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KiRa [710]
3 years ago
15

If 2 objects have the same mass of 100 grams and object A has a volume of 50cm 3 and object B has a volume of 200 cm 3, which of

the 2 objects have the greater density?
Chemistry
1 answer:
kenny6666 [7]3 years ago
4 0

Answer:

The object A will be having the greater density compared to object B.

Explanation:

It is known that density of any object is defined as the mass of any object occupying a given volume. So the ratio of mass and volume will help to determine the density of any object. Density=Mass/Volume

From the above equation, it can be seen that the density of any object is directly proportional to the mass of the object and inversely proportional to the volume occupied by the object.

So in the present context, the mass of objects A and B are same and it is 100 g. Thus, the density of object A and object B will be influenced by their volume. As it is given that the volume of object A is 50 cm3 and object B is 100 cm3, then depending upon the relationship of volume and density, the density of both the objects can be determined. As the object with higher volume will be having lesser density as volume is inversely proportional to density. Thus, in the given case the volume of object B is greater than object A and so the object A will be having greater density compared to object B.

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A sample of gas occupies 2.5 liters of space. How many moles of gas are there if its pressure is 1.2 atm and temperature is 25 C
S_A_V [24]

Moles of gas = 0.123

<h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm , N/m²

V = volume, liter  

n = number of moles  

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m2, v= m³)

T = temperature, Kelvin  

Volume(V) =2.5 L

Pressure(P) = 1.2 atm

Temperature(T) = 25 + 273=298 K

\tt n=\dfrac{PV}{RT}\\\\n=\dfrac{1.2\times 2.5}{0.082\times 298}\\\\n=0.123~moles

7 0
3 years ago
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Alex Ar [27]
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3 0
3 years ago
PLEASE HELP ASAP!!!
mamaluj [8]

Answer:The approximate atomic mass of lead is 207.24 amu.

Explanation:

Abundance of isotope (I) = 1.4% = 0.014

Atomic mass of the Isotope (I) = number protons +  number of neutrons = 82 + 122 = 204 amu

Abundance of isotope (II) = 22.1 % = 0.221

Atomic mass of the Isotope (II) = number protons +  number of neutrons = 82 + 125 = 207 amu

Abundance of isotope (III) = 24.1% = 0.241

Atomic mass of the Isotope (III) = number protons +  number of neutrons = 82 + 124 = 206 amu

Abundance of isotope (IV) = 52.4% = 0.524

Atomic mass of the Isotope (IV) = number protons +  number of neutrons = 82 + 126 = 208 amu

Average atomic mass of an element =

\sum(\text{atomic mass of an isotopes}\times (\text{fractional abundance)})

Average atomic mass of lead =

(204 amu\times 0.014+207 amu\times 0.221 amu+206\times 0.241+208 amu\times 0.524)=207.24 amu

The approximate atomic mass of lead is 207.24 amu.

7 0
3 years ago
Give 2 reason why cryolite is used
Firdavs [7]
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7 0
2 years ago
A sample of chlorine gas has a density of g\L At a pressure of 1.04 for ATM and a temperature of 34°C. Assume Ideal behavior
madam [21]

Answer:

2.93 g/L

Explanation:

We can use the <em>Ideal Gas Law </em>to calculate the density of chlorine.

   pV = nRT

      n = m/M                                  Substitute for n

   pV = (m/M)RT                            Multiply both sides by M

pVM = mRT                                   Divide both sides by V

  pM = (m/V) RT

     ρ = m/V                                   Substitute for m/V

 pM = ρRT                                    Divide each side by RT

    ρ = (pM)/(RT)

===============

p = 1.04 atm

M = 70.91 g/mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 34 °C = 307.15 K                     Solve for ρ

===============

ρ = (1.04 × 70.91)/(0.082 06 × 307.15)

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ρ = 2.93 g/L

6 0
3 years ago
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