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3 years ago

List five physical properties?

2 answers:

5 0

Some examples of physical properties are:

color (intensive)

density (intensive)

volume (extensive)

mass (extensive)

boiling point (intensive): the temperature at which a substance boils.

melting point (intensive): the temperature at which a substance melts.

Leokris [45]3 years ago

5 0

Physical Properties are used to describe how something is. Examples include:

1. Color
2. Shape
3. Size
4. Density
5. Melting/ Boiling point
6. Ordor

Physical changes: occur when matter changes form but not composition (Ex: ice is still H2O). Examples of physical change are:

1. Phase Change (Boiling, melting, freezing, etc.)
2. Size and/ or shape change
3. Dissolving
4. No Color Change (Except dyes)

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Name the element described in each of the following:

musickatia [10]

Period 4 transition element that forms 2+ ion with a half‐filled d sub level is Manganese (Mn)

What is the half-filled d sub-level?

Transition metals are an interesting and challenging group of elements. They have perplexing patterns of electron distribution that don’t always follow the electron-filling rules. Predicting how they will form ions is also not always obvious.

Transition metals belong to the d block, meaning that the d sublevel of electrons is in the process of being filled with up to ten electrons. Many transition metals cannot lose enough electrons to attain a noble-gas electron configuration. In addition, the majority of transition metals are capable of adopting ions with different charges. Iron, which forms either the Fe2+ or Fe3+ ions, loses electrons as shown below.

Some transition metals that have relatively few d electrons may attain a noble-gas electron configuration. Scandium is an example. Others may attain configurations with a full d sublevel, such as zinc and copper.

to know more about half-filled d sub-level

brainly.com/question/24780241

#SPJ4

6 0

2 years ago

The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C

artcher [175]

Explanation:

The given data is as follows.

$P_{1}$ = 100 mm Hg or $\frac{100}{760}atm$ = 0.13157 atm

$T_{1}$ = $1080 ^{o}C$ = (1080 + 273) K = 1357 K

$T_{2}$ = $1220 ^{o}C$ = (1220 + 273) K = 1493 K

$P_{2}$ = 600 mm Hg or $\frac{600}{760}atm$ = 0.7895 atm

R = 8.314 J/K mol

According to Clasius-Clapeyron equation,

$log(\frac{P_{2}}{P_{1}}) = \frac{\Delta H_{vap}}{2.303R}[\frac{1}{T_{1}} - \frac{1}{T_{2}}$

$log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]$

$log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]$

0.77815 = $\frac{\Delta H_{vap}}{19.147J/K mol} \times 6.713 \times 10^{-5} K$

$\Delta H_{vap}$ = $2.219 \times 10^{5}$ J/mol

= $2.219 \times 10^{5}J/mol \times 10^{-3}\frac{kJ}{1 J}$

= 221.9 kJ/mol

Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.

4 0

3 years ago

What effect does temperature have on reaction rate?

barxatty [35]

The answer for this multiple choice question is C

7 0

2 years ago

An electrolytic cell is based upon a nonspontaneous reaction; the Eocell for the reaction is _____.

Likurg_2 [28]

In an electrochemical cell in which the oxidation reaction is nonspontaneous the oxidation will not occur spontaneously at the anode and the reduction will not be spontaneous at the cathode. And according to the law for the calculation of the voltage potential of the electrochemical cell (Ecell):

Ecell = Eox. - Ere. where (Eox. is the potential of the oxidation at the anode and Ere. is the potential of the reduction at the cathode). The standard potential for an electrolytic cell is negative, because of the Ere. which is greater than Eox.

The answer is : less than zero.

6 0

3 years ago

An organic compound is 61.5% C, 2.56% H and 35.9% N by mass. 2.00 grams of this gas is entered into a 300.0 mL flask and heated

pashok25 [27]

Answer:

C₄H₂N₂

Explanation:

First we calculate the moles of the gas, using PV=nRT:

P = 2670 torr ⇒ 2670/760 = 3.51 atm

V = 300 mL ⇒ 300/1000 = 0.3 L

T = 228 °C ⇒ 228 + 273.16 = 501.16 K

3.51 atm * 0.3 L = n * 0.082atm·L·mol⁻¹·K⁻¹ * 501.16 K

n = 0.0256 mol

Now we calculate the molar mass of the compound:

2.00 g / 0.0256 mol = 78 g/mol

Finally we use the percentages given to calculate the empirical formula:

C ⇒ 78 g/mol * 61.5/100 ÷ 12g/mol = 4

H ⇒ 78 g/mol * 2.56/100 ÷ 1g/mol = 2

N ⇒ 78 g/mol * 35.9/100 ÷ 14g/mol = 2

So the empirical formula is C₄H₂N₂

6 0

3 years ago