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nlexa [21]
2 years ago
5

Which of the following refers to a chemical property?. . a. At room temperature, mercury is a liquid, but gold is a solid.. . b.

Water will not burn, but gasoline is flammable.. . c. A diamond is harder than chalk.. . d. Butter melts at room temperature, but gelatin does not.
Chemistry
1 answer:
luda_lava [24]2 years ago
6 0
Out of the choices given, the answer which refers to a chemical property is going to be B. Water will not burn, but gasoline is flammable. The reason why is because burning and flammability are both chemical properties. 
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In two or more complete sentences, explain how to calculate the amount of energy that is transferred when 150 g of cooper cools
ella [17]

Answer:

-2,438 J energy is released

Heat energy is released

Explanation:

The formula for energy absorbed or released is Q = mT(Cp)

Get temperature in degrees T = 21 - 63  = -42°C

Mass of the cooling cooper m = 150g

Specific heat of copper Cp = 0.387J/g-°C

Put all the above parameters into the formula for energy absorbed or released

Q = mT(Cp)

    = 150 x -42(0.387)

    = -2438J  same as -2.438 KJ

-2438J energy is released

8 0
2 years ago
A certain compound with a molar mass of 120.0 g/mol crystallizes with the sodium chloride (rock salt) structure. The length of a
vlada-n [284]

The density of the compound is 4.64 g/cm^{3}.

Explanation:

Sodium chloride crystallize as FCC face centered cubic structure. It consist of four atoms of NaCl. So the unknown compound which crytallizes like NaCl wil have volume as:

radius in fcc is calculated as

\sqrt{2a} = 4r

putting the value where a is length of the edge.

\sqrt{2X461}

= 651.9 = 4r  so r= 162.98 pm

volume is calculated by the formula:

\frac{4}{3} πr^{3}

putting the values,

\frac{4}{3} x 3.14 x (1.62 X 10^{-8})^3

= 4.25X 10 ^-24

1.7 X 10^{-23} cm^{3} = volume of the fcc cell.

mass of the substance = 4 x 120 x 1.6605 x 10^(-24) g/u) (as it contains four atoms)

                                     = 7.9x 10^{-23}

density = \frac{mass}{volume}

putting the values

density = \frac{ 7.9x10^-23 }{1.7 x 10^{-23} }

density = 4.64 g/ cm^{3}

5 0
2 years ago
0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. A
elena-14-01-66 [18.8K]

Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = \frac{0.9975 g}{56 g/mol}=0.01781 mol

The enthlapy of solution :

\frac{Q'}{moles}

=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol

The enthlapy of solution is -55.23 kJ/mol.

8 0
2 years ago
when a volume of a gas is changed from __ L to 4.50 L, the temperature will change from 38.1 C to 15.0C
kow [346]
Assuming that the change of volumen was done at constant pressure and the quantity of gas did not change, you use Charles' Law of gases, which is valid for ideal gases:

V / T = constant => V1 / T1 = V2 / T2 => V1 = [V2 / T2] * T1.

Now plug in the numbers ,where T1 and T2 have to be in absolute scale.

T1 = 38.1 + 273.15 K = 311.25K

T2 = 15.0 + 273.15 K = 288.15K

V1 = 4.5L * 311.25K / 288.15 K = 4.86L.

Answer: 4.86
3 0
2 years ago
A 5.00 mL sample of H3PO4 solution of unknown concentration is titrated with a 0.1090 M NaOH solution. A volume of 7.12 mL of th
babunello [35]
The answer is that first you add 5.00+7.12=12.12
6 0
2 years ago
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