Answer:
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
Explanation:
Water potential = Pressure potential + solute potential
We have :
C = 0.15 M, T = 273.15 K
i = 1
The water potential of a solution of 0.15 m sucrose= 
(At standard temperature)
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
C6H6 + 02 forms CO2 + H20
Complete with factors to stabilize
C6H6 + 15/2 O2 forms 6 CO2 + 3 H2O
You take away only the info you need
C6H6 15/2 O2
1 mol 15/2 or 7,5 mol
15/2 or 7,5 mol of O2 are required.
;)
Answer:
Explanation:
Hello,
In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:
So the total mole of chloride ions:
And the total volume by adding the volume of each solution in L:
Finally, the molarity turns out:
Best regards.
Answer:
1.315x10⁻³M = [Ca²⁺]
Explanation:
Based in the reaction:
Ca₁₀(PO₄)₆(OH)₂(s) ⇄ 10Ca²⁺(aq) + 6PO₄³⁻(aq) + 2OH⁻(aq)
Solubility product, ksp, is defined as:
ksp = [Ca²⁺]¹⁰ [PO₄³⁻]⁶ [OH⁻]²
From 1 mole of hydroxyapatite are produced 10 moles of Ca²⁺ and 6 moles of PO₄³⁻. That means moles of PO₄³⁻ are:
6/10 Ca²⁺ = PO₄³⁻
Replacing in ksp formula:
ksp = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [OH⁻]²
As [OH⁻] is 2.50x10⁻⁶M and ksp is 2.34x10⁻⁵⁹:
2.34x10⁻⁵⁹ = [Ca²⁺]¹⁰ [0.6Ca²⁺]⁶ [2.50x10⁻⁶]²
3.744x10⁻⁴⁸ = 0.046656[Ca²⁺]¹⁶
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<em>1.315x10⁻³M = [Ca²⁺]</em>
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I hope it helps!
Answer:
Explanation:
q= mc theta
where,
Q = heat gained
m = mass of the substance = 670g
c = heat capacity of water= 4.1 J/g°C
theta =Change in temperature=(
66-25.7)
Now put all the given values in the above formula, we get the amount of heat needed.
q= mctheta
q=670*4.1*(66-25.7)
=670*4.1*40.3
=110704.1
