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Triss [41]
3 years ago
5

What amount of ice must be added to 250 g of water at 15°c to cool the water to 0°c and have no ice left

Chemistry
1 answer:
Juliette [100K]3 years ago
7 0

Answer:

\mathbf{m= 47 \ g}

Explanation:

The heat needed to melt ice (q_1) + heat needed to cool water (q_2) = 0

m\Delta H_{fusion} + mc \Delta T = 0

m \times 333.55 J g^{-1} + 250 \ g \times 4.184 \ Jg^{-1 \ 0 } \times (-15.0 \ ^0 C) = 0

m \times 333.55 J g^{-1} =15690

m=\dfrac{15690 \ J }{333.55 J g^{-1} }

\mathbf{m= 47 \ g}

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A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its
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In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
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Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
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2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0
3 years ago
The last element in any period always has:
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Answer: The correct option is The properties of a noble gas.

Explanation: There are 7 periods in the periodic table.

The last element of each period are Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn) and Ununoctium (Uuo).

  • The electronic configuration for Helium is 1s^2. For He, The outermost electrons are 2.
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All these elements have stable electronic configuration and are not reactive in nature. Hence, they are considered as noble gases.

Therefore, the last element of each period always have the properties of a noble gas.

4 0
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3 0
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ra1l [238]

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8 0
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Explanation:

Here I send you all the 3 elements that you are asking for.

Notice that first on the left you will find lewis structure, then condensed and finally chemical formula of each of the compound you enlisted.

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