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ehidna [41]
3 years ago
14

Please help me ..If you don't have a question or something to say just comment don't steal my points . It's not fair. Here are t

he questions:

Chemistry
2 answers:
astra-53 [7]3 years ago
3 0

1.

1.20 \: mol \times  \frac{6.02 \times  {10}^{23} }{1 \: mol}  = 7.224 \times  {10}^{23} atoms

2.

9.25 \times  {10}^{25 \:} molecules \times  \frac{1 \: mol}{6.02 \times  {10}^{23} }

= 153.654485 \: moles

3.

molar mass = 3(24.31) + 127.60 + 5(32.07)

molar mass = 360.88 g

4.

molar mass = 3(58.93) + 2(30.97) + 8(16.00)

molar mass = 366.73 g

5.

molar mass = 98.09

125 \: g \times  \frac{1 \: mol}{98.09 \: g}  = 1.274339892 \: moles

//

I'm too lazy to do the rest but you should get a general idea of how to do it.

Also, round as you wish!

madreJ [45]3 years ago
3 0

Ayo, thanks for my points

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A change that produces one or more ___________________ is a chemical change or chemical reaction.
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Answer:

A change that produces one or more new substances is a chemical change or chemical reaction.

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Which low-energy state of condensed matter is characterized by structural rigidity and resistance to changes of shape or volume?
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Solid is the answer. 

When matter is in its solid state, its molecules are more rigid and move less. they are also more resistant to changes in shape. Think of water in its solid form, ice. Ice is much harder to change the shape of than liquid water, and is much more rigid in its structure. 
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Which group of elements are the most reactive
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Answer:

The most reactive group of elements are found in group 1 which are alkali metals.

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4 years ago
In the phase diagram below, point B describes the critical point.<br><br><br><br> TRUE<br><br> FALSE
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4 0
3 years ago
Three of the reactions that occur when the paraffin of a candle (typical formula C21H44) burns are as follows:
Katarina [22]

Answer and Explanation:

The 3 reactions represented are

C₂₁H₄₄ + 32O₂ -----> 21CO₂ + 22H₂O

C₂₁H₄₄ + (43/2)O₂ -----> 21CO + 22H₂O

C₂₁H₄₄ + 11O₂ -----> 21C + 22H₂O

ΔH°(C₂₁H₄₄) = -476 KJ/mol, ΔH°(O₂) = 0KJ/mol, ΔH°(CO₂) = -393.5 KJ/mol, ΔH°(CO) = -99 KJ/mol, ΔH°(H₂O) = -292.74 KJ/mol, ΔH°(C) = 0KJ/mol

ΔH°f = ΔH°(products) - ΔH°(reactants)

For reaction 1,

ΔH°(products) = 21(ΔH°(CO₂)) + 22(ΔH°(H₂O)) = 21(-393.5) + 22(-292.74) = -14703.78 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -14703.78 - (-476) = - 14227.78 KJ/mol

For reaction 2,

ΔH°(products) = 21(ΔH°(CO)) + 22(ΔH°(H₂O)) = 21(-99) + 22(-292.74) = -8519.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -8519.28 - (-476) = - 8043.28 KJ/mol

For reaction 3,

ΔH°(products) = 21(ΔH°(C)) - 22(ΔH°(H₂O)) = 21(0) + 22(-292.74) = -6440.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -6440.28 - (-476) = - 5968.28 KJ/mol

b) To obtain the q for the combustion of 254g of paraffin, we convert the mass to moles.

Number of moles = mass/molar mass; molar mass of C₂₁H₄₄ = 296 g/mol

Number of moles = 254/296 = 0.858 mole

heat of reaction for the combustion of C₂₁H₄₄ when it is complete combustion, q = ΔH°(complete combustion, i.e. reaction 1) × number of moles = -14227.78 × 0.858 = -12207.435 KJ/mol

c) 8% of the mass of C₂₁H₄₄ undergoes incomplete combustion = 8% × 254 = 20.32g, in number of moles = 20.32/296 = 0.0686 mole

5% of the mass of C₂₁H₄₄ becomes soot = 5% × 254 = 12.7g, in number of moles = 12.7/296 = 0.0429 mole

The remaining paraffin undergoes complete combustion = 87% of 254 = 220.98g, in number of moles = 220.98 = 0.747 mole

q = sum of all the heat of reactions = (0.747 × -14227.78) + (0.0686 × -8043.28) + ( 0.0429 × -5968.28) = -11435.377 KJ

QED!!!

8 0
3 years ago
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