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OlgaM077 [116]
3 years ago
14

< please help!!! >

Chemistry
1 answer:
svlad2 [7]3 years ago
8 0

Answer:

n = 0.573mol

Explanation:

PV = nRT => n = PV/RT

P = 1.5atm

V = 8.56L

R = 0.08206Latm/molK

T = 0°C = 273K

n = (1.5atm)(8.56L)/(0.08206Latm/molK)(273K) = 0.573mol

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Arlecino [84]

Answer:

89°C

Explanation:

Combined Gas Law (P₁V₁)/T₁ = (P₂V₂)/T₂

(1.25 atm)(450 mL)/(65°C) = (0.89 atm)(865 mL)/T₂

8.653846154 = 769.85/T₂

T₂ = 769.85/8.653846154

T₂ = 88.96044444 = 89°C

6 0
3 years ago
In terms of their electron configurations, why is cesium more likely to lose its valence electron than potassium?
Harlamova29_29 [7]

Explanation:

use the term electron sheilding, the more electrons between the valence el3ctron and nucleus the easier to lose the valence electron (more sheilding = easier to lose)

5 0
3 years ago
Determine the mass density, specific volume and specific weight of a liquid whose
irakobra [83]

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5 0
2 years ago
How many cm in 725 km?
tigry1 [53]
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5 0
3 years ago
Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for
abruzzese [7]

Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

Ca^{2+ + y^{4- ⇄  CaY^{2-

Formation constant Kf

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) = 5.0 × 10¹⁰

Now,

[y^{4-] = \alpha _4CH_4Y; ∝₄ = 0.35

so the equilibrium is;

Ca^{2+ + H_4Y ⇄  CaY^{2- + 4H⁺

Given that; CH_4Y = Ca^{2+     { 1 mol Ca^{2+  reacts with 1 mol H_4Y  }

so at equilibrium, CH_4Y = Ca^{2+ = x

∴

Ca^{2+ + y^{4- ⇄  CaY^{2-

x        + x         0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) =  CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] )  = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x²  = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

8 0
3 years ago
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