I need alot of help.

An electric range burner weighing 699.0 grams is turned off after reaching a temperature of 482.0°C, and is allowed to cool down

jasenka [17]

Answer:

0.42 J/gºC

Explanation:

We'll begin by calculating the heat energy used to heat up the water. This can be obtained as follow:

Mass (M) of water = 560 g

Initial temperature (T₁) = 22.7 °C

Final temperature (T₂) = 80.3 °C.

Specific heat capacity (C) of water = 4.18 J/gºC

Heat (Q) absorbed =?

Q = MC(T₂ – T₁)

Q = 560 × 4.18 (80.3 – 22.7)

Q = 2340.8 × 57.6

Q = 134830.08 J

Finally, we shall determine the specific heat capacity of the burner. This can be obtained as follow:

Mass (M) of burner = 699 g

Initial temperature (T₁) = 482.0°C

Final temperature (T₂) = 22.7 °C

Heat (Q) evolved = – 134830.08 J

Specific heat capacity (C) of the burner =?

Q = MC(T₂ – T₁)

–134830.08 = 699 × C (22.7 – 482.0)

–134830.08 = 699 × C × –459.3

–134830.08 = –321050.7 × C

Divide both side by –321050.7

C = –134830.08 / –321050.7

C = 0.42 J/gºC

Therefore, the specific heat capacity of the burner is 0.42 J/gºC

8 0

3 years ago

1)

Delicious77 [7]

Answer:

C) How much energy was added to the substance to increase molecule motion?

Explanation:

The most relevant question to ask regarding this change must take into account the physical knowledge about matter.

When matter changes from liquid state to gaseous state, a physical change called evaporation, the particles (molecules or atoms) of the pure substance will separate from each other, take up more space and move faster.

Condensation is the opposite to evaporation, thus the option A) is not the most relevant question.

The charge of the particles does not change; so the option B) is not relevant at all.

The particles should gain energy from the surroundings to increase their motion (kinetic energy) when they pass from liquid state, where they move slower, to gas state, where they move faster. Hence, the option C), How much energy was added to the substance to increase molecule motion? , is totally relevant.

Since this is an increase in the kinetic energy of the molecules, the option D) is not relevant.

6 0

3 years ago

Select all of the following equations that represent an exothermic reaction.

masya89 [10]

Answer:

C(graphite) → C(diamond), ΔH = - 0.45 kcal

CH4 + 2O2 → CO2 + 2H2O + 212,800 cal

Explanation:

C(graphite) → C(diamond), ΔH = - 0.45 kcal

CH4 + 2O2 → CO2 + 2H2O + 212,800 cal

These reactions are exothermic reaction because heat is evolved.

The energy changes occur during the bonds formation and bonds breaking.

There are two types of reaction endothermic and exothermic reaction.

Endothermic reactions:

The type of reactions in which energy is absorbed are called endothermic reactions.

In this type of reaction energy needed to break the bond are higher than the energy released during bond formation.

For example:

C + H₂O → CO + H₂

ΔH = +131 kj/mol

it can be written as,

C + H₂O + 131 kj/mol → CO + H₂

Exothermic reaction:

The type of reactions in which energy is released are called exothermic reactions.

In this type of reaction energy needed to break the bonds are less than the energy released during the bond formation.

For example:

Chemical equation:

C + O₂ → CO₂

ΔH = -393 Kj/mol

it can be written as,

C + O₂ → CO₂ + 393 Kj/mol

6 0

3 years ago

What is bedrock layer

Nuetrik [128]

Bed rock is a layer underneath minerals

6 0

3 years ago

Calculate and explain (in words) how you would make 90.0 mL of 2.0 M MgSO4 solution from a solid solute MgSO4.

lozanna [386]

Answer:

See Below.

Explanation:

Recall that molarity is defined by moles of solute over liters of solution (mol/L).

Therefore, to make 90.0 mL of 2.0 M MgSO₄, we will need:

$\displaystyle 90.0\text{ mL} \cdot \frac{2.0\text{ mol MgSO$_4$}}{1\text{ L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} = 0.18\text{ mol MgSO$_4$}$

Convert this amount to grams. The molecular weight of MgSO₄ is 120.38 g/mol:

$\displaystyle 0.18\text{ mol MgSO$_4$} \cdot \frac{120.38\text{ g MgSO$_4$}}{1\text{ mol MgSO$_4$}} = 22\text{ g MgSO$_4$}$

Therefore, to make the solution, we can add 22 grams of MgSO₄ into a graduated cylinder, then mix and dilute the solution with distilled water until we reach 90.0 mL.

5 0

2 years ago

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