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2 years ago

When an electron moves from a lower energy level to a higher energy level, the atom:

Chemistry

1 answer:

Ede4ka [16]2 years ago

5 0

Answer:

A. absorbs light

Explanation:

For an electron to move from a lower energy level to a higher energy level, energy needs to be added to the atom.

In most cases this required energy will be light: The electron of the atom will absorb the energy of the light and move to a higher energy level.

When said electron comes back to the lower energy level, the atom will conversely give off light.

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What comes last in the scientific method

azamat

Hypothesis

Therefore the last step in the scientific method is proposing a hypothesis or obtaining a conclusion

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3 years ago

Read 2 more answers

1. How much heat (in calories) is needed to raise 20 g of H2O from 5°C to 40°C? (c = 1.0 cal/g °C)

KIM [24]

Answer:

700 calories

Explanation:

Using the formula below:

Q = m × c × ∆T

Where;

Q = amount of heat required (calories)

m = mass of substance (g)

c = specific heat of substance (cal/g°C)

∆T = change in temperature (°C)

According to this question, the following information was provided;

Q = ?

m = 20g

c = 1.0 cal/g °C

∆T = 40°C - 5°C = 35°C

Using the formula; Q = m × c × ∆T

Q = 20 × 1 × 35

Q = 700 calories

Hence, 700 cal of heat energy is needed to raise 20 g of H2O from 5°C to 40°C.

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2 years ago

In which type of chemical reaction are electrons transferred?

kupik [55]

The correct answer is 2.

In oxidation-reduction reactions (or redox reactions), a number of electrons are transferred between two species. During this process, the oxidation number of an atom (or molecule or ion) changes by receiving or losing an electron.

7 0

3 years ago

Read 2 more answers

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t

Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

For process 1 :

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

For process 2 :

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

5 0

3 years ago