Answer:
1.47 atm
Explanation:
Step 1: Given data
- Initial volume (V₁): 32.4 L
- Initial pressure (P₁): 1 atm (standard pressure)
- Initial temperature (T₁): 273 K (standard temperature)
- Final volume (V₂): 28.4 L
- Final temperature (T₂): 352 K
Step 2: Calculate the final pressure of the gas
We can calculate the final pressure of the gas using the combined gas law.
P₁ × V₁ / T₁ = P₂ × V₂ / T₂
P₂ = P₁ × V₁ × T₂ / T₁ × V₂
P₂ = 1 atm × 32.4 L × 352 K / 273 K × 28.4 L = 1.47 atm
The first thing we need to do here is to recognize the unit of molarity and the units of the given percentage of nitric acid.
Molarity is mol HNO3 / L of solution. This is our aim
The given percentage is 0.68 g HNO3/ g solution
multiplying this with density to convert g solution into mL solution and dividing with the molecular weight of HNO3 (63 g/mol) to convert g HNO3 to mol. Therefore we obtain
0.016 mol/ mL or 16.23 mol/ L (M)
Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture
Na₂SO₄ ---> 2 Na⁺ + SO₄³⁻
1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions
the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
= 0.08 mol
therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol
NaCl ----> Na⁺ + Cl⁻
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
= 0.24 mol
number of Na⁺ ions from NaCl = 0.24 mol
total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive,
therefore total volume = 100 ml + 200 ml = 300 ml
the concentration of Na⁺ ions = number of moles / volume
= 0.4 mol/ 0.3 dm³
concentration of Na⁺ = 1.33 mol/dm³
Answer:
Ag+ = 47 electrons - 1 electron = 46 electrons. Finally: 47 protons , 61 neutrons and 46 electrons.
Explanation:
