2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O
v(H₂SO₄)=0.25 L
c(H₂SO₄)=2.00 mol/L
v(NaOH)=2.00 L
n(H₂SO₄)=c(H₂SO₄)v(H₂SO₄)
n(NaOH)=2n(H₂SO₄)=2c(H₂SO₄)v(H₂SO₄)
c(NaOH)=n(NaOH)/v(NaOH)=2c(H₂SO₄)v(H₂SO₄)/v(NaOH)
c(NaOH)=2*2.00*0.25/2.00=0.5 mol/L
the concentration of the NaOH solution is 0.5 mol/L
Answer:
The forward reaction will keep occurring until all NO or all NO₂ is consumed.
Clarification:
- According to Le Châtelier's principle, when a system at equilibrium experiences a disturbance from an outside source, the system will adjust to counteract this disturbance and restore equilibrium.
- Thus, removing the product (N₂O₃) from the system effectively lowers the product concentration, prompting the reaction to shift forward and generate additional product in order to alleviate the strain caused by the removal of N₂O₃.
- Consequently, the reaction will proceed forward until all of either NO or NO₂ is depleted.
Solution:
The gas's new temperature is 604K
Justification:
Assuming standard temperature and pressure, we can determine the gas's temperature using the ideal gas law;
Step 1: Formulate the general gas law equation
P1V1/T1 = P2V2/T2
Step 2: Insert the values, converting as needed to standard units.
P1 = 0.800 atm
V1 = 0.180 L
T1 = 29°C = 273 + 29 = 302K
P2 = 3.20 atm
V2 = 90 mL = 90 * 10^-3 L = 0.09 L
Step 3: Solve for T2
The new gas temperature T2 is calculated as:
T2 = P2V2T1/(P1V1)
T2 = 3.20 * 0.09 * 302 / (0.800 * 0.180)
T2 = 86.976 / 0.144
T2 = 604K
The gas's new temperature is 604K.
Response:
Sulfate- SO4^2-
Sulfite- SO3^2-
Permanganate- MnO4
Carbonate- CO3^2
Clarification:
KEEP GOING WITH YOUR STUDIES!
684 kcal. One mole of glucose weighs roughly 180g. Given that 1g of glucose releases 3.8 kcal, we calculate for 1 mole of glucose: 180g -> 180g * 3.8 kcal/g = 684 kcal.
