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Virty [35]
4 years ago
14

Which of the following fractions can be used in the conversion of 1000 mL to the unit m3? 1L/1ML 1CM^3/1ML 1CM^-3/1ML 1M^3/1L

Chemistry
1 answer:
Flauer [41]4 years ago
6 0
The conversion that can be used is 1 cm³ / mL
After multiplying by this conversion, we get:
1000 mL * (1 cm³ / mL) = 1000 cm³

Next, we may convert directly from cm³ to m³ by:
1 m = 100 cm
1 m³ = 100³ cm³

Using this conversion, we get the volume in m³ to be:
0.001 m³
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Please help me with this homework I really need help with this homework
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Answer:

For the first question its C, Gas

For the second one table

Explanation:

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During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of
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Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate \rm CaCO_3?

Refer to a modern periodic table for relative atomic mass data:

  • Ca: 40.078;
  • C: 12.011;
  • O: 15.999.

Formula mass of \rm CaCO_3:

M(\mathrm{CaCO_3})  = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}.

\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol.

How many moles of \rm CaCl_2 will be produced?

The coefficient in front of \rm CaCO_3 in the chemical equation is the same as that in front of \rm CaCl_2. That is:

\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1.

\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol.

What's the theoretical yield of calcium chloride? In other words, what's the mass of \rm 0.949184\;mol of \rm CaCl_2?

Again, refer to a periodic table for relative atomic data:

  • Ca: 40.078;
  • Cl: 35.45.

M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}.

\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}.

What's the actual yield of calcium chloride?

\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%.

\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}.

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Explanation:

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  Information about helium:

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  To write an atom we use this format:

                                     ₙᵇGˣ

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   b is the mass number

    x is the charge on the atom

Using the periodic table as guide

    Symbol of helium is He

     Atomic number of helium is 2

     mass number of helium is 4

     charge on helium atom is 0

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Every atom contains protons, neutrons and electrons;

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   Protons are positively charge particles in an atom. The atomic number is the number of protons in an atom.

   neutrons do not have charges.

 

   Helium has:

        Number of protons = 2

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        Number of neutrons = 2

Mass of each subatomic particle:

      1 electron = 9.11 x 10⁻³¹kg

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     Protons and neutrons have the same mass:

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           2 protons = 3.34 x 10⁻²⁷kg

           2 neutrons = 3.34 x 10⁻²⁷kg

Learn more:

helium brainly.com/question/2439349

#learnwithBrainly

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More pollutants accumulate
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