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vichka [17]
3 years ago
11

According to the unbalanced equation: KCIO3 → KCI + O2 how many

Chemistry
1 answer:
Finger [1]3 years ago
5 0

Answer:

06

Explanation:

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A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
V125BC [204]

Taking into account the definition of calorimetry, the specific heat of metal is 0.165 \frac{cal}{gC}.

<h3>Definition of calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

  • Q is the heat exchanged by a body of mass m.
  • C is the specific heat substance.
  • ΔT is the temperature variation.

<h3>Specific heat capacity of the metal</h3>

In this case, you know:

For metal:

  • Mass of metal = 50 g
  • Initial temperature of metal= 45 °C
  • Final temperature of metal= 11.08 ºC
  • Specific heat of metal= ?

For water:

  • Mass of water = 250 g
  • Initial temperature of water= 10 ºC
  • Final temperature of water= 11.08 ºC
  • Specific heat of water = 1.035 \frac{cal}{gC}

Replacing in the expression to calculate heat exchanges:

For metal: Qmetal= Specific heat of metal× 50 g× (11.08 C - 45 C)

For water: Qwater=  1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- Specific heat of metal× 50 g× (11.08 C - 45 C)= 1.035 \frac{cal}{gC} × 250 g× (11.08 C - 10 C)

Solving:

- Specific heat of metal× 50 g× (-33.92 C)= 1.035 \frac{cal}{gC} × 250 g× 1.08 C

Specific heat of metal× 1696 g×C= 279.45 cal

Specific heat of metal= \frac{279.45 cal}{1696 gC}

<u><em>Specific heat of metal= 0.165 </em></u>\frac{cal}{gC}

Finally, the specific heat of metal is 0.165 \frac{cal}{gC}.

Learn more about calorimetry:

brainly.com/question/11586486

brainly.com/question/24724338

brainly.com/question/14057615

brainly.com/question/24988785

#SPJ1

7 0
2 years ago
How do I find the slope of this graph?​ Is it even possible?
nasty-shy [4]

Answer:

Not exactly But you can take the slope of the curved portion and the slope of the flatline.

It wont do you much good since your working for absorbance but if you ever see something like a temperature change you can use the slope(s) to find freezing points/melting

Explanation:

If you need to submit a slope you could use a best fit which is just point to point or you could break it up like i mentioned

4 0
3 years ago
Which is an example of an organism? A. an octopus B. sand granules C. water D. the colors of a feather
denpristay [2]
A. An octopus

An organism is an individual animal, plant, or single-celled life form.

The octopus is the only living thing listed so that is why it is the example of an organism
5 0
3 years ago
Periods on the periodic table are
Roman55 [17]

Answer:

Periods on the periodic table of elements

Explanation:

In the periodic table of elements, there are seven horizontal rows of elements called periods. The vertical columns of elements are called groups, or families. (See also The Periodic Table: Metals, Nonmetals, and Metalloids.) In each period (horizontal row), the atomic numbers increase from left to right.

6 0
3 years ago
The change in entropy, Δ S ∘ rxn , is related to the the change in the number of moles of gas molecules, Δ n gas . Determine the
rewona [7]

Answer:

The entropy decreases.

Explanation:

The change in the standard entropy of a reaction (ΔS°rxn) is related to the change in the number of gaseous moles (Δngas), where

Δngas = n(gaseous products) - n(gaseous reactants)

  • If Δngas > 0, the entropy increases
  • If Δngas < 0, the entropy decreases.
  • If Δngas = 0, there is little or no change in the entropy.

Let's consider the following reaction.

2 H₂(g) + O₂(g) ⟶ 2 H₂O(l)

Δngas = 0 - 3 = -3, so the entropy decreases.

8 0
3 years ago
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