The new volume when pressure increases to 2,030 kPa is 0.8L
BOYLE'S LAW:
The new volume of a gas can be calculated using Boyle's law equation:
P1V1 = P2V2
Where;
- P1 = initial pressure (kPa)
- P2 = final pressure (kPa)
- V1 = initial volume (L)
- V2 = final volume (L)
According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:
406 × 4 = 2030 × V2
1624 = 2030V2
V2 = 1624 ÷ 2030
V2 = 0.8L
Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.
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Answer:
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Explanation:
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Answer: 36.53g
Explanation:
First we need to find the amount of NaCl that dissolves in 1L of the solution that produced 5M of NaCl
Molarity = 5M
MM of NaCl = 58.45
Molarity = Mass conc (g/L) / MM
Mass conc. (g/L) of NaCl = Molarity x MM
= 5 x 58.45 = 292.25g
Next, we need to find the amount that will dissolve in 125mL(i.e 0.125L)
From the calculations above,
292.25g of NaCl dissolved in 1L
Therefore Xg of NaCl will dissolve in 0.125L of the solution i.e
Xg of NaCl = 292.25 x 0.125 = 36.53g.
Therefore 36.53g of NaCl will dissolve in 125mL of the solution
Answer:
ΔHorxn = - 11.79 KJ
Explanation:
2 SO 2 ( g ) + O 2 ( g ) ⟶ 2 SO 3 ( g )
The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol
From the reaction above, 2 mol of SO2 reacts to produce 2 mol of SO3. Assuming ideal gas behaviour,
1 mol = 22.4l
x mol = 2.67l
Upon cross multiplication and solving for x;
x = 2.67 / 22.4 = 0.1192 mol
0.1192 mol of SO2 would react to produce 0.1192 mol of SO3.
Amount of heat is given as;
ΔHorxn = ∑mΔHof(products) − ∑nΔHof(reactants)
Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol.
ΔHorxn = 0.1192 mol * (− 395.7 kJ / mol) - 0.1192 mol * ( − 296.8 kJ / mol)
ΔHorxn = - 47.17kj + 35.38kj
ΔHorxn = - 11.79 KJ
