An iron ball has a diameter of 12cm and is 0.02 cm too large to pass through a hole in a brass plate when the ball and plate are at the same temperature of 25°C. At what temperature the same for both ball and plate would the ball just pass through the hole. (linear expansivties for iron and brass are 1.32×10^-5 K ^-1 and 2.0×10^-5 K^-1​

270°C

Explanation:

Since αB > αI

We want LB - LI = 0.02 cm at a given temperature t

But

ΔLB= LBαBΔt

ΔLI = LIαIΔt

Where;

LI = length of the iron

LB= length of he brass

αB = Linear expansivity of brass

αI = Linear expansivity of iron

Δt = Change in temperature

ΔLB - ΔLI =   (LBαB - LIαI)Δt = 0.02

LB = LI = 12 cm

Δt = 0.02/12( 2.0×10^-5 -  1.32×10^-5)

Δt = 0.02/12(0.68 ×10^-5)

Δt = 245°C

25°C + 245°C = 270°C

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

answer: a. same unit

explanation: