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Answer:
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Explanation:
Approximately 2 mL of Solution A (on the left) is added to a sample of Solution B (on the right) with a dropping pipet. If a precipitate forms, the resulting precipitate is suspended in the mixture. The mixture is then stirred with a glass stirring rod and the precipitate is allowed to settle for about a minute.
Solution A: 0.5 M sodium hydroxide, colorless
Solution B: 0.2 M nickel(II) nitrate, green
Precipitate: light green
Ni(NO3)2(aq) + 2 NaOH(aq) —> Ni(OH)2(s) + 2 NaNO3(aq)
Credits:
Design
Kenneth R. Magnell Central Michigan University, Mt. Pleasant, MI 48859
John W. Moore University of Wisconsin - Madison, Madison, WI 53706
Video
Jerrold J. Jacobsen University of Wisconsin - Madison, Madison, WI 53706
Text
Kenneth R. Magnell Central Michigan University, Mt. Pleasant, MI 48859
Assuming that the number of mols are constant for both conditions:
![\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%20%5Cfrac%7BP_1V_1%7D%7BT_1%7D%20%3D%20%20%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
Now you plug in the given values. V_1 is the unknown.
![\frac{0.50 atm*V_1}{325K} = \frac{1.2 atm* 48 L}{230K} ](https://tex.z-dn.net/?f=%5Cfrac%7B0.50%20atm%2AV_1%7D%7B325K%7D%20%3D%20%5Cfrac%7B1.2%20atm%2A%2048%20L%7D%7B230K%7D%0A)
Separate V_1
![V_1= \frac{1.2 atm* 48 L * 325K }{230K*0.50 atm }](https://tex.z-dn.net/?f=V_1%3D%20%5Cfrac%7B1.2%20atm%2A%2048%20L%20%2A%20325K%20%7D%7B230K%2A0.50%20atm%20%7D)
V= 162.782608696 L
There are 2 sig figs
V= 160 L