In a food chain we arrange the energy in the form of a pyramid.
The producers are on the base of pyramid and then consumers are towards peak.
in the given food chain grass is being eaten by grasshopper which are food of birds.
Grasshoppers are also eaten up by Hawks. so both brids and hawks are feeding upon grasshoppers thus the amount of energy transferred from the grass to the grasshopper is the same as the amount of energy transferred from the grasshopper to the bird.
Answer: D. Slow down the chain reaction by absorbing free neutrons
Explanation: just got it right on the quiz A P E X
Answer:In determining the energy of activation, why was it prudent to run the slowest trial done at room temperature in the hot water bath and the fastest trial done at room temperature in the cold water bath?
Explanation:
Answer:
6 x 10⁵ kg Hg
Explanation:
The mass of mercury in the entire lake is found by multiplying the concentration of the mercury by the volume of the lake.
The volume of the lake is calculated in cubic feet:
V = (SA)x(depth) = (100mi²)(5280ft/mi)² x (20ft) = 5.57568 x 10¹⁰ ft³
Cubic feet are then converted to mL (1cm³=1mL)
(5.57568 x 10¹⁰ ft³) x (12in/ft)³ x (2.54cm/in)³ = 1.578856752 x 10¹⁵ mL
The mass of mercury is then found:
m = CV = (0.4μg/mL)(1g/10⁶μg)(1kg/1000g) x (1.578856752 x 10¹⁵ mL) = 6 x 10⁵ kg Hg
Answer:
0.88 g
Explanation:
Using ideal gas equation to calculate the moles of chlorine gas produced as:-
where,
P = pressure of the gas = 805 Torr
V = Volume of the gas = 235 mL = 0.235 L
T = Temperature of the gas = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
R = Gas constant = 
n = number of moles of chlorine gas = ?
Putting values in above equation, we get:
According to the reaction:-
1 mole of chlorine gas is produced when 1 mole of manganese dioxide undergoes reaction.
So,
0.01017 mole of chlorine gas is produced when 0.01017 mole of manganese dioxide undergoes reaction.
Moles of 
= 0.01017 moles
Molar mass of = 86.93685 g/mol
So,
Applying values, we get that:-
<u>0.88 g of 
should be added to excess HCl (aq) to obtain 235 mL of 
at 25 degrees C and 805 Torr.</u>
