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2 years ago

How many atoms are in 35 grams of table salt

1 answer:

4 0

Adopting the number of avogrado 6.02 * 10²³ / mol
<span>Sodium chloride (table salt)</span> Molar Mass = 58.44 g / mol
We will first have to find the number of moles in 35 grams of the element, like this:

1 mol ----------------- 58.44 g
X ---------------------- 35 g

58.44 * x = 35 * 1
58.44x = 35
$X = \frac{35}{58.44}$
X = 0.598904...
X ≈ 0.60<span> mol </span>

Now we will find how many atoms there are in 0.60 mol of this element, like this:

1 mol -------------------- 6.02 * 10²³ atoms
0.60 mol ----------------- X

X = 0.60 * 6.02 * 10²³
$\boxed{\boxed{x \approx 3.612 * 10^{23}\:atoms}}\end{array}}\qquad\quad\checkmark$

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Write the lewis structure for mgi2. draw the lewis dot structure for mgi2. include all lone pairs of electrons.

viktelen [127]

Lewis Structure is drawn in following steps,

1) Calculate Number of Valence Electrons:

# of Valence electrons in Mg = 2
# of Valence electrons in I     = 7
# of Valence electrons in I     = 7
---------
Total Valence electrons = 16

2) Draw Mg as a central atom surround it by two atoms of Iodine.

3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.

Total Valence electrons 16
- Four electrons - 4
----------
    12

4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.

The final lewis structure formed is as follow,

4 0

2 years ago

What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?

Sergeu [11.5K]

Answer:

$\large \boxed{\text{b. 1.5 h}}$

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

$\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt$

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

$\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}$

2. Calculate the half-life

$t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463 h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}$

4 0

2 years ago

a fixed amount of gas at 25.0 degrees Celsius occupies a volume of 10.0 L when the pressure is 629 torr. Use Charles law to calc

kirill [66]

Explanation:

Since pressure remained constant, we can eliminate P from the equation

$\frac{pv }{t} = \frac{pv}{t}$

Doing some algebra and converting temperature to Kevin by adding 273, you should obtain the same result.

8 0

2 years ago

Which concept describes the formation of four equivalent, single, covalent bonds by carbon in its compounds that resembles metha

Montano1993 [528]

Carbon is special and unique because it is able to form different compounds with a lot of elements, including itself. When it bonds with itself, this is possible because of the concept of hybridization. It is the mixing of atomic orbitals into a new hybrid orbital. In this case, methane is formed through the sp³ hybridization.

4 0

2 years ago

131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___

inn [45]

For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,
8.04 days = ln 2 / λ
λ = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ = ln 2 / 8.04 days = 0.693 / 8.04 days
t = 13.0 days

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

8 0

2 years ago