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Sonja [21]
3 years ago
5

Convert from 1.56×1030 particles of sodium chloride (NaCl) to grams of sodium chloride.

Chemistry
1 answer:
Anika [276]3 years ago
8 0

Answer:

15.14×10⁷ g

Explanation:

Given data:

Number of particles of NaCl = 1.56×10³⁰ particles

Mass of sodium chloride = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ particles

1.56×10³⁰ particles × 1 mol / 6.022 × 10²³ particles

0.259 ×10⁷ mol

Mass in gram:

Mass = number of moles × molar mass

Mass = 0.259 ×10⁷ mol × 58.44 g/mol

Mass = 15.14×10⁷ g

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2 years ago
Predict the products of the thermal decomposition of strontium nitrate and show the reaction as a word equation.
Schach [20]

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Explanation:

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2 years ago
What is the atomic mass/mass number of the atom in the diagram above?
Lelechka [254]

Answer:

the atomic number is 5

the atomic mass is 11

Explanation:

The atomic number is the amount of protons inside the nucleus, and this number also equals the amount of electrons. Since it shows you the nucleus and the electrons, all you need to do is count the protons (positive charge inside the nucleus) or count all the electrons (negative charge outside the nucleus, in the rings) and you should have your atomic number.

As for mass, all you need to do is count all the protons and neutrons inside the nucleus and add them up. Protons = 5, Neutrons = 6. (you add them since the equation for atomic mass is Atomic Mass = Protons + neutrons. This works every time)

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3 0
1 year ago
Use the data given below to construct a Born-Haber cycle to determine the heat of formation of KCl. Δ H°(kJ) K(s) → K(g) 89 K(g)
AURORKA [14]

Explanation:

The net equation will be as follows.

          K(s) + Cl_{2}(g) \rightarrow KCl(s)

So, we are required to find \Delta H_{formation} for this reaction.

Therefore, steps involved for the above process are as follows.

Step 1:  Convert K from solid state to gaseous state

          K(s) \rightarrow K(g),    \Delta H_{1} = 89 kJ

Step 2:  Ionization of gaseous K

           K(g) \rightarrow K^{+}(g) + e^{-},    H_{2} = 418 KJ

Step 3:  Dissociation of Cl_{2} gas into chlorine atom .

            \frac{1}{2} Cl_{2}(g) \rightarrow Cl(g),   \Delta H_{3} = \frac{244}{2} = 122 KJ

Step 4: Iozination of chlorine atom.

              Cl(g) + e^{-} \rightarro Cl^{-}(g),      H_{4} = -349 KJ

Step 5:  Add K^{+} ion and Cl^{-} ion formed above to get KCl .

              K^{+}(g) + Cl^{-}(g) \rightarrow KCl(s),   H_{5} = -717 KJ

Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.

      \Delta H_{f} = \DeltaH_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} + \Delta H_{5}

                  = 89 + 418 + 122 - 349 - 717

                  = - 437 KJ/mol

Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.

5 0
3 years ago
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