Answer: 16 atm
Explanation:
P1V1 = P2V2
P2 = P1V1/V2
=4 atm x 8.00 L/2.00L = 16 atm
Answer:
See the answer below
Explanation:
The chaparral biome is a temperate biome with a characteristic high temperature and dryness during summer and mild rainy winters and springs. The biome can be found in relatively small amounts in the major continents of the world with its rich plant and animal diversity who have successfully adapted to the conditions of the biome.
Due to the high biodiversity of the chaparral biome, <u>one would expect it to be resilient to the loss of a single species.</u> <em>The more the biodiversity of a biome or community, the more resilient such biome or community would be to the loss of species and lower the biodiversity, the more sensitive the community would be to the loss of species. </em>
Answer:
This question is incomplete, here's the complete question:
<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>
Explanation:
Reaction :-
K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4
Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol
Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol
Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L
Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L
Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol
Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.
0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+
Final concentration of potassium cation
= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M
The power used by Alex to drag the log across the yard is determined as 2,656 W.
<h3>Mass of the log</h3>
The mass of the log is calculated as follows;
W = mg
m = W/g
m = (400)/9.8
m = 40.82 kg
<h3>Velocity of the log</h3>
K.E = ¹/₂mv²
v² = 2K.E/m
v² = (2 x 900)/(40.82)
v² = 44.096
v = 6.64 m/s
<h3>Power used by Alex</h3>
P = Fv
P = 400 x 6.64
P = 2,656 W
Learn more about power here: brainly.com/question/13881533
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