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3 years ago

How many molecules of liquid methanol are in 10.00 mL of methanol, given its density is 0.7918 g/mL?

1 answer:

7 0

$d=0,7918\frac{g}{mL}\\ V=10mL \ \ \ \ \ \ \ \ \ \ \ m=dV=0,7918\frac{g}{mL}*10mL=7,918g\\\\ 6,02*10^{23} \ \ \ \ \ \ \ \Rightarrow \ \ \ \ 32g\\ N_{x} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ 7,918g\\\\ N_{x}=\frac{6,02*10^{23}*7,918g}{32g}\approx1,49*10^{23}$

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A. The person weighs 56 pounds more on Mars than on the moon

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4 years ago

For each of the following sublevels, give the n and l values and the number of orbitals: (a) 6g; (b) 4s; (c) 3d.

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Answer:

(a) 6g. Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 9

(b) 4s. Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 1

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n, n = 1, 2, 3...

2. Subshell number, 0 ≤ l ≤ n − 1, orbital s - 0, p - 1, d - 2, f - 3

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So,

(a) 6g. Shell 6, n = 6. Subshell g, l = 4. Number of orbitals in sublevel = 2l+1 = 9

(b) 4s. Shell 4, n = 4. Subshell s, l = 0. Number of orbitals in sublevel = 2l+1 = 1

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3 years ago

A 0.175 M solution of an enantiomerically pure chiral compound D has an observed rotation of +0.27° in a 1-dm sample

-BARSIC- [3]

Answer:

The specific rotation of D is 11.60° mL/g dm

Explanation:

Given that:

The path length (l) = 1 dm

Observed rotation (∝) = + 0.27°

Molarity = 0.175 M

Molar mass = 133.0 g/mol

Concentration in (g/mL) = 0.175 mol/L × 133.0 g/mol

Concentration in (g/mL) = 23.275 g/L

Since 1 L = 1000 mL

Concentration in (g/mL) = 0.023275 g/mL

The specific rotation [∝] = ∝/(1×c)

= 0.27°/( 1 dm × 0.023275 g/mL )

= 11.60° mL/g dm

Thus, the specific rotation of D is 11.60° mL/g dm

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3 years ago