Answer:
It kind of is logical so my answer is yes
Ideal gas law is a combination of three gas laws, which are Boyle's law, Charles' law and Avogadro's law. Ideal gas law states that PV = nRT, where:
P = pressure of the gas
V = volume of the gas
n = no of moles of the gas
R = universal gas constant
T = absolute temperature in Kelvin
Answer:
Barium has the same number of valence electrons as calcium
Explanation:
Valence electrons is the number of electrons of an atom on the outer shell.
Those valence electrons can participate in the formation of a chemical bond (if the outer shell is not closed); in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair.
<u>Calcium</u> is an atom, part of group 2, called the alkaline earth metals. The alkaline earth metals have 2 valence electrons.
<u>Sulfur </u>is part of a group 16, called the chalcogens or oxygen family. Those atoms have 6 valence electrons. They can form a bound with atoms of group 2 such as calcium, but do not have the same number of valence electrons.
<u>Potassium</u> is part of group 1, called the alkali metals or lithium family. Those atoms have 1 valence electrons. That means Potassium do not have the same number of valence electrons like calcium.
<u>Neon</u> is part of group 18, the noble gasses. Those are stable atoms, which means they have 8 valence electrons. They do not have the same number of valence electrons like Calcium.
<u>Barium</u> an atom, part of group 2, called the alkaline earth metals. The alkaline earth metals have 2 valence electrons. Calcium is also part of this group.
This means barium has the same number of valence electrons as Calcium.
Answer:
this doesnt make sence ezxplain the subject
Explanation:
The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)
<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
- Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
- Initial temperature (T₁) = –25 °C
- Final temperature (T₂) = 0 °
- Change in temperature (ΔT) = 0 – (–38) = 38 °C
- Specific heat capacity (C) = 2050 J/(kg·°C)
- Heat (Q₁) =?
Q = MCΔT
Q₁ = 0.4 × 2050 × 38
Q₁ = 31160 J
<h3>How to determine the heat required to melt the ice at 0 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
- Heat (Q₂) =?
Q = mL
Q₂ = 0.4 × 334000
Q₂ = 133600 J
<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 0 °C
- Final temperature (T₂) = 100 °C
- Change in temperature (ΔT) = 100 – 0 = 100 °C
- Specific heat capacity (C) = 4180 J/(kg·°C)
- Heat (Q₃) =?
Q = MCΔT
Q₃ = 0.4 × 4180 × 100
Q₃ = 167200 J
<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
- Heat (Q₄) =?
Q = mHv
Q₄ = 0.4 × 2260000
Q₄ = 904000 J
<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 100 °C
- Final temperature (T₂) = 160 °C
- Change in temperature (ΔT) = 160 – 100 = 60 °C
- Specific heat capacity (C) = 1996 J/(kg·°C)
- Heat (Q₅) =?
Q = MCΔT
Q₅ = 0.4 × 1996 × 60
Q₅ = 47904 J
<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
- Heat for –38 °C to 0°C (Q₁) = 31160 J
- Heat for melting (Q₂) = 133600 J
- Heat for 0 °C to 100 °C (Q₃) = 167200 J
- Heat for vaporization (Q₄) = 904000 J
- Heat for 100 °C to 160 °C (Q₅) = 47904 J
- Heat for –38 °C to 160 °C (Qₜ) =?
Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Qₜ = 31160 + 133600 + 167200 + 904000 + 47904
Qₜ = 1.28×10⁶ J
Learn more about heat transfer:
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