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Nastasia [14]
2 years ago
12

1 mole of no2(g) has a greater entropy than 1 mole of n2o4(g). true or false

Chemistry
1 answer:
bija089 [108]2 years ago
7 0
In order to answer this, you need to find the empirical data for the standard entropies. Please refer to this link: http://www.mrbigler.com/misc/energy-of-formation.PDF

For NO₂ gas, the entropy is 240 J/mol-K. For N₂O₄ gas, the entropy is 304.2 J/mol-K. Therefore, <em>the statement is false.</em>
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4.45 kcal of heat was added to increase the temperature of a sample of water from 23.0 °C to 57.8 °C. Calculate
Alona [7]

Answer:

m = 4450 g

Explanation:

Given data:

Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)

Initial temperature = 23.0°C

Final temperature = 57.8°C

Specific heat capacity of water = 1 cal/g.°C

Mass of water in gram = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 57.8°C - 23.0°C

ΔT = 34.8°C

4450 cal = m × 1 cal/g.°C × 34.8°C

m = 4450 cal / 1 cal/g

m = 4450 g

4 0
3 years ago
Carbon 14 decays to Carbon 12 and has a half-life of 5730 years. If a fossil is analysed and it has 5 grams of Carbon 14 and 15
Katen [24]

Answer:

im sorry but this question doesnt make sense

Explanation:

7 0
2 years ago
For a tablet containing 500. mg of vitamin C, calculate how many ml of 0.095 M NaOH is required to reach the equivalence point.
Gala2k [10]

Answer:

mL of NaOH required =29.9mL

Explanation:

Let us calculate the moles of vitamin C in the tablet:

The molar mass of Vitamin C is 176.14 g/mole

moles=\frac{mass}{molarmass}=\frac{500mg}{176.14}=\frac{0.5}{176.14}=0.0028

Thus we need same number of moles of NaOH to reach the equivalence point.

For NaOH solution:

moles=MolarityXvolume=0.095Xvolume

0.00283=0.095Xvolume

volume=0.0299L=29.9mL

7 0
3 years ago
A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the
Masteriza [31]

Answer:- C. Hafnium.

Solution:- Mass of the sample is 46.0 g and it's volume is 3.5cm^3 .

From mass and volume, we can calculate it's density using the formula:

density=\frac{mass}{volume}

density=\frac{46.9g}{3.5cm^3}

density=\frac{13.4g}{cm^3}

On the basis of the density, this substance could either be mercury or hafnium. Since the substance is a solid at room temperature where as mercury is liquid. So, it can't be mercury.

The right choice is C) Hafnium.

3 0
3 years ago
Consider the half reaction below.
Radda [10]
Answer: the second option: <span>Iron is being oxidized
</span>

Explanation:

1) Oxidation is the increase of the oxidation state (number) due to the loss of electrons.

2) In the given reaction, you can see that in the left side the atom is Fe. 

When an element (atom) is not combined (or combined with it self) its oxidation state is 0. 

3) In the right side of the given equation you that iron is now in form of cation with charge 2+: Fe²⁺.

That means that the new oxidation state of the element is 2+.

4) This change in the oxidation state, of course, is accompanied by the loss of the two electrons: 2e⁻.

5) Conclusion: the iron has oxidized by losing two electrons and increasing its oxidation state from 0 to 2+.
<span></span>
7 0
3 years ago
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