A mass of <span>58.44 g hope this helps and plz rate below and thank me on my profile plz</span>
Barium chloride + sodium sulphate --> barium sulphate + sodium chloride
BaCl2 + Na2SO4 ---> BaSO4 + 2 NaCl
The barium sulphate appears as a white precipitate
Silver nitrate + Sodium chloride ---> Silver Chloride + sodium nitrate
AgNO3 + NaCl ----> AgCl + NaNO3
The silver chloride appears as a white precipitate.
These are sometimes called double decomposition reactions.
Answer:
The flow rate would be 22.5ml/hr
Explanation:
Volumetric flow rate = Mass flow rate ÷ density
Mass flow rate = 3mg/min = 3mg/min × 60min/1hr = 180mg/hr
Density = mass/volume = 2g/250ml = 0.008g/ml = 0.008g/ml × 1000mg/1g = 8mg/ml
Volumetric flow rate = 180mg/hr ÷ 8mg/ml = 22.5ml/hr
Answer:
HClO₃ /chloric acid /suffix -ic/ ClO₃⁻ (chlorate)
HClO₂/ chlorous acid/ suffix -ous/ ClO₂⁻ (chlorite)
HNO₃ /nitric acid /suffix -ic/ NO₃⁻ (nitrate)
HNO₂/ nitrous acid/ suffix -ous/ NO₂⁻ (nitrite)
Explanation:
Chlorine has 4 positive oxidation numbers to form oxyacids: +1, +3, +5 and +7.
- When it uses the oxidation number +5, it forms HClO₃, which is named chloric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion ClO₃⁻ (chlorate).
- When it uses the oxidation number +3, it forms HClO₂, which is named chlorous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion ClO₂⁻ (chlorite).
Nitrogen has 2 positive oxidation numbers to form oxyacids: +3 and +5.
- When it uses the oxidation number +5, it forms HNO₃, which is named nitric acid, with the suffix -ic. When it loses an H⁺, it forms the oxyanion NO₃⁻ (nitrate).
- When it uses the oxidation number +3, it forms HNO₂, which is named nitrous acid, with the suffix -ous. When it loses an H⁺, it forms the oxyanion NO₂⁻ (nitrite).
This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:
pH = pKa + log[(salt/acid]
Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid).
Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L
Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M
Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48
