Yes, o-toluic acid is soluble in ether as ether is slightly polar and it is soluble in NaOH because it is likely to form soluble compounds with it.
Naphthalene is insoluble in NaOH.
Answer:
The new temperature of the water bath 32.0°C.
Explanation:
Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)
Initial temperature of the water = 
Final temperature of the water = 
Specific heat capacity of water under these conditions = c = 4.18 J/gK
Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J
( 1kJ=1000 J)




The new temperature of the water bath 32.0°C.
The volume of the piece of aluminum is 1.96 mL
Explanation:
Density is the relationship of the mass of a substance and its volume.
In this case, the mass of aluminum is 5.30 g and the density is 2.70 g/mL
The formula to apply is;
D=M/V where D is density in g/mL, M is mass in g and V is volume in mL
2.70=5.30/V
V=5.30/2.70 =1.96 mL
Learn More
Density of a substance:brainly.com/question/12605423
Keywords: volume, aluminum,density
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Hmm, friction maybe? I guess it depends on how fast she stopped?
Answer:
The formula of the compound is:
N2H2
Explanation:
Data obtained from the question:
Nitrogen (N) = 93.28%
Hydrogen (H) = 6.72%
Next, we shall determine the empirical formula for the unknown compound. This is illustrated below:
N = 93.28%
H = 6.72%
Divide by their molar mass
N = 93.28 /14 = 6.663
H = 6.72 /1 = 6.7
Divide by the smallest
N = 6.663 / 6.663 = 1
H = 6.72 /6.663 = 1
Therefore, the empirical formula is NH.
Now, we can obtain the formula of the compound as follow:
The formula of a compound is simply a multiple of the empirical formula.
[NH]n = 30.04
[14 + 1]n = 30.04
15n = 30.04
Divide both side by 15
n = 30.04/15
n = 2
Therefore, the formula of the compound is:
[NH]n => [NH]2 => N2H2