3C2H402 atoms in formula

Which of the following mixture types canNOT be filtered to remove solutes?

iVinArrow [24]

1. A) Colloids only

2. C) Molecules of the dispersion medium colliding with dispered phase particles

Hope this helps!

3 0

3 years ago

jin has four colorful glass bottles on his windowsill. one is red, one is violet, one is green, and one is yellow. the light tha

Over [174]

The red bottle would have the lowest frequency because red light has the longest wavelengths. The light passing through the violet would have the highest frequency because its wavelengths are the shortest.

4 0

3 years ago

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What is the wavelength of a photon with an energy of 3.38 x 10-19 J?

irina1246 [14]

Answer:

C. 588 nm

Explanation:

Given parameters:

Energy of the photon = 3.38 x 10⁻¹⁹J

Unknown:

Wavelength of the photon = ?

Solution:

The energy of a photon can be expressed as;

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

hc = E x wavelength

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

h is the Planck's constant = 6.63 x 10⁻³⁴m²kg/s

c is the speed of light = 3 x 10⁸m/s

E is the energy

Wavelength = $\frac{6.63 x 10^{-34} x 3 x 10^{8} }{3.38 x 10^{-19} }$ = 5.89 x 10⁻⁷m

4 0

3 years ago

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What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

Now put all the given values in the above formula, we get:

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

3 years ago

When 0.105 mol propane, C3H8 is burned in an excess of oxygen, how many moles of oxygen are consumed? The products are carbon di

torisob [31]

Answer: The amount of oxygen gas consumed is 0.525 moles

Explanation:

We are given:

Moles of propane burned = 0.150 moles

The chemical equation for the combustion of propane follows:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

By Stoichiometry of the reaction:

1 mole of propane reacts with 5 moles of oxygen gas

So, 0.150 moles of propane will react with = $\frac{5}{1}\times 0.105=0.525mol$ of oxygen gas

Hence, the amount of oxygen gas consumed is 0.525 moles

6 0

3 years ago