How many millilitres of water at 23 degree centigrade with the density of 1.00 g / ml must be mixed with 180 ml (about 6 oz) of coffee at 95 degree centigrade so that the resulting combination will have a temperature of 60 degree centigrade assume that coffee and water have the same density and the same specific heat

141.7475 milliliters of water must be added.

Explanation:

Heat gained by water will be equal to heat loss by the coffee solution.

Mass of water =

Specific heat capacity of water=

Initial temperature of the water=

Final temperature of water coffee solution= = 60 °C

Mass of coffee solution=

1 oz = 28.3495 grams

Specific heat capacity of coffee=

Initial temperature of the coffee solution=

Final temperature of water coffee solution= =T=60 °C

On substituting all values:

we get,

Density of the water = 1.00 g/mL

Volume of the water =

141.7475 milliliters of water must be added.

it's c. elements are placed in rows based on how may valencies electrons have.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3