When the heat lost by water = heat gained by ethanol
∴( M* C * ΔT )w = (M*C*ΔT ) eth
when Mw mass of water = 40 g
C specific heat of water = 4.18
ΔT = (70- Tf)
and M(eth) mass (ethanol) = 40 g
C specific heat of ethanol = 2.44
ΔT = (Tf - 10 )
by substitution:
40* 4.18 * (70 - Tf) = 40 * 2.44 * (Tf-10)
∴Tf = 47.9 °C
<span>C) decrease, increase, increasing
</span>
Hello there!
The correct answer is D. Slice of Cheese.
Hope This Helps You!
Good Luck :)
To solve the problem we will use the following equation:
n(Li) = N(Li) / NA
n - the number of moles
N - the number of particles
NA - Avogadro's number that has following value: 6.022×10²³ mol⁻¹
From the upper relation follows:
N(Li) = n(Li) × NA
N(Li) = 1mole × 6.022×10²³ mol⁻¹ = 6.022×10²³ atoms of <span>lithium</span>
