6 16.34) Show that 5000 cal are required to increase the temperature of 50 g of water from 0°C to 100°C. The specific heat capacity for
water is 1 cal/g °C

Detail is given below.

Explanation:

Given data:

Mass of water = 50 g

Initial temperature = 0°C

Final temperature = 100°C

Specific heat capacity of water = 1 cal/g.°C

Calories absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 100°C - 0°C =  100°C

Now we will put the values in formula.

Q = 50 g×1 cal/g.°C ×100°C

Q = 5000 cal

Hence, it is proved that 5000 calories required to increase the temperature of 50 g of water from 0°C to 100°C.

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

i don't know people like to fight everywhere rip.