Question

Cate grame de impuritati contine un aliaj format din 127 g argint de puritate 85% si 492,5 g aur de puritate 80%?

Answer 1

Răspuns:

m,impur=127g

m,impuritate ?

m,impuritate-=m,Ag impur-m,Ag pur

p=m,purx100/m,impur---> m,pur= 127x85/100g=107,95g

m,impuritate= 127-107,95=>19,05g

m,Au pur= 492,5x80/100g=394g

m,impuritate= 492,5-394=>98,5g

m,impuritate, total= 19,05+98,5=>117,55g

Explicație: