Answer is: The zooplankton can be used to help control algal growth in some coral communities.
Hypothesis is a suggested explanation for an observable phenomenon.
Steps of the scientific method:
1) ask a question about something that is observed.
2) do background research.
3) construct a hypothesis, an attempt to answer questions with an explanation that can be tested. In this example, does the presence of a certain type of zooplankton inhibited the algae growth
4) test of hypothesis.
5) analyze collected data and draw a conclusion. In this example, the zooplankton can be used to help control algal growth.
2 Li(s) +Cl₂→ 2 Li⁺ (aq) + 2Cl⁻ (aq)
The cell potential of the reaction above is +4.40V
<em><u>calculation</u></em>
Cell potential =∈° red - ∈° oxidation
in reaction above Li is oxidized from oxidation state 0 to +1 therefore the∈° oxid = -3.04
Cl is reduce from oxidation state 0 to -1 therefore the ∈°red = +1.36 V
cell potential is therefore = +1.36 v -- 3.04 = + 4.40 V
Answer:
Percent yield = 50%
Explanation:
Given data:
Mass of CH₄ = 16 g
Mass of O₂ = 32 g
Mass of CO₂ = 11 g
Percent yield of CO₂ = ?
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Number of moles of CH₄:
Number of moles = mass/ molar mass
Number of moles = 16 g /16 g/mol
Number of moles = 1 mol
Number of moles of O₂:
Number of moles = mass/ molar mass
Number of moles = 32 g /32 g/mol
Number of moles = 1 mol
Now we will compare the moles of CO₂ with both reactant.
O₂ : CO₂
2 : 1
1 : 1/2×1= 0.5 mol
CH₄ : CO₂
1 : 1
Number of moles of CO₂ produced by oxygen are less so it will limiting reactant.
Theoretical yield:
Mass of CO₂:
Mass = number of moles × molar mass
Mass = 0.5 mol × 44 g/mol
Mass = 22 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 11 g/ 22 g × 100
Percent yield = 50%
I'll assume that the car is a Honda Insight with <span>EPA gas mileage rating of 57mi/gal in the city.
First, we will convert all units into Km and mL:
1 mile = </span><span>1.609344 Km
</span><span>1gal = 3785.411 mL
</span>
Then, we will calculate the distance the car can move by multiplying the mileage rating and the the volume available as follows (note that I will be converting units in the same step):
distance = <span>(57mi/gal) x (1.609344 km/1mi) x (1gal/3785.411mL) x 355mL
= 8.6027 Km</span>
