Answer:
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The temperature of the container containing the 4.47 g of Ar is 358.16 K
<h3>How to determine the mole of Ar</h3>
- Mass of Ar = 4.47 g
- Molar mass of Ar = 40 g/mol
- Mole of Ar =?
Mole = mass / molar mass
Mole of Ar = 4.47 / 40
Mole of Ar = 0.11175 mole
<h3>How to determine the temperature </h3>
- Number of mole (n) = 0.11175 moles
- Volume = 2.65 L
- Pressure (P) = 1.24 atm
- Gas constant (R) = 0.0821 atm.L/Kmol
The temperature of the container can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
Divide both side by nR
T = PV / nR
T = (1.24 × 2.65) / (0.11175 × 0.0821)
T = 358.16 K
Learn more about ideal gas equation:
brainly.com/question/4147359
Answer:
5.0
Explanation:
We have a buffer system formed by a weak base (C₅H₅N) and its conjugate acid (C₅H₆N⁺). We can calculate the pOH using the Henderson-Hasselbach's equation.
pOH = pKb + log [acid]/[base]
pOH = -log 1.8 × 10⁻⁹ + log 0.02/0.01
pOH = 9.0
Then, we will calculate the pH.
pH + pOH = 14
pH = 14 - pOH = 14 - 9.0 = 5.0
Answer:
The correct answer is 0.165 g NaCl.
Explanation:
The following is the precipitation reaction taking place between sodium chloride and silver nitrate:
NaCl (aq) + AgNO₂ (aq) ⇒ NaNO₃ (aq) + AgCl (s)
The complete ionic reaction of the reaction will be,
Na⁺ + Cl⁻ + Ag⁺ + NO₃⁻ ⇒ AgCl (s) + Na⁺ + NO₃⁻
Hence, the net ionic equation for the mentioned reaction is:
Ag⁺ (aq) + Cl⁻ (aq) ⇒ AgCl (s)
Thus, it can be witnessed that one mole of chloride ion is needed so that one mole of Ag⁺ ion get neutralized. There is a need to find the moles of silver ions present in the solution of AgNO₃ and then transform these moles to the moles of chloride ion. Ultimately, these moles can be converted to the concentration of sodium chloride needed.
The no. of moles of silver ions found in silver nitrate solution is,
(2.50 × 10² mL) × (0.0113 mol Ag⁺/1000 ml solution) = 2.83 × 10⁻³mol Ag⁺
Now the moles of chloride ions needed to precipitate the silver ions is,
(2.83 × 10⁻³ mol Ag⁺ ) × (1 mol Cl⁻/1 mol Ag⁺) = 2.825 × 10⁻³mol Cl⁻
The mass of sodium chloride needed for precipitating the silver ions will be,
mass of NaCl = (2.83 × 10⁻³ mol Cl⁻) × (1 mol NaCl / 1 mol Cl⁻) × (58.44 grams / 1 mol NaCl)
= 0.165 gram NaCl.
Answer:
the complete question is found in the attachment
Explanation:
the complete explanation is found in the attachment
