Answer:
1. V2.
2. 299K.
3. 451K
4. 0.25 x 451 = V2 x 299
Explanation:
1. The data obtained from the question include:
Initial volume (V1) = 0.25mL
Initial temperature (T1) = 26°C
Final temperature (T2) = 178°C
Final volume (V2) =.?
2. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K
3. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Final temperature (T2) = 178°C
Final temperature (T1) = 178°C + 273 = 451K
4. Initial volume (V1) = 0.25mL
Initial temperature (T1) = 299K
Final temperature (T2) = 451K
Final volume (V2) =.?
V1 x T2 = V2 x T1
0.25 x 451 = V2 x 299
D. Drop in barometric pressure, warm ocean water, humid air. The low pressure brings in a cool air mass causing collision of two different masses.
My father rode out a typhoon near Okinawa WWII, onboard the battleship USS Missouri BB-63.
Violent pitching, alarms going off for approaching capsize pitch. The captain came on loudspeaker “ don’t worry men, land is near... about a mile straight down”.
Answer:
CuSO4
Explanation:
Na2S + CuSO4 → Na2SO4 + CuS
The reaction is balanced (same number of elements in each side)
To determine limiting reagent you need to know the moles you have of each.
Molar mass Na2S = 23 * 2 + 32 = 78
Molar mass CuSO4 = 63.5 + 32 + 16 * 4 = 159.5
Na2S mole = 15.5 / 78 = 0.2
CuSO4 mole = 12.1/159.5 = 0.076
*Remember mole = mass / MM
With that information now you have to divide each moles by its respective stoichiometric coefficient
Na2S stoichiometric coefficient : 1
Na2S : 0.2 / 1 = 0.2
CuSO4 stoichiometric coefficient: 1
CuSO4: 0.076 / 1 = 0.076
The smaller number between them its the limiting reagent, CuSO4
Answer:
2H⁺ + NO₃⁻ + 1e⁻ → NO₂ + H₂O
Explanation:
NO₃⁻ → NO₂
In left side, Nitrogen acts with +5 by oxidation number
In right side, the oxidation number is +4
This is a reduction reaction, because the oxidation number has decreased. So the N has gained electrons.
NO₃⁻ + 1e⁻ → NO₂
In acidic medium, we have to add water, where there are less oxygens to ballance the amount. We have 2 O in left side, and 3 O in right side, so we have to add 1 H₂O on left side.
NO₃⁻ + 1e⁻ → NO₂ + H₂O
Now that oxygens are ballanced, we have to ballance the hydrogens by adding protons in the opposite side
2H⁺ + NO₃⁻ + 1e⁻ → NO₂ + H₂O