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3 years ago

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The molar mass of sucrose (C12H22O11) is 342.3 g/mol. A chemist has a 0.500-mol sample of sucrose. What is the mass, in grams, o

f this sample?

2 answers:

Gekata [30.6K]3 years ago

8 0

171 is the answer that I got after my calculations.

Alenkasestr [34]3 years ago

4 0

Answer:

171 grams.

Explanation:

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Condensation is the change of water from its gaseous form (water vapor) into liquid water. Condensation generally occurs in the atmosphere when warm air rises, cools and looses its capacity to hold water vapor. As a result, excess water vapor condenses to form cloud droplets.

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2 years ago

Compound a has a pka of 7 and compound b has a pka of 10. compound a is how many times more acidic than compound b?

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Ka is the acid dissociation equilibrium constant. The larger the value of the Ka, the stronger is the acid. To find Ka from pKa, the equation is:

pKa = -log[Ka]

@pKa = 7
7 = -log[Ka]
Ka = 1×10⁻⁷

@pKa = 10
10 = -log[Ka]
Ka = 1×10⁻¹⁰

This, pKa 7 is more acidic than pKa 10. The scale factor would be:
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<em>Therefore, Compound A is 1,000 times more acidic than Compound B.</em>

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3 years ago

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If the molecular weight of a semiconductor is 27.9 grams/mole and the diamond lattice constant is 0.503 nm, what is the density

adelina 88 [10]

Explanation:

The given data is as follows.

Mass = 27.9 g/mol

As we know that according to Avogadro's number there are $6.023 \times 10^{26}$ atom present in 1 mole. Therefore, weight of 1 atom will be as follows.

1 atoms weight = $\frac{38}{6.023 \times 10^{26}}$

In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.

Therefore, total weight of atoms in a unit cell will be as follows.

= $\frac{8 \times 27.9 g/mol}{6.023 \times 10^{26}}$

= $37.06 \times 10^{-26}$

Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.

= $a^{3}$

= $(0.503 \times 10^{-9})^{3}$

= $0.127 \times 10^{-27} m^{3}$

Formula to calculate density of diamond cell is as follows.

Density = $\frac{mass}{volume}$

= $\frac{37.06 \times 10^{-26}}{0.127 \times 10^{-27} m^{3}}$

= 2918.1 $g/m^{3}$

or, = 0.0029 g/cc (as 1 $m^{3} = 10^{6} cm^{3}$ )

Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.

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3 years ago