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Dmitry [639]
4 years ago
7

An atom is the simplest substance known and provides the building block from which all matter is composed.

Chemistry
1 answer:
Westkost [7]4 years ago
4 0
It is true but jejdjdjenbebebrjrjjrbbr
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Is it A,B,C,D please explain why you choosed that?
Verdich [7]

Answer:

C.

Explanation:

The hand was touching the pan, which was hot, so the human reacted to it by putting the hand away from the pan.

4 0
3 years ago
Vincent combines sodium and chlorine in two different beakers (beaker X and beaker Y). In both beakers, the chemical reaction sh
Elza [17]

Answer: A. the chemical change will occur faster in beaker X.

Explanation:

Temperature is one of the factors that affect the rates of chemical reactions. Increase in temperature increases the rates of reaction by increasing the kinetic energy of the reacting particles so that energetic collisions occur and more bonds in the reactants will be broken and; atoms and ions recombine to form new compounds. Beaker X which is at room temperature has higher temperature than beaker Y which is kept in the refrigerator, thus reacting particles in beaker X has more kinetic energy than the ones in beakerA. the chemical change will occur faster in beaker X. Y.

8 0
4 years ago
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1 Evaporation

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7 0
3 years ago
What types of intermolecular forces are found in CH<br> 4
Sphinxa [80]
London dispersion forces
8 0
4 years ago
At 298 K the standard enthalpy of combustion of sucrose is -5645 kJ/mol and the standard reaction Gibbs energy is -5798 kJ/mol.
natka813 [3]

Explanation:

The given data is as follows.

             T = 298 K,          \Delta H^{o} = -5645 kJ/mol

          \Delta G^{o} = -5798 kJ/mol

Relation between \Delta H and \Delta G are as follows.

          \Delta G^{o} = \Delta H^{o} - T \Delta S^{o}    

             -5798 kJ/mol = -5645 kJ/mol - 298 \times \Delta S^{o}

                       -153 kJ/mol = -298 \times \Delta S^{o}

                    \Delta S^{o} = 0.513 kJ/mol K

Now, temperature is 37^{o}C = (37 + 273) K = 310 K

Since,        \Delta G = \Delta H^{o} - T \Delta S^{o}

                            = -5645 kJ/mol - 310 K \times 0.513 kJ/mol K

                            = (-5645 kJ/mol - 159.03 kJ/mol)

                            = -5804.03 kJ/mol

As, change in Gibb's free energy = maximum non-expansion work

            \Delta G = \Delta G_{310 K} - \Delta G_{298 K}

                           = -5804.03 kJ/mol - (-5798 kJ/mol)

                           = -6.03 kJ/mol

Therefore, we can conclude that the additional non-expansion work is -6.03 kJ/mol.

5 0
3 years ago
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