Answer:
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Explanation:
The translated question is:
<em>What maximum amount of grams of potassium nitrate (V) can be dissolved in 300g of water at 90 °C</em>
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<h2>Solution</h2>
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To answer the question you need to consultate the solubiity information for potassium nitrate (V), KNO₃.
The attached table contains the solutibility table for KNO₃ at different temperatures.
At 90ºC it is 203g / 100g water.
Then, to calculate the <em>maximum amount of grams of potassium nitrate (V) that can be dissolved in 300g of water at 90 °C</em>, just multiply by the amount of water:
- 203g / 100g water × 300 g water = 609g ← answer
Answer:
a. pH = 13.50
b. pH = 13.15
Explanation:
Hello!
In this case, since the undergoing chemical reaction between KOH and HBr is:
As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:
![pOH=-log([OH^-])=-log(0.320)=0.50](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.320%29%3D0.50)
Thus, the pH is:
Which is the same answer for a and b as they ask the same.
Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:
It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:
Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):
![[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5BKOH%5D%3D%5Cfrac%7B0.00425mol%7D%7B0.03000L%7D%3D0.142M)
So the pOH and the pH turn out:
Best regards!
Answer:
10.52g KOH
Explanation:
250.0 ml X 1L/1000ml X 0.75 mol KOH/1L X 56.105gKOH/1 mol KOH =10.52g KOH
<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
Answer:
None of the options are correct. The correct answer is 30020J
Explanation:
Data obtained from the question include:
Mass (M) = 2000g
Initial temperature (T1) = 10°C
Final temperature (T2) = 29°C
Change in temperature (ΔT) = T2 - T1 = 29°C - 10°C = 19°C
Heat (Q) =?
Specific heat capacity (C) of granite = 0.79J/g°C
Applying the equation Q = MCΔT, the heat absorbed by the granite rock can be obtained as follow:
Q = MCΔT
Q = 2000 x 0.79 x 19
Q = 30020J
