The coil of a galvanometer has a resistance of 18.2 Ω, and its meter deflects full scale when a current of 6.78 mA passes through it. To make the galvanometer into an ammeter, a 26.9-mΩ shunt resistor is added to it. What is the maximum current that this ammeter can read?

Explanation:

Ammeter is used in series in a circuit .It has two resistances in parallel . They are 18.2 ohm  and .0269 ohm . For full scale deflection ,  current of .00678 A passes in 18.2 ohm .

potential over it

= 118.2 x .00678

= .801396 V

Same volt will act on .0269 ohm shunt which is in parallel

current in it

.801396 / .0269 A

= 29.79 A

Total current

29.79 + .00678

= 29.79678

=

  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

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