Which of these belongs in the outermost shell (energy level) of an atom?

An ionic bond forms between sodium chorine when the sodium atoms _______ electrons.

Natalija [7]

is 7 cus the valance electron for sodium is Ne 3s1 and for chlorine is Ne 3s13p5

4 0

4 years ago

5. How can you tell the difference between CuS and Cu2S

puteri [66]

Answer:

Density, melting point. and magnetic properties

Explanation:

I can think of three ways.

1. Density

The density of Cu₂S is 5.6 g/cm³; that of CuS is 4.76 g/cm³.

It should be possible to distinguish these even with high school equipment.

2. Melting point

Cu₂S melts at 1130 °C (yellowish-red); CuS decomposes at 500 °C (faint red).

A Bunsen burner can easily reach these temperatures.

3. Magnetic properties

You can use a Gouy balance to measure the magnetic susceptibilities.

In Cu₂S the Cu⁺ ion has a d¹⁰ electron configuration, so all the electrons are paired and the solid is diamagnetic.

In CuS the Cu²⁺ ion has a d⁹ electron configuration, so all there is an unpaired electron and the solid is paramagnetic.

A sample of Cu₂S will be repelled by the magnetic field and show a decrease in weight.

A sample of CuS will be attracted by the magnetic field and show an increase in weight.

In the picture below, you can see the sample partially suspended between the poles of an electromagnet.

6 0

3 years ago

What is the ph of a 0.0055 m ha (weak acid) solution that is 8.2% ionized?

Trava [24]

Answer is: pH value of weak is 3.35.
Chemical reaction (dissociation): HA(aq) → H⁺(aq) + A⁻(aq).
c(HA) = 0.0055 M.
α = 8.2% ÷ 100% = 0.082.
[H⁺] = c(HA) · α.
[H⁺] = 0.0055 M · 0.082.
[H⁺] = 0.000451 M.
pH = -log[H⁺].
pH = -log(0.000451 M).
pH = 3.35.
pH (potential of hydrogen) is a numeric scale used to specify the acidity or basicity <span>an aqueous solution.</span>

3 0

3 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$