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geniusboy [140]
4 years ago
5

How heat moves from one end of a solid to another

Chemistry
1 answer:
pantera1 [17]4 years ago
3 0
In conduction, the thermal energy of a particle is transferred to other particles throughout the solid. The particles with more energy are transferred to those with less.
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Which organelle in the table is correctly matched with its function?<br> which one is it??
Papessa [141]

Answer:

\huge\boxed{\sf Ribosomes}

Explanation:

<h3>Organelles and their function:</h3><h3><u>Lysosomes:</u></h3>
  • Lysosomes functions in the digestion of food of the cell.
  • It contains hydrolytic enzymes.
<h3><u>Vacuole:</u></h3>
  • Vacuole mostly functions in storage.
<h3><u>Mitochondrion:</u></h3>
  • Mitochondrion is the power house of the cell.
<h3><u>Ribosome:</u></h3>
  • Ribosome functions in protein synthesis.

\rule[225]{225}{2}

4 0
2 years ago
What is the sequence of coefficients that will balance the following decomposition reaction of dihydrogen monosulfide? H₂S ----&
RUDIKE [14]

8 S

In this case you must start balancing the sulfur to have 8 on each side of the reaction. So:

Answer:

Explanation:

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

Then, you must balance the chemical equation. For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts.  

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.

Then, taking into account all of the above, you can determine the amount of elements on each side of the equation:

Left side: 2 H and 1 S

Right side: 2 H and 8 S

In this case you must start balancing the sulfur to have 8 on each side of the reaction. So:

8 H₂S → H₂ + S₈

Now the amount of elements on each side of the equation:

Left side: 16 H and 8  

Right side: 2 H and 8 S

Now you need to start balancing the hydrogen to get the same amount from each side of the reaction. So:

8 H₂S → 8 H₂ + S₈

Now the amount of elements on each side of the equation:

Left side: 16 H and 8  

Right side: 16 H and 8 S

<u><em>The balanced reaction is:</em></u>

<u><em>8 H₂S → 8 H₂ + S₈</em></u>

3 0
3 years ago
The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen
tino4ka555 [31]

Answer: The mass percentage of Tl_2SO_4 is 5.86%

Explanation:

To calculate the mass percentage of Tl_2SO_4 in the sample it is necessary to know the mass of the solute (Tl_2SO_4 in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the TlI precipitate.  We can establish a relation between the mass of TlI and Tl_2SO_4 using the stoichiometry of the compounds:

moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.

Since for every mole of Tl in TlI there are two moles of Tl in Tl_2SO_4, we have:

moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol

Using the molar mass of Tl_2SO_4 we have:

mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g

Finally, we can use the mass percentage formula:

mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%

6 0
3 years ago
What is the top of the way called
Mademuasel [1]
The top of a wave is called the crest. 

Hope I helped! :D
6 0
3 years ago
Which type of silk is the strongest? (1 po o nonbiodegradable
vesna_86 [32]

Answer:

d. major ampullate

6 0
2 years ago
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