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2 years ago

Is spilling cake batter a physical or a chemical change?Explain why?

2 answers:

5 0

Physical because even thought the cake batter has been spilt it is still cake batter, its just on the floor now.Nothing has changed the chemical composition of it.

Ket [755]2 years ago

5 0

pyshical ye ey eye eyde8uiqe wdcjoewkje wjenckjn2 kj2 kjdn2k w

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Aluminum oxide (Al2O3) is a widely used industrial abrasive (known as emery or corundum), for which the specific application dep

ivanzaharov [21]

Answer:

On the Moh's scale of hardness, aluminum oxide is positioned just below to diamond due to which it is considered as one of the hardest known compounds. This also shows that the compound exhibit an enormous amount of lattice energy, as to transform the oxide into its constituent ions, the energy is required to overcome.

Based on the chemical formula of the compound, that is, Al2O3, it is shown that the ions of Al3+ and O2- are kept close due to the activity of the strong electrostatic ionic bonds. The electrostatic forces and the ionic bonding between the ions are extremely robust due to the presence of the ions high charge density. Therefore, to dissociate the bonds, an enormous amount of energy is needed, and at the same time, a high amount of lattice energy is present.

7 0

3 years ago

How does the atmosphere make conditions on earth suitable for living things

hram777 [196]

The atmosphere provides C02 for us, it protects us from the Sun's UV rays, it protects us from many objects in space that could crash into earth without it, and it holds the moisture in the air.

8 0

2 years ago

Be sure to answer all parts. Consider the following balanced redox reaction (do not include state of matter in your answers): 2C

timurjin [86]

Answer:

The specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$2CrO_2^- + 6ClO^- + 2H_2O\rightarrow 2CrO_4^{2-} + 3Cl_2 + 4OH^-$

The half cell reactions for the above reaction follows:

Oxidation half reaction: $CrO_2^- + 2H_2O + 4OH^-\rightarrow CrO_4^{2-} + 4H_2O + 3e^-$

Reduction half reaction: $2ClO^- + 4H_2O + 2e^-\rightarrow Cl_2 + 2H_2O + 4OH^-$

Thus, the specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

8 0

3 years ago

Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

Answer:

For a: The total heat required is 36621.5 J

For b: The total heat required is 58944.5 J

Explanation:

For a:

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

For process 2:

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

For b:

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K$

Putting values in equation 1, we get:

$q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J$

For process 2:

We are given:

$m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J$

For process 3:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K$

Putting values in equation 1, we get:

$q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J$

For process 4:

We are given:

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J$

Total heat required = $[q_1+q_2+q_3+q_4]$

Total heat required = $[448.14+4583.5+21445.2+32467.7]J=58944.5J$

Hence, the total heat required is 58944.5 J

8 0

3 years ago

What does the root “thermal” mean? Please Help I will give you Brainliest

Nezavi [6.7K]

Answer:

If it has to do with heat, it's thermal. ... The Greek word therme, meaning “heat,” is the origin of the adjective thermal. Something that is thermal is hot, retains heat, or has a warming effect.

Explanation:

hope it helps :)

7 0

2 years ago