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Evgesh-ka [11]
2 years ago
14

In a calorimetry experiment 2.50 g of methane is burnt in excess oxygen. 30% of the

Chemistry
1 answer:
zzz [600]2 years ago
8 0

The total energy released per gram of methane in the experiment is 119941.33 J/g

<h3>How to determine the change in the temperature of water</h3>
  • Initial temperature of water (T₁) = 25 °C
  • Final temperature of water (T₂) = 68 °C
  • Change in temperature (ΔT) = ?

Change in temperature (ΔT) = T₂ – T₁

Change in temperature (ΔT) = 68 – 25

Change in temperature (ΔT) = 43 °C

<h3>How to determine the heat absorbed by the water</h3>

The absorbed by the water can be obtained as illustrated below:

  • Mass of water (M) = s00 g
  • Change in temperature (ΔT) = 43 °C
  • Specific heat capacity of the water (C) = 4.184 J/gºC
  • Heat (Q) =?

Q = MCΔT

Q = 500 × 4.184 × 43

Q = 89956 J

<h3>How to determine the energy released by methane in the experiment</h3>
  • Heat absorbed by water = 89956 J
  • Percentage of heat absorbed by water = 30%
  • Heat released by methane =?

Heat absorbed by water = 30% of heat released by methane

89956 = 30% × heat released by methane

89956 = 0.3 × heat released by methane

Divide both sides by 0.3

Heat released by methane = 89956 / 0.3

Heat released by methane = 299853.33 J

<h3>How to determine the heat released per gram of methane</h3>
  • Heat released by methane (Q) = 299853.33 J
  • Mass of methane (m) = 2.5 g
  • Heat per gram (ΔH) =?

Q = m × ΔH

Divide both sides by m

ΔH = Q / m

ΔH = 299853.33 / 2.5

ΔH =119941.33 J/g

Learn more about heat transfer:

brainly.com/question/6363778

#SPJ1

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Lab Report
Diano4ka-milaya [45]

Answer:

The purpose of the experiment is to see how water of different temperature and salinity affect the density.

Explanation:

Temperature and salinity directly affect the density of the water. Water of low temperature is more dense than water of high temperature, BUT, (fresh)water with no salt is less dense than (sea)water with more salt, so temperature and salinity change density of water.

3 0
2 years ago
Which equation represents the correct net ionic equation for the reaction between Ca(OH)2 and H2SO4?
Georgia [21]

<u>Answer:</u> The correct net ionic equation for the reaction is 2H^{+}(aq)+2OH^{-}(aq)\rightarrow 2H_2O(l)

<u>Explanation:</u>

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.  

The balanced molecular equation is:

Ca(OH)_2(aq)+H_2SO_4(aq)\rightarrow CaSO_4(aq)+2H_2O(l)

The complete ionic equation follows:

Ca^{2+}(aq)+2OH^-(aq)+2H^+(aq)+SO_4^{2-}(aq)\rightarrow 2Ca^{2+}(aq)+SO_4^{2-}(aq)+H_2O(l)

As calcium and sulfate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

2H^{+}(aq)+2OH^{-}(aq)\rightarrow 2H_2O(l)

Hence, the correct net ionic equation for the reaction is 2H^{+}(aq)+2OH^{-}(aq)\rightarrow 2H_2O(l)

7 0
3 years ago
what happened when potassium chloride is added to silver nitrate solutions . there no reaction B two soluble salts are formed C
Mashutka [201]

Answer:

D. One soluble and one insoluble salts are formed

Explanation:

8 0
3 years ago
The half-life of Pa-234 is 6.75 hours. If a sample of Pa-234 contains 112.0 g, 1 point
umka2103 [35]

Answer:

28g remain after 13.5 hours

Explanation:

Element decayment follows first order kinetics law:

ln[Pa-234] = -kt + ln [Pa-234]₀ <em>(1)</em>

<em>Where [Pa-234] is concentration after t time, k is rate constant in time, and [Pa-234]₀ is initial concentration</em>

Half-life formula is:

t_{1/2} =  \frac{ln2}{k}

6.75 = ln2 / k

<em>k = 0.1027hours⁻¹</em>

Using rate constant in (1):

ln[Pa-234] = -0.1027hours⁻¹×13.5hours + ln [112.0g]

ln[Pa-234] = 3.332

[Pa-234] = <em>28g after 13.5 hours</em>

<em />

5 0
3 years ago
Sixty-two months is equivalent to how many seconds
Olenka [21]

Answer: 163,060,000

Explanation:

6 0
3 years ago
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