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mash [69]
3 years ago
5

How many mol of hydrogen in 2.8 mol of caffeine.

Chemistry
2 answers:
pochemuha3 years ago
5 0
Empirical formula of caffeine is C₈H₁₀N₄O₂.

In 1 mol of caffeine we have 10 mol of hydrogen.
In 2,8 mol of caffeine we have x mol of hydrogen.

x = 2,8 mol * 10 mol / 1 mol = 28 mol

ANSWER: There are 28 mol of hydrogen.

:-) ;-)
Rama09 [41]3 years ago
5 0

Answer:

Empirical formula of caffeine is C₈H₁₀N₄O₂. In 1 mol of caffeine we have 10 mol of hydrogen. In 2,8 mol of caffeine we have x mol of hydrogen. ANSWER: There are 28 mol of hydrogen.

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Convert 685K to Fahrenheit
joja [24]

773.33 degrees F
F=Fahrenheit
K=Kelvin
F=K x 9/5 - 459.67
773.33 = 685 x 9/5 - 459.67

8 0
3 years ago
The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma
IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

A=6R^{2}\sqrt{3}

Here, R is radius and is related to a as follows:

R=\frac{a}{2}

Putting the value in expression for area,

A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}

The value of a is 0.4961 nm

Since, 1 nm=10^{-7}cm

Thus, 0.4961 nm=4.961\times 10^{-8} cm

Putting the value,

Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}

Now, volume can be calculated as follows:

V=Area\times c

The value of c is 1.360 nm or 1.360\times 10^{-7} cm

Putting the value,

V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}

now, number of atom in unit cell can be calculated by using the following formula:

n=\frac{\rho N_{A}V_{c}}{A}

Here, A is atomic mass of Cr_{2}O_{3} is 151.99 g/mol.

Putting all the values,

n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18

Thus, there will be 18 Cr_{2}O_{3} units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of Cr^{3+} and O^{2-} is 62 pm and 140 pm respectively.

Converting them into cm:

1 pm=10^{-10}cm

Thus,

r_{Cr^{3+}}=6.2\times 10^{-9}cm

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

Volume of sphere will be sum of volume of total number of cations and anions thus,

V_{S}=V_{Cr^{3+}}+V_{O^{2-}}

Since, volume of sphere is V=\frac{4}{3}\pi r^{3},

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

Putting the values,

V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758

Thus, atomic packing factor is 0.758.

6 0
3 years ago
Read 2 more answers
Drag the item from the idea bank to its corresponding match
babymother [125]
Hope this is helpful

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Enzymes in the small intestine break down proteins, carbohydrates, and lipids. These enzymes are named after the substrate they
svetoff [14.1K]

The enzymes and their respective substrates are as follows:

  • Protease enzymes such as trypsin and chymotrypsin break down proteins
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  • Lipase enzyme breaks down lipids.

In the small intestine, a protease enzyme known as chymotrypsin breaks down protein, pancreatic amylase breaks down carbohydrates, while pancreatic lipase breaks down lipids.

More on biological enzymes can be found here: brainly.com/question/12194042

7 0
2 years ago
Liquid Q is a polar solvent and liquid R is a nonpolar solvent. On the basis of this information, you would expect:
bearhunter [10]
4) is correct
This is because water is polar and it will mix with a polar solvent. A good rule for remembering the behavior of non-polar and polar compounds when it comes to being miscible is that "like dissolves like."
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