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DedPeter [7]
3 years ago
11

What evidence supports the law of conservation of energy?

Chemistry
1 answer:
Travka [436]3 years ago
3 0

Answer:

light energy is converted to chemical energy during photosynthesis.

Explanation:

Law of conservation of energy says: Energy can neither be created nor destroyed-only converted from one form of energy to another.

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a rock that originally had a mass of 1.00g of uranium_238 now only has .125g left. how old is the rock if the half life of urani
Romashka [77]

Answer:

no it would be 2.5 mill your welcome

Explanation:

6 0
3 years ago
One solution has a formula C (n) H (2n) O (n) If this material weighs 288 grams, dissolves in weight 90 grams, the solution will
saw5 [17]

Explanation:

The given data is as follows.

Boiling point of water (T^{o}_{b}) = 100^{o}C = (100 + 273) K = 323 K,

Boiling point of solution (T_{b}) = 101.24^{o}C = (101.24 + 273) K = 374.24 K

Hence, change in temperature will be calculated as follows.

              \Delta T_{b} = (T_{b} - T^{o}_{b})

                           = 374.24 K - 323 K

                           = 1.24 K

As molality is defined as the moles of solute present in kg of solvent.

            Molality = \frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}

Let molar mass of the solute is x grams.

Therefore,   Molality = \frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}

                        m = \frac{288 g \times 1000}{x g \times 90}              

                          = \frac{3200}{x}

As,    \Delta T_{b} = k_{b} \times molality

                 1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}

                       x = \frac{0.512 ^{o}C/m \times 3200}{1.24}

                          = 1321.29 g

This means that the molar mass of the given compound is 1321.29 g.

It is given that molecular formula is C_{n}H_{2n}O_{n}.

As, its empirical formula is CH_{2}O and mass is 30 g/mol. Hence, calculate the value of n as follows.

                n = \frac{\text{Molecular mass}}{\text{Empirical mass}}

                   = \frac{1321.29 g}{30 g/mol}

                   = 44 mol

Thus, we can conclude that the formula of given material is C_{44}H_{88}O_{44}.

4 0
3 years ago
During the nuclear fission of plutonium-244, barium-144 produced, along with an unknown nucleus and three neutrons. What must be
Marta_Voda [28]

Answer:

Explanation: In the previous section we listed four characteristics of radioactivity and nuclear decay that form the basis for the use of radioisotopes in the health and biological sciences. A fifth characteristic of nuclear reactions is that they release enormous amounts of energy. The first nuclear reactor to achieve controlled nuclear disintegration was built in the early 1940s by Enrico Fermi and his colleagues at the University of Chicago. Since that time, a great deal of effort and expense has gone into developing nuclear reactors as a source of energy. The nuclear reactions presently used or studied by the nuclear power industry fall into two categories: fission reactions and fusion reactions

6 0
2 years ago
1) The heat of combustion for the gases hydrogen, methane and ethane are −285.8, −890.4 and −1559.9 kJ/mol respectively at 298K.
Morgarella [4.7K]

Answer:

The enthalpy of the reaction is 64.9 kJ/mol.

Explanation:

H_2 + \frac{1}{2}O_2\rightarrow H_2O,\Delta H_1 =-285.8 kJ..[1]

CH_4 + 2O_2\rightarrow CO_2 + 2H_2O,\Delta H_2 =-890.4 kJ..[2]

C_2H_6 + \frac{7}{2}O_2\rightarrow 2CO_2 + 3H_2O,\Delta H_3= -1559.9 kJ..[3]

2CH_4(g)\rightarrow C_2H_6(g) + H_2(g),\Delta H_4=?..[4]

2 × [2] - [1]- [3] = [4]  (Using Hess's law)

\Delta H_4=2\times \Delta H_2 -\Delta H_1 -\Delta H_3

\Delta H_4=2\times (-890.4 kJ)-(-285.8 kJ) -(-1559.9 kJ)

\Delta H_4=64.9 kJ/mol

The enthalpy of the reaction is 64.9 kJ/mol.

3 0
3 years ago
Brainleist if correct<br> ___________is part of the Milky Way Galaxy.
lana [24]

Answer:

Our Sun (a star) and all the planets around it are part of a galaxy known as the Milky Way Galaxy.

4 0
3 years ago
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