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Musya8 [376]
3 years ago
9

Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH=1.06×10

−10m d H = 1.06 × 10 − 10 m , into the answer box.
Chemistry
1 answer:
kkurt [141]3 years ago
4 0

Answer:

The diameter of the hydrogen \mathbf{d =1.0605 \times 10^{-10}\ m}

Explanation:

From the given information:

Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

L = \dfrac{nh}{2 \pi}

Where the generic expression for angular momentum is:

L = mvr.

replacing the value of L into the previous equation, we have:

mvr= \dfrac{nh}{2 \pi}

v= \dfrac{nh}{2 \pi mr} ----- (1)

The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.

\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}   ----- (2)

replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:

\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2

ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}

r =\dfrac{n^2h^2}{4 \pi^2 mke^2}

For ground-state n = 1

h = (6.625 \times 10^{-34} \ J.s)^2

m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)

Ke = (1.6 \times 10^{-19} \ C)^2

r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}

r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}

\mathbf{r = 5.3025 \times 10^{-11} \ m}

Therefore, the diameter of hydrogen d = 2r

\mathbf{d = ( 2 \times  5.3025 \times 10^{-11} \ m})}

\mathbf{d =1.0605 \times 10^{-10}\ m}}

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Sodium Chloride is an ionic compound. Its molar mass is 58.44g. One formula unit of NaCl consists of one____, whose chemical sym
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Answer: One formula unit of NaCl consists of one cation, whose chemical symbol is Na^+ and one anion whose chemical symbol is Cl^-

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Tin (II) fluoride, formerly found in many kinds of toothpaste, is formed in this reaction: Sn (s) + 2HF (g) ——> SnF2 (s) + H2
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1.34 L of HF

Explanation:

We have the following chemical reaction:

Sn (s) + 2 HF (g) → SnF₂ (s) + H₂ (g)

First we calculate the number of moles of SnF₂:

number of moles = mass / molecular weight

number of moles of SnF₂ = 5 / 157 = 0.03 moles

From the chemical reaction we see that 1 mole of SnF₂ are produced from 2 moles of SnF₂. This will mean that 0.03 moles of SnF₂ are produced from 0.06 moles of HF.

Now at standard temperature and pressure (STP) we can use the following formula to calculate the volume of HF:

number of moles = volume / 22.4 (L/mole)

volume of HF = number of moles × 22.4

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When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas write the balanved equati
babunello [35]

Answer: When maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}.

Explanation:

The word equation is given as maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas.

Now, in terms of chemical formulae this reaction equation will be as follows.

MnO_{2} + HCl \rightarrow H_{2}O + MnCl_{2} + Cl_{2}

Here, number of atoms on reactant side are as follows.

  • Mn = 1
  • O = 2
  • H = 1
  • Cl = 1

Number of atoms on product side are as follows.

  • Mn = 1
  • O = 1
  • H = 2
  • Cl = 4

To balance this equation, multiply HCl by 4 on reactant side and multiply H_{2}O by 2 on product side. Therefore, the equation can be rewritten as follows.

MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}

Hence, number of atoms on reactant side are as follows.

  • Mn = 1
  • O = 2
  • H = 4
  • Cl = 4

Number of atoms on product side are as follows.

  • Mn = 1
  • O = 2
  • H = 4
  • Cl = 4

Since, this equation contains same number of atoms on both reactant and product side. Therefore, this equation is now balanced equation.

Thus, we can conclude that when maganese dioxide is added to hydrogen chloride you get water maganese dichloride and chlorine gas then balanced equation is MnO_{2} + 4HCl \rightarrow 2H_{2}O + MnCl_{2} + Cl_{2}.

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3 years ago
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