If the specific heat of water is 4,186 J/kg∙°C, how much heat is required to increase the temperature of 1.2 kg of water from 23 °C to 39 °C?

8.0 × 10⁴ J

Explanation:

Step 1: Given data

Step 2: Calculate the change in the temperature

ΔT = 39 °C - 23 °C = 16 °C

Step 3: Calculate the heat required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 4,186 J/kg.°C × 1.2 kg × 16 °C

Q = 8.0 × 10⁴ J

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.