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guajiro [1.7K]
1 year ago
15

A(n) _____________ is a mixture of fuel and oxygen for which the masses of these two components are exactly those needed for com

plete combustion, with no residue.
Chemistry
1 answer:
aev [14]1 year ago
6 0

A stoichiometric mixture is a mixture of fuel and oxygen for which the masses of these two components are exactly those needed for complete combustion.

A stoichiometric mixture is a balanced mixture of fuel and oxygen.

The fuel and the oxygen react completely without the excesses of either.

The opposite of a stoichiometric mixture is called feeding an excess, when minimum one reactant is an excess amount.

Balanced chemical equation for reaction of combustion one type of a fuel: C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

Stoichiometric mixture for this example is when fuel (C₈H₁₈) and oxygen(O₂) react in proportion 1 : 12.5.

More about a stoichiometric mixture: brainly.com/question/19585982

#SPJ4

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In your own words, describe chromosomes, genes, and DNA.
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3 years ago
What happens when gasoline is used to power a vehicle?
Romashka [77]
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3 0
3 years ago
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

3 0
3 years ago
What type of bonds do compounds formed from non metal consist of?​
GuDViN [60]

Compounds formed from non-metals consist of molecules. The atoms in a molecule are joined together by covalent bonds. These bonds form when atoms share pairs of electrons.

7 0
3 years ago
How many liters of water must be added to 0.291 g of NaCl to create a salt water solution that has a concentration of 2.41 g/L?
dmitriy555 [2]
C=2.41 g/L
m(NaCl)=0.291 g

c=m(NaCl)/v

v=m(NaCl)/c

v=0.291/2.41= 0.1207 L = 120.7 mL
4 0
3 years ago
Read 2 more answers
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