The molarity of the diluted solution is 0.32 M
Considering the question given above, the following data were obtained:
Volume of stock solution (V₁) = 500 mL
Molarity of stock solution (M₁) = 2.1 M
Volume of diluted solution (V₂) = 3.25 L = 3.25 × 1000 = 3250 mL
<h3>Molarity of diluted solution (M₂) =....? </h3>
The molarity of the diluted solution can be obtained as follow:
<h3>M₁V₁ = M₂V₂</h3>
2.1 × 500 = M₂ × 3250
1050 = M₂ × 3250
<h3>Divide both side by 3250</h3><h3 />
M₂ = 1050 / 3250
<h3>M₂ = 0.32 M</h3>
Therefore, the molarity of the diluted solution is 0.32 M
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Given the data from the question, the final temperature is 200 K, while pressure remains constant.
<h3>Basic concepts </h3>
To obtain the correct answer to the question, we shall consider two conditions:
- Case 1 (temperature is constant)
- Case 2 (pressure is constant)
<h3>Case 1 (Temperature is constant) </h3>
We shall determine the new pressure by using the combined gas equation (P₁V₁ / T₁ = P₂V₂ / T₂) as illustrated below:
- Initial volume (V₁) = 3 L
- Initial pressure (P₁) = 1 atm
- Temperature = constant
- New Volume (V₂) = 2 L
- New pressure (P₂) =?
P₁V₁ / T₁ = P₂V₂ / T₂
Since temperature is constant, we have:
P₁V₁ = P₂V₂
3 × 1 = P₂ × 2
3 = P₂ × 2
Divide both side by 2
P₂ = 3 / 2
P₂ = 1.5 atm
<h3>Case 2 ( pressure is constant) </h3>
We shall determine the new temperature by using the combined gas equation (P₁V₁ / T₁ = P₂V₂ / T₂) as illustrated below:
- Initial volume (V₁) = 3 L
- Initial pressure (T₁) = 300 K
- Pressure = constant
- New Volume (V₂) = 2 L
- New pressure (T₂) =?
P₁V₁ / T₁ = P₂V₂ / T₂
Since pressure is constant, we have:
V₁ / T₁ = V₂ / T₂
3 / 300 = 2 / T₂
1 / 100 = 2 / T₂
Cross multiply
T₂ = 100 × 2
T₂ = 200 K
SUMMARY
- when the temperature is constant, the new pressure is 1.5 atm
- When the pressure is constant, the new temperature is 200 K
From the calculations made above, we can conclude that the correct answer is:
The final temperature is 200 K, while pressure remains constant.
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Answer:
![V_{NH_3}=39.6L](https://tex.z-dn.net/?f=V_%7BNH_3%7D%3D39.6L)
Explanation:
Hello,
By assuming STP conditions (0°C and 1atm), we first compute the reacting moles of both hydrogen and nitrogen as shown below via the ideal gas equation:
![n_{N_2}=\frac{PV}{RT}=\frac{1atm*36.7L}{0.082 \frac{atm*L}{mol*K}*273.15K} =1.64molN_2\\n_{H_2}=\frac{PV}{RT}=\frac{1atm*59.4L}{0.082 \frac{atm*L}{mol*K}*273.15K} =2.65molH_2](https://tex.z-dn.net/?f=n_%7BN_2%7D%3D%5Cfrac%7BPV%7D%7BRT%7D%3D%5Cfrac%7B1atm%2A36.7L%7D%7B0.082%20%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A273.15K%7D%20%3D1.64molN_2%5C%5Cn_%7BH_2%7D%3D%5Cfrac%7BPV%7D%7BRT%7D%3D%5Cfrac%7B1atm%2A59.4L%7D%7B0.082%20%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A273.15K%7D%20%3D2.65molH_2)
Next, one identifies the limiting reagent by computing the moles of hydrogen that completely react with 1.64mol of nitrogen as follows:
![n_{H_2}^{reacting}=1.64molN_2*\frac{3molH_2}{1molN_2}=4.92molH_2](https://tex.z-dn.net/?f=n_%7BH_2%7D%5E%7Breacting%7D%3D1.64molN_2%2A%5Cfrac%7B3molH_2%7D%7B1molN_2%7D%3D4.92molH_2)
In such a way, as there are just 2.65 available moles of hydrogen one states that we have spare nitrogen and the hydrogen is the limiting reagent, thus, the yielded moles of ammonia are computed as:
![n_{NH_3}=2.65molH_2*\frac{2molNH_3}{3molH_2}=1.767molNH_3](https://tex.z-dn.net/?f=n_%7BNH_3%7D%3D2.65molH_2%2A%5Cfrac%7B2molNH_3%7D%7B3molH_2%7D%3D1.767molNH_3)
Finally, one computes the required volume in liters as:
![V_{NH_3}=\frac{nRT}{P}=\frac{1.767mol*0.082 \frac{atm*L}{mol*K} *273.15K}{1atm}\\V_{NH_3}=39.6L](https://tex.z-dn.net/?f=V_%7BNH_3%7D%3D%5Cfrac%7BnRT%7D%7BP%7D%3D%5Cfrac%7B1.767mol%2A0.082%20%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%20%2A273.15K%7D%7B1atm%7D%5C%5CV_%7BNH_3%7D%3D39.6L)
Best regards.
Alkanes represent one family of organic compounds that is composed of carbon and hydrogen.
Alkanes are organic compounds that solely have hydrogen and carbon atoms in single bonds and do not have any additional functional groups. Alkanes can be divided into three groups: cycloalkanes, branched alkanes, and linear straight-chain alkanes.
The general formula for alkanes is CnH2n+2. Alkanes are also saturable hydrocarbons. Cyclic hydrocarbons like cycloalkanes have rings-shaped arrangements of their carbon atoms.
Therefore, Alkanes are an example of an organic chemical family made of carbon and hydrogen.
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