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pentagon [3]
4 years ago
12

Calculate the number of C, H, and O atoms in 1.50 g of glucose, a sugar

Chemistry
1 answer:
DIA [1.3K]4 years ago
4 0
Chemical formula of the glucose: C₆H₁₂O₆

We calculate the molar mass:
atomic mass (C)=12 u
atomic mass (H)=1 u
atomic mass (O)=16 u

atomic weight (C₆H₁₂O₆)=6(12 u)+12(1u)+6(16 u)=72 u+12u+96 u=180 u.
Therefore : 1 mol of glucose will be 180 g
The molar mass would be: 180 g/ mol


2) we calculate the number of moles of 1.5 g.
180 g---------------------1 mol
1.5 g----------------------  x

x=(1.5 g * 1 mol) / 180 g≈8.33*10⁻³ moles

we knows that:
1 mol = 6.022 * 10²³ particles (atoms or molecules)

3)We calculate the number of molecules:

Therefore:
1 mol-----------------------6.022*10²³ molecules of glucose
8.33*10⁻³ moles--------        x

x=(8.33*10⁻³ moles * 6.022*10²³ molecules)/1 mol≈5.0183*10²¹ molecules.

4)We calculate the number of C, H and O atoms:
A molecule of glucose have 6 atoms of C, 12 atoms of H, and 6 atoms of O,
number of atoms of C=(6 atoms/1 molecule)(5.0183*10²¹molecules)≈
3.011*10²²

number of atoms of H=(12 atoms/1 molecule)(5.0183*10²¹ molecules)≈
6.022*10²² .

number of atoms of O=(6 atoms/1 molecule)(5.0183*10²¹ molecules)≈
3.011*10²²

Answer: we have 3.011*10²² atoms of C, 6.022*10²² atoms of H, and 3.011*10²² atoms of O.
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Calculate the energy of attraction between a cation with a valence of 1 and an anion with a valence of 3 the centers of which ar
Marrrta [24]

Answer:

The energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J

Explanation:

Let the charge on the cation be q₁

Also let the charge on the anion be q₂

A cation q₁ with a valence of 1, has a charge of 1 X 1.602×10⁻¹⁹C = 1.602×10⁻¹⁹C

An anion q₂ with a valence of 3, has a charge of 3 X 1.602×10⁻¹⁹C = 4.806 ×10⁻¹⁹C

The distance between the two charges is 7.5nm = 7.5 X10⁻⁹m

Energy of attraction = \frac{Kq_1q_2}{r^2}

Where k is coulomb's constant = 8.99 X 10⁹ Nm₂/C₂

Energy of attraction = \frac{8.99 X 10^9 (1.602X10^{-19})(4.806 X10^{-19})}{(7.5X10^{-9})^2}

Energy of attraction = 1.231 X 10⁻¹¹ J

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3 years ago
An orbital with n = 4 and ℓ = 2 would be described as
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Answer:

4d

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