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lara31 [8.8K]
4 years ago
14

Why does fingernail polish float on top of water?​

Chemistry
1 answer:
7nadin3 [17]4 years ago
5 0
Difference in density between the two liquids
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You have 15.0 L of nitrogen gas at 100°C and 200 kPa. What volume of gas will you have at STP?
Pavel [41]
Actually, you do NOT need to calculate the number of moles, so you do NOT need to convert the volume and pressure. (You DO have to convert the temperature.) That's because for this problem, pV/T is constant, so the units cancel (but the temperature scale matters). 

<span>200kPa * 15L / 373K = 101kPa * V / 273K </span>
<span>V = 21.7 L </span>
7 0
4 years ago
A student does not observe a change when holding a test tube in a flame. However, a change is expected. What is the most likely
dalvyx [7]

The answer is: The reactants were not heated long enough.

For all chemical reaction some energy is required and that energy is called activation energy (energy that needs to be absorbed for a chemical reaction to start).

Activation energy is the minimum energy colliding particles must have in order to react.

By lowering activation energy, reaction need less heat.

In this example, there is not enough heat.

3 0
3 years ago
Read 2 more answers
A particular flask has a mass of 17.4916 g when empty. When filled with ordinary water at 20.0°c (density = 0.9982 g/ml), the ma
LUCKY_DIMON [66]

The mass of the empty flask is 17.4916 g. Now after feeling the ordinary water the mass of the flask is 43.9616 g. Thus the change of weight due to addition of ordinary water is (43.9616 - 17.4916) = 26.47 g.

Now as the density of the ordinary water at 20°C is 0.9982 g/ml, so 26.47 g is equivalent to \frac{26.47}{0.9982}=26.5177 mL of water. Thus the capacity of the flask is 26.5177 mL.

Now the density of heavy water is 1.1053 g/mL at 20°C. Thus 26.5177 mL of heavy water is equivalent to (1.1053×26.5177) = 29.310 g.

Thus the total weight of the flask filled with heavy water will be (17.4916 + 29.310) = 46.8016 g at 20°C.  

4 0
3 years ago
Read 2 more answers
CdF2(s)⇄Cd2+(aq)+2F−(aq)A saturated aqueous solution of CdF2 is prepared. The equilibrium in the solution is represented above.
kupik [55]

Answer:

The correct answer is option a.

Explanation:

CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)

Equilibrium concentration cadmium ions = [Cd^{2+}]=0.0585 M

Equilibrium concentration fluoride ions = [F^{-}]=0.117 M

Molar solubility is the maximum concentration of salt present in water in ionic form beyond that no more salt will exist in its ionic form and will settle down in bottom of the solution.

The molar solubility of the solid cadmium fluoride = 0.0585 M

CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)..[1]

NaF(s)\rightleftharpoons Na^{+}(aq)+F^-(aq)

Due to addition of sodium fluoride will increase concentration of fluoride in the solution.And due to common ion effect the equilibrium will shift in backward direction in [1], that is precipitation of more cadmium fluoride.

Hence, decrease in solubility will be observed.

8 0
3 years ago
Calculate empirical formula 24.5 g nitrogen 70 g oxygen
LenKa [72]
24.5 / 14 = 1.75
70 / 16 = 4.375


4.375 / 1.75 = 2.5

empirical formula = N2O5
3 0
4 years ago
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