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kati45 [8]
3 years ago
5

Water has a higher boiling point than expected because

Chemistry
1 answer:
faust18 [17]3 years ago
3 0
Because of the strong attractions between polar water molecules.
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Which of the following is a pure substance? <br> brass<br> sodium<br> rocks<br> steel
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Homogeneous mixture<span>steel

</span>element<span>sodium


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Arrange the following molecules in order of increasing bond polarity (highest bond polarity at the bottom):
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NF3– 0.94– third
NCl3–0.12– second
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CF4–1.43– fourth

NBr3—NCl3—NF3—CF4
Lowest. Highest
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Any two objects in the universe, without exception, will do what
o-na [289]
Any two objects in the universe, without exception, attract each other.
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3 years ago
A teapot with a surface area of 785 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic
lina2011 [118]

Answer:

Given info: The surface area of teapot plated with silver is 700 cm2700 cm2, the cell is powered by 12.0-V12.0-V, the resistance of the cell is 1.80 Ω1.80 Ω, thickness of silver layer is 0.133-mm0.133-mm and density of silver is 10.5×103 kg/m310.5×103 kg/m3.

Write the expression for the mass of the silver.

m=ρAdm=ρAd (1)

Here,

ρρ is the density of silver.

AA is the surface area.

dd is the thickness of the silver layer.

Substitute 10.5×103 kg/m310.5×103 kg/m3 for ρρ, 700 cm2700 cm2 for AA and 0.133-mm0.133-mm for dd in equation (1) to find mass of the silver.

m=(10.5×103 kg/m3)(700 cm2(10−21 cm)2)(0.133-mm(10−3 m1 mm))=0.0978 kg(103 g1 kg)=97.8 gm=(10.5×103 kg/m3)(700 cm2(10−21 cm)2)(0.133-mm(10−3 m1 mm))=0.0978 kg(103 g1 kg)=97.8 g

Thus, the mass of silver is 97.8 g97.8 g.

Write the expression for number of moles of silver.

n=mWan=mWa (2)

Here,

mm is the mass of silver.

WaWa is the atomic weight.

The atomic weight of silver is 107.8 g/mole107.8 g/mole

Substitute 97.8 g97.8 g for mm, and 107.8 g/mole107.8 g/mole for WaWa in equation (2) to find number of moles of silver.

n=97.8 g107.8 g/mole=0.907 moln=97.8 g107.8 g/mole=0.907 mol

Thus, the number of moles of silver is 0.907 mol0.907 mol.

7 0
3 years ago
The rate constant k of the second-order reaction CH3CHO→CH4+CO is 6.73×10−3Lmol s. The concentration of CH3CHO at t=50.0 seconds
Tanzania [10]

Answer:

The initial concentration of ethanal was 0.1590 mol/L.

Explanation:

Integrated rate law for second order kinetic:

k=\frac{1}{t}(\frac{1}{[A]}-\frac{1}{[A]_o})

k = Rate constant =6.73\times 10^{-3} L mol s

t = Time elapsed  = 50.0 s

[A]_o =initial concentration of ethanal

[A] = Concentration of ethanal left after time t = 0.151 mol/L

On substituting the value:

6.73\times 10^{-3} L mol s=\frac{1}{50.0 s}(\frac{1}{0.151 mol/L}-\frac{1}{[A_o]})

[A]_o=0.1590 mol/L

The initial concentration of ethanal was 0.1590 mol/L.

5 0
3 years ago
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