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Both of these questions can be solved using the equation M1V1 = M2V2, where M is concentration anf V is volume.
For the first case, M2 = 0.2 mol/L, M1 = 3 mol/L, and V2 = 250mL. So now you want V1. Solving for V1, V1 = (M2 / M1)V2 =
(0.2 / 3)(250) = 16.7 mL. So what that means is that you need 16.7 mL of 3M HCl, and the rest of the 250 mL (which would be 250 - 16.7 = 233.3 mL) would be water, with which you're diluting the HCl.
Same principle for the second problem, except now we have percentages and not mol/L. You can treat the percentages as concentrations. Since you're starting with pure isopropyl alcohol, M1 = 100%. You want a final volume of 500 mL and a final concentration of 70%. To find the volume of isopropyl alcohol you need to start with, solve for V1. So V1 = (M2 / M1)V2 = (70 / 100)(500) = 350 mL. So you need 350 mL of isopropyl alcohol and the rest of the 500 mL (that is, 150 mL) you can fill with water.
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I think it is maintaining certain internal condition
Answer:
No, the metal is not pure gold because density is equal to mass divided by volume and in this case we end up with about 15.3, making this metal less dense than gold.
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