"Observable" is the one characteristic of event that must exist when <span>using science to investigate physical phenomena. </span>
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:
ooh sorry, but will this help you now:
Ocean dynamics define and describe the motion of water within the oceans. Ocean temperature and motion fields can be separated into three distinct layers: mixed (surface) layer, upper ocean (above the thermocline), and deep ocean. Ocean currents are measured in sverdrup (sv), where 1 sv is equivalent to a volume flow rate of 1,000,000 m (35,000,000 cu ft) per second.
Surface currents, which make up only 8% of all water in the ocean, are generally restricted to the upper 4…
Explanation:
Hope this helps :)
Answer:
E = 1.602v
Explanation:
Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …
Zn⁰(s) => Zn⁺²(aq) + 2 eˉ
2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)
_____________________________
Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)
Given E⁰ = 1.562v
Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044
E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v
Answer:
Project 3.
Explanation:
Project 3's anticipated cost is 12 to 17 million dollars. It is a <em>lower </em>anticipated cost than Project 2 and Project 4, but <em>higher</em> than Project 1 by one million dollars at maximum cost anticipation. Additionally, the percentage of wildlife to benefit is 70-80%, which is <em>second</em> to the most wildlife to benefit which is Project 4 at 75-80%.
And finally, for community support for Project 3 - the chart lists it as high. This outclasses Project 2 and Project 4, but balances with Project 1. However, Project 1 costs 13 to 16 million dollars and <em>only</em> benefits 15-25% of wildlife.
