A 60 mAs results in an exposure of 85 mGya, with all factors remaining the same, what would the new exposure be if 120 mAs is used?

d₂ = 170 mGya

Explanation:

the relationship between absonbed 'd' and exposure 'E' is given as;

D(Gv) = F . x (AS/xB)

F is a conversion coefficient depending on medium

so we can simply write

d₁/d₂ = x₁/x₂

Given that;

our x₁ = 60 mAs, x₂ = 120 mAs,  d₁ = 85 mGya, d₂ = ?

from the given formula,

d₂ = (x₂d₁ / x₁)

now we substitute

d₂ = (120 × 85) / 60

d₂ = 170 mGya

∴ if 120 mAa is used,  the new exposure will be 170 mGya

  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

seeing the graph would