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Nostrana [21]
3 years ago
11

The density of liquid mercury is 13.69 g/cm^3. How many atoms of mercury are in a 15.0 cm^3 sample? Use "E" for "x10" and use si

g figs.
Chemistry
1 answer:
balu736 [363]3 years ago
6 0
The formula is  m = D x V
D = <span>13.69 g/cm^3.
</span>V = <span>15.0 cm^3 
the mass of the liquid mercury is m= </span>13.69 g/cm^3 x 15.0 cm^3 = 195g
the molar mass of Hg is 200,
1 mole of Hg = 200g Hg, so #mole of Hg= 195 / 200 = 0.97 mol
but we know that
 1 mole  = 6.022 E23 atoms
0.97 mole=?

6.022 E23 atoms x 0.97 / 1 mole = 5.84 E23 atoms
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13.0 g NaHSO4 is dissolved in water to make a 2.00 L solution what is the molarité
Sati [7]

Answer:

0.054 M

Explanation:

1 mol NaHSO4 -> 120 g

                   x    ->13 g

x= 0.108 mol NaHSO4

M= mol solute/ L solution

M= 0.108 mol NaHSO4/ 2.00L

M= 0.054 M

8 0
3 years ago
The number of nucleons in an atom is the atom's (atomic, mass) number.
ELEN [110]
The number of neutrons in an atom is the number of particles present in its nucleus.
The atomic number is the number of protons whereas the mass number is the number of protons and number of neutrons together
This implies that the number of neutrons is the atom's mass number
3 0
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Write a balanced equation for its combustion reaction.
laila [671]
<h3>Answer:</h3>

C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)

<h3>Explanation:</h3>

The balanced chemical equation for the combustion of the hydrocarbon in question is;

C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)

  • A balanced chemical equation is one in which the number of atoms of each element is equal on both sides of the equation.
  • In this case;
  • Reactant side has; 5 carbon atoms, 12 hydrogen atoms and 16 Oxygen atoms
  • Product side has; 5 carbon atoms, 12 hydrogen atoms and 16 Oxygen atoms
  • An equation is balanced by putting appropriate coefficients on reactants and products involved in the reaction.
  • An equation is balanced so as to obey the law of conservation of mass.
5 0
3 years ago
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xenn [34]
1 is b 2 is a 3 is d 4 is a 5 is c
5 0
3 years ago
onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0
3 years ago
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