and each one of them has detrimental effects on the environment, not to mention the chemical waste created by the continuous upkeep required. The cumulative effects of these ingredients on the environmental render their harmful actions more potent and difficult to reverse.
Answer:
666,480 Joules or 669.48 kJ
Explanation:
We are given;
- Volume of water as 2.0L or 2000 ml
but, density of water is 1 g/ml
- Therefore, mass of water is 2000 g
- Initial temperature as 20 °C
- Final temperature as 99.7° C
Required to determine the heat change
We know that ;
Heat change = Mass × Temperature change × specific heat
In this case;
Specific heat of water is 4.2 J/g°C
Temperature change is 79.7 °C
Therefore;
Heat change = 2000 g × 79.7 °C × 4.2 J/g°C
= 669,480 Joules 0r 669.48 kJ
Thus, the heat change involved is 666,480 Joules or 669.48 kJ
Answer:
a. NH3+ HCl → NH4Cl
c. 2O3 → 3O2
Explanation:
A redox reaction is a reaction in which there is loss or gain of electrons. As a result of that , there is a change in the oxidation number of the species involved in the reaction.
If we look at the species shown in the answer, there isn't any change in oxidation number as we move from left to right hence they are not redox reactions.
Redox reactions lead to change in oxidation number of species from left to right.
Answer:
The answer to your question is [H₃O⁺] = 0.025 [OH⁻] = 3.98 x 10⁻¹³
Explanation:
Data
[H⁺] = ?
[OH⁻] = ?
pH = 1.6
Process
Use the pH formula to calculate the [H₃O⁺], then calculate the pOH and with this value, calculate the [OH⁻].
pH formula
pH = -log[H₃O⁺]
-Substitution
1.6 = -log[H₃O⁺]
-Simplification
[H₃O⁺] = antilog (-1,6)
-Result
[H₃O⁺] = 0.025
-Calculate the pOH
pOH = 14 - pH
-Substitution
pOH = 14 - 1.6
-Result
pOH = 12.4
-Calculate the [OH⁻]
12.4 = -log[OH⁻]
-Simplification
[OH⁻] = antilog(-12.4)
-Result
[OH⁻] = 3.98 x 10⁻¹³
