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Charra [1.4K]
3 years ago
11

What methods of heat transfer will occur in your solar oven?

Chemistry
1 answer:
Arlecino [84]3 years ago
6 0

Answer: Convection, Conduction, and radiation

***If you found my answer helpful, please give me the brainliest. :) ****

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4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o
Vilka [71]

<u>Answer:</u> The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = \frac{1}{1}\times 0.0711=0.0711mol of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g

To calculate the percentage yield of 3-hexanol, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%

Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

4 0
3 years ago
How can you distinguish between a solution, colloid, and suspension?
RSB [31]

Answer:

solution is clear solution while colloidal is between the solution and suspension. And in suspension particles are suspended.

Explanation:

In solution light can be passed without any scattering of light from solute particles while suspension is cloudy and having larger particle size than colloids, if suspension stands for a while particles will settle down easily.

In colloids light  will scattered and dispersed by reflecting with large particles.

3 0
3 years ago
The ion MnOis often used to analyze for the Fe2+ content of an aqueous solution by using the (unbalanced) reaction Mno+Fe2+ + Fe
dusya [7]

Answer:

B) 0.230 M

Explanation:

The first step is to <u>balance the reaction</u> between the Ferrous ion and the permanganate ion:

5~Fe^+^2~+~MnO_4^-^1~+~8H^+~->~5Fe^+^3~+~Mn^+^2~+4H_2O

Then we have to <u>calculate the moles</u> of MnO_4^-:

M~=~\frac{mol}{L}

mol~=~M*L

mol~=~0.033~M*0.0633L=~0.002088~mol~MnO_4^-

Then using the <u>molar ratio</u> we can find the moles of Fe^+^2:

0.002088~mol~MnO_4^-\frac{5~mol~Fe^+^2}{1~mol~MnO_4^-}=0.01044~mol~Fe^+^2

Finally we can calculate the molarity:

M=\frac{0.01044~mol~Fe^+^2}{0.0455~L}=0.230~M

5 0
3 years ago
How many atoms are there in 5.00 mol of sulphur (S)?
Mumz [18]

Answer:

3.01 × 10²⁴ atoms S

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

5.00 mol S

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

<u />5.00 \ mol \ S(\frac{6.022 \cdot 10^{23} \ atoms \ S}{1 \ mol \ S} ) = 3.011 × 10²⁴ atoms S

<u />

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.011 × 10²⁴ atoms S ≈ 3.01 × 10²⁴ atoms S

6 0
3 years ago
Astronomers often scan the sky using devices called radio
sukhopar [10]

Answer:

A radio telescope is simply a telescope that is designed to receive radio waves from space.

radio telescopes helps to study naturally occurring radio light from stars, galaxies, black holes, and other astronomical objects. We can also use them to transmit and reflect radio light off of planetary bodies in our solar system.

6 0
3 years ago
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