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gogolik [260]
3 years ago
5

The density of water is about 1.0 g/mL at room temperature. Briefly explain how the density of an aqueous solution at room tempe

rature can be significantly less than 1.0 g/mL. Give an example of such a solution.
Chemistry
1 answer:
maksim [4K]3 years ago
7 0

Answer:

The density of water is about 1.0 g/mL at room temperature.

Briefly explain how the density of an aqueous solution at room temperature can be significantly less than 1.0 g/mL.

Give an example of such a solution.

Explanation:

That means 1.0mL of water weighs ---- 1.0g

If any other aqueous solution which has mass less than 1.0g will have density less than 1.0g.

For example aqueous solution of sulfuric acid has density 0.98g/mL.

That means 1mL of sulfuric acid has mass 0.98g.

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Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ
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To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows 
time         0         40        80       120       160
moles    0.100   0.067  0.045    0.030    0.020


Q1)
for the first 40 s change of moles ;
      = -d[A] / t
      = - (0.067-0.100)/40s
      = 8.25 x 10⁻⁴ mol/s
for the next 40 s
      = -(0.045-0.067)/40
      = 5.5 x 10⁻⁴ mol/s
the 40 s after that
      = -(0.030-0.045)/40 s 
     = 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ   -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation. 
Since this is a first order reaction,
b = 1
therefore the reaction is 
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
   = 8.25 x 10⁻³ s⁻¹

7 0
3 years ago
A standard 1.00 kg mass is to be cut from a bar of steel having an equilateral triangular cross section with sides equal to 2.50
Elden [556K]

Answer:

The section of the bar is 2.92 inches.

Explanation:

Mass of the steel cut ,m = 1.00 kg = 1000 g

Volume of the steel bar = V = Area × height

Height of the of the  section of bar = h

Area of  Equilateral triangular = \frac{\sqrt{3}}{4}a^2

a = 2.50 inches

Cross sectional area of the steel mass = A

A=\frac{\sqrt{3}}{4}(2.50 inches)^2=2.71 inches^2

V = 2.71 inches^2\times h

Density of the steel = d =7.70 g/cm^3

1cm^3 = 0.0610237 inches^3

d=\frac{m}{v}

\frac{7.70 g}{0.0610237 inches^3}=\frac{ 1000 g}{2.71 inches^2\times h}

h=\frac{ 1000 g\times 0.0610237 inches^3}{2.71 inches^2\times 7.70 g}

h = 2.92 inches

The section of the bar is 2.92 inches.

3 0
3 years ago
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