Answer:
Sample response:
The costs of using both renewable and nonrenewable resources depend on the extent of the use. If renewable resources are managed wisely, the use of the resource will not exceed the rate at which it is replenished. In this instance the cost of using renewable resources can be minimized, if not entirely eliminated. The cost of using nonrenewable resources is harder to minimize because nonrenewable resources cannot be replenished at the rate at which they are used. The environmental impact of using nonrenewable resources such as fossil fuels is greater than just the loss of the resource itself. Other impacts such as acid rain, global warming, and atmospheric pollution can result from the use of nonrenewable resources.
Explanation:
2021 edge
have a nice day
Answer:
MnO4- + 5 VO +2 + 11 H2O = 5 V(OH)4+ + Mn+2 + 2 H+
Explanation:
Answer:
Mass = 14.3 g
Explanation:
Given data:
Mass of Mg(OH)₂ = 16.0 g
Mass of HCl = 11.0 g
Mass of MgCl₂ = ?
Solution:
Chemical equation:
Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O
Number of moles of Mg(OH)₂ :
Number of moles = mass/ molar mass
Number of moles = 16.0 g/ 58.3 g/mol
Number of moles = 0.274 mol
Number of moles of HCl :
Number of moles = mass/ molar mass
Number of moles = 11.0 g/ 36.5 g/mol
Number of moles = 0.301 mol
Now we will compare the moles of Mg(OH)₂ and HCl with MgCl₂.
Mg(OH)₂ : MgCl₂
1 : 1
0.274 : 0.274
HCl : MgCl₂
2 : 1
0.301 : 1/2×0.301 = 0.150
The number of moles of MgCl₂ produced by HCl are less so it will limiting reactant.
Mass of MgCl₂:
Mass = number of moles × molar mass
Mass = 0.150 × 95 g/mol
Mass = 14.3 g
Answer: C. ethanol
The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.
<u>The enthalpy of combustion of the unknown compound is</u>
ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol
<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,
e% = ( | ΔHx - ΔH | / ΔHx ) x 100%
where ΔHx is the enthalpy of combustion of the probable compound.
The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol
Compound Enthalpy of combustion (kJ/mol) Deviation
Methane - 890.7 43.8%
Ehylene -1411.2 9.3%
Ethanol -1368.6 6.5%
According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.
