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Andreas93 [3]
3 years ago
13

An example of a double replacement chemical reaction takes place in batteries. In an alkaline battery, Manganese oxide reacts wi

th water to form a different manganese oxide and hydroxide. Which is the correct balanced equation for this?
Chemistry
1 answer:
ozzi3 years ago
6 0

Answer:

2MnO2 + H2O => Mn2O3 + 2OH is the correct balanced equation.

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What happens if I overshoot the endpoint in titration?
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In the periodic table hydrogen is placed in Group 1A (group 1) and helium is placed in Group 8A (group 18). The most likely reas
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This question appears incomplete

Explanation:

This question appears incomplete because of the absence of options. However, hydrogen is placed in group 1 because it has just one electron in it's outermost shell (which happens to be the only shell it has) just like every other group 1A/group 1 element. While helium is placed in group 8A/group 18 because it has a completely filled outermost shell (which is also the only shell it has) just like every other element in group 8A/group 18.

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What does the pH of a solution depend on?
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Read 2 more answers
Calculate the amount of heat in kJ that is required to heat 25.0 g of ice from -25 °C to 105 °C in a closed vessel and sketch a
kolezko [41]

Answer:

The total amount of heat required for the process is 76.86 KJ

Explanation:

We can divide the process in 5 parts, in which we can calcule each amount of heat required (see attached Heating curve):

(1) Ice is heated from -25ºC to 0ºC. We can calculate the heat of this part of the process as follows. Note that we must convert J in KJ (1 KJ= 1000 J).

Heat (1) = mass ice x Specific heat ice x (Final temperature - Initial Temperature)

Heat (1) =25 g x 2.11 J/g.ºC x \frac{1 KJ}{1000 J} x (0ºC-(-25º)

Heat (1) = 1.32 KJ

(2) Ice melts at ºC (it becomes liquid water). This is heating at constant temperature (ºC), so we use the melting enthalphy (ΔHmelt) and we must use the molecular weight of water (1 mol H₂O = 18 g):

Heat (2) = mass ice x ΔHmelt

Heat (2)= 25 g  x  \frac{6.01KJ} {1 mol H2O} x \frac{1 mol H2O}{18 g}

Heat (2)= 8.35 KJ

(3) Liquid water is heated from 0ºC to 100 ºC:

Heat (3)= mass liquid water x Specific heat water x (Final T - Initial T)

Heat (3)= 25 g x 4.18 J/gºC x 1 KJ/1000 J x (100ºC - 0ºC)

Heat (3)= 10.45 KJ

(4) Liquid water evaporates at 100ºC (it becomes water vapor). This is a process at constant temperature (100ºC), and we use boiling enthalpy:

Heat (4)= mass water x ΔH boiling

Heat (4)= 25 g x \frac{40.67 KJ}{mol H20} x \frac{1 mol H20}{18 g}

Heat (4)= 56.49 KJ

(5) Water vapor is heated from 100ºC to 105ºC. We use the specific capacity of water vapor:

Heat (5)= mass water vapor x Specific capacity vapor x (Final T - Initial T)

Heat (5)= 25 g x 2.00 J/g ºC x 1 KJ/1000 J x (105ºC - 100ºC)

Heat (5)= 0.25 KJ

Finally, we calculate the total heat involved in the overall process:

Total heat= Heat(1) + (Heat(2) + Heat(3) + Heat(4) + Heat(5)

Total heat= 1.32 KJ + 8.35 KJ + 10.45 KJ + 56.49 KJ + 0.25 KJ

Total heat= 76.86 KJ

3 0
4 years ago
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