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Studentka2010 [4]
2 years ago
15

Elements can be described by various properties, and identified by their boiling and melting points. For example, gold melts at

1,064ºC and boils at 2,856ºC. Does boiling point depend on the mass present?
A. No; chemical properties stay the same regardless of mass.
B. No; physical properties stay the same regardless of mass.
C. Yes; physical properties can change when mass increases or decreases.
D. No; qualitative properties like boiling point stay the same regardless of mass.
Chemistry
2 answers:
jekas [21]2 years ago
7 0

Answer:

B i supposed.

Explanation:

aleksandr82 [10.1K]2 years ago
5 0

Answer:

Explanation:

Melting and boiling point variations are not clear (do not have uniform pattern) in periodic table. But we can see, some elements have higher melting and boiling points and some have less. Here we study melting and boiling points of s, p, d blocks elements. IVAth group elements (C,Si) show high melting and boiling points because they have covalent gigantic lattice structures.

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K+ and NO3- ions

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2 years ago
Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

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