Answer:
P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)
Explanation:
Given:
P₁ = 90 atm P₂ = ?
V₁ = 18 Liters(L) L₂ = 12 Liters(L)
=> decrease volume => increase pressure
=> volume ratio that will increase 90 atm is (18L/12L)
T₁ = 272 Kelvin(K) T₂ = 274 Kelvin(K)
=> increase temperature => increase pressure
=> temperature ratio that will increase 90 atm is (274K/272K)
n₁ = moles = constant n₂ = n₁ = constant
P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)
By rule of sig. figs., the final answer should be rounded to an accuracy equal to the 'measured' data value having the least number of sig. figs. This means P₂ ≅ 100 atm based on the given value of P₁ = 90 atm.
Answer:
B. 8
Explanation:
because I learned that in 6th
<span>Mass=25.7gm
dimensions,
l=2.3cm
b=4.01cm
h=1.82cm
Thus,Density=Mass/Volume=m/lbh=25.7/2.3*4.01*1.82 g cm-3</span>
Answer:
the molarity is 3.68 moles/L
Explanation:
the molality of the solution of sucrose is
m= moles of glucose / Kg of solvent (water)= 6.81 ,
since the molecular weight of glucose is 180.156 gr/mole , then per each kilogram of solvent there is
6.81 moles*180.156 gr/mole + 1000 gr of water = 2226.86 gr of solution
from the density
volume of solution = mass of solution/density = 2286.86 gr / 1.2 gr/ml = 1855.71 ml
therefore there is 1000 gr of water in 1855.71 ml
then the molarity M is
M= moles of glucose / L of solution = (moles of glucose / Kg of solvent) * (Kg of solvent/L of solution) = 6.81 moles/Kg * 1Kg/1.85 L = 3.68 moles/L
M= 3.68 moles/L
Note:
- Would be wrong in this case to assume density of water = 1 Kg/L since the solution is heavily concentrated in glucose and therefore the density of water deviates from its pure value.
Answer:
A certain portion of the land will be set aside as wildlife habitat
