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MariettaO [177]
3 years ago
15

The melting point of table salt is 801 degrees celsius. is this a physical or chemical property?

Chemistry
1 answer:
Valentin [98]3 years ago
5 0
It is a Physical property because like water it changes into a ice cube but it can be melted into water or be turned back into an ice cube.
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What is the density of a solid that has a mass of 9.2 grams and a volume of 12.3 cm3 ? Would this object float or sink in water?
cluponka [151]
  • Mass=9.2g
  • Volume=12.3

\boxed{\sf Density=\dfrac{Mass}{Volume}}

\\ \sf\longmapsto Density=\dfrac{9.2}{12.3}

\\ \sf\longmapsto Density=0.74g/cm^3

  • Density of water =1g/cm^3

It will float in water

7 0
3 years ago
What aspect of metallic bonding is responsible for malleability ductility and conductivity of metals?
ANTONII [103]
Bonds are forces of attractions between atoms formed by the transfer of electrons or sharing of electrons. Metallic bond is a type bond that exist in metallic structures where the atoms of the metals attracts the sea of electrons in the structure.It is these metallic bonds that results to the malleability , ductility and conductivity of metals because in that the sea of electrons makes them conduct electricity. In addition the atoms of metals in the structure are ions which can slide past each other in the sea of electrons.
4 0
3 years ago
PLEASE HELP I WILL GIVE BRAINLIEST
stellarik [79]

Answer: YES!

Explanation:

Ox2 Co2 Hydrogen and Corossion 3

3 0
3 years ago
For a trip to the Moon, a rocket must lift off from
VLD [36.1K]

Answer:

20 m/s^2

Explanation:

given,

final velocity (v) = 6000m/s

initial velocity (u) = 0m/s

time taken (t) = 5 minutes

= 5×60second

= 300second

acceleration(a) = ?

we know that,

a = (v-u)/t

= (6000-0)/300

= 20 m/s^2

5 0
3 years ago
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
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