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vampirchik [111]
3 years ago
9

Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ

Chemistry
1 answer:
Leokris [45]3 years ago
6 0
Given teh equation adn the heat of reaction, reaction 2's heat of reaction can be obtained by simply multiplying teh heat of reaction of 1 by 3. The final answer is -6129 kJ. 
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Guys please help me I have a final exam
Natasha2012 [34]

Explanation:

ok that is ans k............

4 0
3 years ago
How can you make a saturated solution from a supersaturated solution?
Lerok [7]

Answer:

warming up the solution.

Explanation:

  • A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent. ​The additional solute will not dissolve in a saturated solution.

  • Supersaturated solution is a solution that contains more of the dissolved material than could be dissolved by the solvent under normal circumstances.

  • A saturated solution can become supersaturated when it is cooled.

  • Also, we can form a saturated solution from a supersaturated solution via warming up the solution:

<em>The solubility of solid solutes in liquid solvents increases as the solvent is warmed up.</em>

7 0
3 years ago
What are at least 2 differences mitosis and meiosis?
attashe74 [19]

Mitosis has one round of cellular division and genetic separation whereas meiosis has two rounds. The two processes are also different because in mitosis the daughter cells are exactly identical to the parent cells compared to meiosis where the daughter cells are not genetically identical to the parent cells


Hope this helped

3 0
4 years ago
At 25◦C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by)2545(2121x
lions [1.4K]

The question is incomplete. The complete question is:

At 25◦C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by the equation:

ΔV = x1x2(45x1 + 25x2)

Where ΔV is  in cm3-mol-1. At these conditions, the molar volumes of pure liquid 1 and 2 are V1= 110 and V2= 90 cm3-mol-1. Determine the partial molar volumes 1VE and 2VE in a mixture containing 40 mole percent of specie 1.

Answer:

1VE = 117.92 cm³.mol⁻¹, 2VE = 97.92 cm³.mol⁻¹

Explanation:

In the equation given, x represents the molar fraction of each substance, thus x1 = 0.4 and x2 = 0.6. Because of the mixture, the molar partial volume of each substance will change by a same amount, which will be:

ΔV = 0.4*0.6(45*0.4+ 25*0.6)

ΔV = 7.92 cm³.mol⁻¹

1VE - V1 = 7.92

1VE = 7.92 + 110

1VE = 117.92 cm³.mol⁻¹

2VE - V2 = 7.92

2VE = 7.92 + 90

2VE = 97.92 cm³.mol⁻¹

3 0
3 years ago
How many mols of chlorine are in 120g of chlorine gas
xz_007 [3.2K]

Answer:

Number of moles of chlorine = 3.38 mol

Explanation:

Given data:

Mass of chlorine = 120 g

Moles of chlorine = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of chlorine = 35.5 g/mol

Now we will put the values in formula.

Number of moles = 120 g/ 35.5 g/mol

Number of moles = 3.38 mol

6 0
3 years ago
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