So we were working on some pulley problems but this one has kinda left me scratching my head, please help! My question is for part C :' )

Explanation:

(c) I assume we're looking for mA.

Sum of forces on B in the -y direction:

∑F = ma

mBg − T = mBa

Sum of forces on A in the +x direction:

∑F = ma

T = mAa

Substitute:

mBg − mAa = mBa

mBg − mBa = mAa

mA = mB (g − a) / a

Plug in values:

mA = (5 kg) (10 m/s² − 0.01 (10 m/s²)) / (0.01 (10 m/s²))

mA = 495 kg

The answer key seems to have a mistake.  It's possible they meant mB = 1 kg, or they changed mB to 5 kg but forgot to change the answer.

  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

increases it