The mass of AgNO₃ present in the given solution will be 5.31grams.
The molality is defined as moles of solute, which in your case is silver nitrate, per kilogram of the solvent. This means that the solution's molality essentially tells about the number of moles of solute present in 1 kg of solvent.
In this case, a 0.250 molal solution would contain 0.250 moles of solute for every 1 kg of solvent. The problem tells that we have 125 g of solvent available. Using the solution's molality as a conversion factor to calculate how many moles of silver nitrate it must contain and we should not forget to convert the mass of solvent from grams to kilograms
Now, we know that silver nitrate has a molar mass of 169.87 g mol⁻¹, which means that the one mole of silver nitrate has a mass of 169.87 g
We can thus say that 0.03125 moles of silver nitrate will have a mass of 0.03125 moles of AgNO₃ × 169.87/ 1 mole of AgNO₃
=5.31g
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Answer:
cover the taste with sugar add two times the amount your supposed to put in
Explanation:
The answer's the last choice. A well tested explanation of a natural phenomenon
Answer:
24.15%
Explanation:
According to the given situation the computation of the percent yield of the reaction is shown below:-
PV = NRT = N = ![\frac{PV}{RT}](https://tex.z-dn.net/?f=%5Cfrac%7BPV%7D%7BRT%7D)
Mole of
= ![\frac{PV}{RT}](https://tex.z-dn.net/?f=%5Cfrac%7BPV%7D%7BRT%7D)
= ![\frac{1\times 0.450}{0.0821\times 295}](https://tex.z-dn.net/?f=%5Cfrac%7B1%5Ctimes%200.450%7D%7B0.0821%5Ctimes%20295%7D)
= ![\frac{0.450}{24.2195}](https://tex.z-dn.net/?f=%5Cfrac%7B0.450%7D%7B24.2195%7D)
= 0.0186
Mole of ![N_2H_4 = \frac{2.45}{32}](https://tex.z-dn.net/?f=N_2H_4%20%3D%20%5Cfrac%7B2.45%7D%7B32%7D)
= 0.077
Now, the percentage of yield is
= ![\frac{Practical\ yield}{Theoretical\ yield}\times 100](https://tex.z-dn.net/?f=%5Cfrac%7BPractical%5C%20yield%7D%7BTheoretical%5C%20yield%7D%5Ctimes%20100)
= ![\frac{0.0186}{0.077}\times 100](https://tex.z-dn.net/?f=%5Cfrac%7B0.0186%7D%7B0.077%7D%5Ctimes%20100)
= 24.15%
Therefore for computing the percentage of yield we simply divide the practical yield by theoretical yield and multiply with 100 so that we can get the result into the percentage form.
Answer:
[Na₂CO₃] = 0.094M
Explanation:
Based on the reaction:
HCO₃⁻(aq) + H₂O(l) ↔ CO₃²⁻(aq) + H₃O⁺(aq)
It is possible to find pH using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [A⁻] / [HA]
Where [A⁻] is concentration of conjugate base, [CO₃²⁻] = [Na₂CO₃] and [HA] is concentration of weak acid, [NaHCO₃] = 0.20M.
pH is desire pH and pKa (<em>10.00</em>) is -log pka = -log 4.7x10⁻¹¹ = <em>10.33</em>
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Replacing these values:
10.00 = 10.33 + log₁₀ [Na₂CO₃] / [0.20]
<em> [Na₂CO₃] = 0.094M</em>
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