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Yuki888 [10]
1 year ago
6

A solution of saline may contain 1.81 mol of sodium chloride (NaCl). How many sodium atoms are present in the solution?

Chemistry
1 answer:
joja [24]1 year ago
5 0

Answer:

1.09 x 10²⁴ atoms Na

Explanation:

In one mole of NaCl, there is also one mole of Na. To convert between moles and atoms, you need to use Avogadro's number. When writing the conversion, the desired unit (atoms) should be placed in the numerator to allow for the cancellation of units (moles). The final answer should have 3 sig figs to match the given number (1.81 mole).

Avogadro's Number:

6.022 x 10²³ atoms = 1 mole

1.81 moles Na           6.022 x 10²³ atoms
---------------------  x  --------------------------------  =  1.09 x 10²⁴ atoms Na
                                       1 mole

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Carbon reacts with steam (h2o) at high temperatures to produce hydrogen and carbon monoxide. if 2 mol of carbon are exposed to 3
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The balanced equation for the above reaction is as follows;
C + H₂O ---> H₂ + CO
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4 0
3 years ago
Calculate the no. of each atoms presents in 10 grams of calcium carbonate.
raketka [301]

Molar Mass of Calcium carbonate:-

\\ \sf\longmapsto CaCO_3

\\ \sf\longmapsto 40u+12u+3(16u)

\\ \sf\longmapsto 52u+48u

\\ \sf\longmapsto 100u

\\ \sf\longmapsto 100g/mol

  • Given Mass=10g

\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}

\\ \sf\longmapsto No\:of\;moles=\dfrac{10}{100}

\\ \sf\longmapsto No\:of\:moles=0.1mol

Now

\boxed{\sf No\:of\:molecules=No\;of\:moles\times Avagadro\: No}

\\ \sf\longmapsto 0.1\times 6.023\times 10^{23}

\\ \sf\longmapsto 6.023\times 10^{22}molecules

8 0
3 years ago
A chemical reaction was used to produce 2.95 moles of copper(II) bicarbonate, Cu(HCO3)2.
BARSIC [14]

Answer:

About 547 grams.

Explanation:

We want to determine the mass of copper (II) bicarbonate produced when a reaction produces 2.95 moles of copper (II) bicarbonate.

To do so, we can use the initial value and convert it to grams using the molar mass.

Find the molar mass of copper (II) bicarbonate by summing the molar mass of each individual atom:

\displaystyle \begin{aligned} \text{MM}_\text{Cu(HCO$_3$)$_2$} &= (63.55 + 2(1.01)+2(12.01)+6(16.00))\text{ g/mol} \\ \\  &=185.59\text{ g/mol} \end{aligned}

Dimensional Analysis:

\displaystyle 2.95\text{ mol Cu(HCO$_3$)$_2$}\cdot \frac{185.59 \text{ g Cu(HCO$_3$)$_2$}}{1 \text{ mol Cu(HCO$_3$)$_2$}} \Rightarrow 547 \text{ g Cu(HCO$_3$)$_2$ }

In conclusion, about 547 grams of copper (II) bicarbonate is produced.

8 0
3 years ago
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