Answer: he gave him a map of somewhere
Explanation:
Answer:
the first one is balanced.
Answer:
The total heat required is 691,026.36 J
Explanation:
Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L
Where Q: amount of heat, m: mass and L: latent heat
On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).
In this case, the total heat required is calculated as:
- Q for liquid water. This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° C
Q= c*m*ΔT

Q=103,763.2 J
- Q for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then
)
Q= m*L

Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)
- Q for temperature change from 100.0
∘
C to 154
∘
C, this is, the sensible heat of steam from 100 °C to 154°C.
Q= c*m*ΔT

Q=25,176.96 J
So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J
<u><em>The total heat required is 691,026.36 J</em></u>
Answer:
- A) pH = 2.42
- B) pH = 12.00
Explanation:
<em>The dissolution of HCl is HCl → H⁺ + Cl⁻</em>
- To solve part A) we need to calculate the concentration of H⁺, to do that we need the moles of H⁺ and the volume.
The problem gives us V=2.5 L, and the moles can be calculated using the molecular weight of HCl, 36.46 g/mol:
= 9.60*10⁻³ mol H⁺
So the concentration of H⁺ is
[H⁺] = 9.60*10⁻³ mol / 2.5 L = 3.84 * 10⁻³ M
pH = -log [H⁺] = -log (3.84 * 10⁻³) = 2.42
- <em>The dissolution of NaOH is NaOH → Na⁺ + OH⁻</em>
- Now we calculate [OH⁻], we already know that V = 2.0 L, and a similar process is used to calculate the moles of OH⁻, keeping in mind the molecular weight of NaOH, 40 g/mol:
= 0.02 mol OH⁻
[OH⁻] = 0.02 mol / 2.0 L = 0.01
pOH = -log [OH⁻] = -log (0.01) = 2.00
With the pOH, we can calculate the pH:
pH + pOH = 14.00
pH + 2.00 = 14.00
pH = 12.00