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Bingel [31]
3 years ago
5

What happens to radio waves when they are exposed to a plant?​

Chemistry
1 answer:
Ymorist [56]3 years ago
5 0

Answer:

,,

sdxxxp

Explanation:

,,xossssssss

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What is precipitate ​
natima [27]

Answer:

cause something (a substance) to be deposited in solid form from a solution

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What is the summary at the end of an experiment that explains the results
jasenka [17]

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The conclusion

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If 6.02×1023 atoms of element Y have a mass of 28.09 g, what is the identity of Y?
Mazyrski [523]

Silicon is the element having a mass of 28.09 g

<u>Explanation</u>:

  • Silicon is the element having an atomic mass of 28.09 g / mol. So 28.09 g of silicon contains 6.023 \times 10^23 atoms. One mole of each element can produce one mole of compound.
  • The Atomic weight of an element can be determined by the number of protons and neutrons present in one atom of that element. So atomic weight expressed in grams always contain the same number of atoms( 6.023 \times10^23).
  • Avagadro number is the number of atoms of 1 mole of any gas at standard temperature and pressure. It has been determined that 6.023 \times 10^23 atoms of an element are equal to the average atomic mass of that element.
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3 years ago
Convert .035 gallons to liters
kvasek [131]
.035 gallons equals 0.1324894 L
8 0
3 years ago
Read 2 more answers
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
masya89 [10]

Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
  • \rm O: \rm 15.999.

Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

6 0
3 years ago
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