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4 years ago

Suppose you had a mixture of sand and small, hollow beads. How might you separate the mixtures

Chemistry

1 answer:

umka21 [38]4 years ago

3 0

Answer:

I am not sure if this is the answer but maybe oil?

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The side chains of amino acids may contain..

salantis [7]

The correct answer is d)all of above

examples for a are tyrosine for b lysine and for c isoleucine

6 0

3 years ago

Read 2 more answers

A 1.10 g sample contains only glucose and sucrose. When the sample is dissolved in water to a total solution volume of 25.0L, th

olga_2 [115]

Answer:

$\large \boxed{79 \, \%}$

Explanation:

I assume the volume is 2.50 L. A volume of 25.0 L gives an impossible answer.

We have two conditions:

(1) Mass of glucose + mass of sucrose = 1.10 g

(2) Osmotic pressure of glucose + osmotic pressure of sucrose = 3.78 atm

Let g = mass of glucose

and s = mass of sucrose. Then

g/180.16 = moles of glucose, and

s/342.30 = moles of sucrose. Also,

g/(180.16×2.50) = g/450.4 = molar concentration of glucose. and

s/(342.30×2.50) = s/855.8 = molar concentration of sucrose.

1. Set up the osmotic pressure condition

Π = cRT, so

$\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}$

Now we can write the two simultaneous equations and solve for the masses.

2. Calculate the masses

$\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4) & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}$

We have 0.229 g of glucose and 0.871 g of sucrose.

3. Calculate the mass percent of sucrose

$\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}$

6 0

3 years ago

The energy in eV for light with a wavelength of 6250 angstroms is _. Note - there are 1.6 x 10-12 erg in 1 eV.

vivado [14]

Answer:

2 eV

Explanation:

The energy of a photon of light is given by the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

In this problem we have:

$h=6.63\cdot 10^{-34} Js$

$c=3.0\cdot 10^8 m/s$

$\lambda=6250 A = 6250\cdot 10^{-10} m$ is the wavelength of the photon

Therefore, the energy in Joules is

$E=\frac{(6.63\cdot 10^{-34})(3.0\cdot 10^8)}{6250\cdot 10^{-10}}=3.2\cdot 10^{-19}J$

We want to convert this energy into electronvolts: we know that the conversion factor is

$1 eV = 1.6\cdot 10^{-19}J$

Therefore,

$E=\frac{3.2\cdot 10^{-19}}{1.6\cdot 10^{-19}}=2 eV$

5 0

3 years ago

The monomer of poly(vinyl chloride) has the formula C2H3Cl. If there are 1,565 repeat units in a single chain of the polymer, wh

monitta

Answer:

$\large \boxed{9.780 \times 10^{4}\text{ u}}$

Explanation:

The molecular mass of a monomer unit is:

C₂H₃Cl = 2×12.01 + 3×1.008 + 35.45 = 24.02 + 3.024 + 35.45 = 62.494 u

For 1565 units,

$\text{Molecular mass} = \text{1565 units} \times \dfrac{\text{62.494 u}}{\text{1 unit }} = \mathbf{9.780 \times 10^{4}}\textbf{ u}\\\\\text{The molecular mass of the chain is $\large \boxed{\mathbf{9.780 \times 10^{4}}\textbf{ u}}$}$