Answer:
81°C.
Justification:
We can arrive at this conclusion using the formula:
Q = m.c.ΔT,
where Q denotes the heat lost by water (Q = - 1200 J).
m represents the mass of water (m = 20.0 g).
c indicates the specific heat of water (c = 4.186 J/g.°C).
ΔT signifies the difference between the starting temperature and the final temperature (ΔT = final T - initial T = final T - 95.0°C).
Given Q = m.c.ΔT
It follows that (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).
(- 1200 J) = 83.72 final T - 7953.
∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.
Consequently, the correct answer is: 81°C.
The atomic number corresponds to the number of protons
Protons are denoted as P and Electrons as E P = E
The atomic mass equals the sum of Neutrons and Protons
Atomic number = atomic mass = neutrons
P = E
Atomic mass - atomic number = Neutrons
Example:
Calcium consists of 20 Protons 20P = 20E
Atomic mass - atomic number = neutron count:)
Answer: The number of sulfur dioxide molecules present is 1.27·10²³.
Calculating: m(SO₂) equals 13.5 g.
Using the formula n(SO₂) = m(SO₂) ÷ M(SO₂).
This gives n(SO₂) = 13.5 g ÷ 64 g/mol.
Resulting in n(SO₂) = 0.21 mol.
Subsequently, N(SO₂) = n(SO₂) ·Na.
Therefore, N(SO₂) = 0.21 mol · 6.022·10²³ 1/mol.
Ultimately, N(SO₂) equals 1.27·10²³.
Where n represents amount of substance.
M refers to molar mass.
Na is Avogadro's number.
Answer:
Explanation:
Given data:
Initial temperature T₁ = 25.2°C = 298.2K
Initial pressure P₁ = 0.6atm
Final temperature = 72.4°C = 345.4K
What we need to find:
Final pressure = ?
To determine this, we apply a modified version of the combined gas law with constant volume. This simplifies our calculations to:
Here, P and T signify pressure and temperatures, 1 refers to initial and 2 to final temperatures.
Now we can substitute the known variables:
P₂ = 0.7atm
Answer:
In all listed reactions, ΔH°rxn does not correspond to the ΔH°f of the resulting product.
Explanation:
The standard enthalpy of formation (ΔH°f) signifies the enthalpy change that occurs when 1 mole of a product is created from its basic elements in their standard states.
1/2 O₂(g) + H₂O(g) ⟶ H₂O₂(g)
ΔH°rxn does not equal ΔH°f of the product, since H₂O(g) is a compound rather than an element.
Na⁺(g) + F⁻(g) ⟶ NaF(s)
ΔH°rxn is not the same as ΔH°f of the product because Na and F are not in their standard states (Na(s); F₂(g)).
K(g) + 1/2 Cl₂(g) ⟶ KCl(s)
ΔH°rxn is not equal to ΔH°f of the product due to K being outside its standard state (K(s)).
O₂(g) + 2 N₂(g) ⟶ 2 N₂O(g)
ΔH°rxn does not match ΔH°f of the product as 2 moles of N₂O are produced.
In none of the above cases does ΔHrxn match ΔHf of the product.
