A1.0 l solution of b e f 2 was electrolyzed for 17.5 h to give 16.178 g of beryllium. assuming the minimum voltage needed was available, what amperage would be needed to complete the electrolysis in the given time?

I = 5.499 A

Explanation:

∴ m Be = 16.178 g

∴ t = 17.5 h

⇒ Q = (m×F×z)/M total electrical charge

∴ F ≅ 96500 C/molFaraday's constant

∴ M Be = 9.012 g/mol

∴ z Be = 2

⇒ Q = (16.178 g Be)(96500 C/mol)(2)/(9.012 g/mol)

⇒ Q = 346466.267 C

⇒ I = Q/tamperage

∴ t = (17.5 h)(60 min/h)(60 s/min) = 63000 s

⇒ I = 346466.267 C / 63000 s

⇒ I = 5.499 A

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

2b2t

explanation:

2b2t