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3 years ago

Would fluorine form a positive or a negative ion?

1 answer:

8 0

It would form a negative ion... it lacks 1 e in its valence shell., it is easier for F to accept an e than to shed all existing 7.
e= electron

You might be interested in

How many electrons in an atom can have each of the following quantum number or sublevel designations?

Karo-lina-s [1.5K]

Answer: (a) 2 (b) 6 (c) 14

Explanation:

In the Azimuthal quantum number(l) electrons in a particular subshell (such as s, p, d, or f) are defined by values of l (0, 1, 2, or 3).

s is l=0, p is l=1, d is l=2, f is l=3.

The magnetic quantum number (ml) The value of ml can range from -l to +l, including zero. Thus the s, p, d, and f subshells contain 1, 3, 5, and 7 orbitals each, with values of m within the ranges 0, ±1, ±2, ±3 respectively. Each shell can have 2 x l + 1 sublevels, and each of these sublevel can accommodate up to two electrons.

(a) n=2, l=1, ml=0. If l=1 then 2 x 1+ 1=3 sublevels, 3*2=6 electrons. When l=1, ml =-1,0,+1, ml=0 accommodate two(2)electrons

(b) 5p. p is l=1 If l=1 then 2 x 1+ 1=3 sublevels, 3*2= electrons. This means in the 5 shell, the p orbital has 3 subshell and accommodate 6 electrons.

(c) n = 4, l = 3 if l=3 then 2 x 3+ 1=7 sublevel 7*2=14 electrons. This means the in the 4 shell, the f orbital has 7 subshell and accomdate 14 elections.

7 0

3 years ago

A student dissolves of glucose in of a solvent with a density of . The student notices that the volume of the solvent does not c

nikitadnepr [17]

Answer:

0.052 M

0.059 m

Explanation:

There is some missing info. I think this is the complete question.

<em>A student dissolves 4.6 g of glucose in 500 mL of a solvent with a density of 0.87 g/mL. The student notices that the volume of the solvent does not change when the glucose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits.</em>

Step 1: Calculate the moles of glucose (solute)

The molar mass of glucose is 180.16 g/mol.

4.6 g × 1 mol/180.16 g = 0.026 mol

Step 2: Calculate the molarity of the solution

0.026 moles of glucose are dissolved in 500 mL (0.500 L) of solution. We will use the definition of molarity.

M = moles of solute / liters of solution

M = 0.026 mol / 0.500 L = 0.052 M

Step 3: Calculate the mass corresponding to 500 mL of the solvent

The solvent has a density of 0.87 g/mL.

500 mL × 0.87 g/mL = 435 g = 0.44 kg

Step 4: Calculate the molality of the solution

We will use the definition of molality.

m = moles of solute / kilograms of solvent

m = 0.026 mol / 0.44 kg = 0.059 m

4 0

3 years ago

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is: