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allochka39001 [22]
3 years ago
15

How many moles of NaOH are present in 27.5 mL of 0.270 M NaOH?

Chemistry
1 answer:
Naddik [55]3 years ago
8 0
The solution for the question above is:

C = 0.270 
<span>V = 0.0275L </span>
<span>n = ? </span>

<span>Use the molar formula which is: C = n/V </span>
<span>Re-arrange it to: n = CV </span>
<span>n = (0.270)*(0.0275) </span>
<span>n = 0.007425 mols </span>
<span>(more precise) n = 7.425 x 10^-3 mols
</span>
7.425 x 10^-3 mols is the answer.
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What is the empirical formula of a compound composed of 30.5 g potassium (k) and 6.24 g oxygen (o)?
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Rhodium crystallizes in a face-centered cubic unit cell. The radius of a rhodium atom is 135 pm. Determine the density of rhodiu
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Answer:

Density of unit cell ( rhodium) = 12.279 g/cm³

Explanation:

Given that:

The radius (r) of a rhodium atom = 135 pm

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r = \dfrac{a}{2\sqrt{2}}

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a = edge length.

Making "a" the subject of the formula:

a = 2 \sqrt{2} \times r

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to cm, we get:

a = 381.8 × 10⁻¹⁰ cm

However, recall that:

density \ of \ unit \ cell = \dfrac{mass \ of \ unit \ cell}{volume \ of \unit \ cell}

where;

mass of unit cell = mass of atom × numbers of atoms per unit cell

Also;

mass\  of\ atom =\dfrac{ atomic \ mass}{Avogadro  \  number}

mass\  of\ atom =\dfrac{ 102.9}{6.023 \times 10^{23}}

Recall also that number of atoms in a unit cell for a  face-centered cubic = 4

So;

mass \ of \ unit \ cell= \dfrac{102.90}{6.023 \times 10^{23}}\times 4

mass of unit cell = 6.83380375 × 10⁻²² g

Density  \ of  \ unit \  cell = \dfrac{6.83380375 \times 10^{-22}}{(381.8\times 10^{-10})^3}

Density of unit cell ( rhodium) = 12.279 g/cm³

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