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aev [14]
3 years ago
9

Determine the relative formula mass of hexasodium difluoride using the periodic table below. A. 138 g/mol B. 176 g/mol C. 20 g/m

ol D. 42 g/mol
Chemistry
2 answers:
Irina-Kira [14]3 years ago
4 0

Answer:

B. 176 g/mol

Explanation:

chemistry ed tell

laila [671]3 years ago
3 0

Answer:

Option B. 176g/mol

Explanation:

We'll begin by writing the chemical formula for hexasodium difluoride. This is given below:

Hexasodium means 6 sodium atom

Difluoride means 2 fluorine atom.

Therefore, the formula for hexasodium difluoride is Na6F2.

The relative formula mass of a compound is obtained by simply adding the atomic masses of the elements present in the compound.

Thus, the relative formula mass of hexasodium difluoride, Na6F2 can be obtained as follow:

Molar mass of Na = 23g/mol

Molar mass of F = 19g/mol

Relative formula mass Na6F2 = (23x6) + (19x2)

= 138 + 38

= 176g/mol

Therefore, the relative formula mass of hexasodium difluoride, Na6F2 is 176g/mol

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The initial activity of 37Ar is 8540 disintegrations per minute. After 10.0 days, the activity is 6990 disintegrations per minut
mylen [45]

Answer:

Approximately 3318 disintegrations per minute.

Explanation:

The activity A of a radioactive decay at time t can be found with the following equation:

\displaystyle A(t) = A_0 \cdot \mathrm{e}^{-\lambda\cdot t}.

In this equation,

  • \mathrm{e} is the natural base. \mathrm{e}\approx 2.71828.
  • A_0 is the initial activity of the decay. For this question, A_0 = \rm 8540\; min^{-1}.
  • The decay constant \lambda of this sample needs to be found.

The decay constant here can be found using the activity after 10 days. As long as both times are in the same unit (days in this case,) conversion will not be necessary.

A(10) = A_0\cdot \mathrm{e}^{\rm -\lambda \times 10\;day}= (\mathrm{8540\; min^{-1}})\cdot \mathrm{e}^{\rm -\lambda \times 10\;day}.

A(10) = \rm 6690\; min^{-1}.

\displaystyle \frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}} = \mathrm{e}^{\rm -\lambda \times 10\;day}

Apply the natural logarithm to both sides of this equation.

\displaystyle \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)} = \ln{\left(\mathrm{e}^{\rm -\lambda \times 10\;day}\right)}.

\displaystyle \rm -\lambda \cdot (10\;day) = \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)}.

\displaystyle \rm \lambda= \rm \frac{\displaystyle \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)}}{-10 \; day} \approx 0.0200280\; day^{-1}.

Note that the unit of the decay constant \lambda is \rm day^{-1} (the reciprocal of days.) The exponent -\lambda \cdot t should be dimensionless. In other words, the unit of t should also be days. This observation confirms that there's no need for unit conversion as long as the two times are in the same unit.

Apply the equation for decay activity at time t to find the decay activity after 47.2 days.

\displaystyle \begin{aligned}A(t)& = A_0 \cdot \mathrm{e}^{-\lambda\cdot t}\\&\approx \rm \left(8540\; min^{-1}\right)\cdot \mathrm{e}^{-0.0200280\; day^{-1}\times 47.2\;day}\\&\approx \rm 3318\; min^{-1}\end{aligned}.

By dimensional analysis, the unit of activity here should also be disintegrations per minute. The activity after 47.2 days will be approximately 3318 disintegrations per minute.

8 0
3 years ago
2Ag (s) + H2S(s) → Ag2S(s) + H2(g)
Advocard [28]

Answer: Oxidation half: 2Ag\rightarrow 2Ag^++2e^-

Reduction half: 2H^++2e^-\rightarrow H_2

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

2Ag(s)+H_2S(s)\rightarrow Ag_2S(s)+H_2(g)

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Oxidation half: 2Ag\rightarrow 2Ag^++2e^-

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Reduction half: 2H^++2e^-\rightarrow H_2

8 0
3 years ago
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