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SVETLANKA909090 [29]
3 years ago
5

What type of map should have a north arrow, key ,and map scales?

Chemistry
1 answer:
KengaRu [80]3 years ago
7 0
Your answer would be C


hope this helped
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Elements and compounds are usually more dense in their solid than in their
sladkih [1.3K]

Explanation:

The molecules in water is more tightly packed than in ice, so water has greater density than ice.

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Which of the following would most likely be spontaneous?
vfiekz [6]
Answer : B because if it requires more energy, it might be a bigger reaction than normal, which could be defined as spontaneous
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Consider the following reaction representing the combustion of propane: ????????3HH8 + O2 → CO2 + H2O a. Balance the equation b.
emmainna [20.7K]

Answer:

a) C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

b) You need 5 moles of O₂ per mole of propane.

c) 181 g of O₂

d) 254 L of O₂ and 1210 L of air

e) 152 L of carbon dioxide

f) The gross heat released is 2354 kJ

Explanation:

C₃H₈ + O₂ → CO₂ + H₂O

a) To balance a combustion reaction you must add CO₂ as carbons of hydrocarbons you have. Then, you should add waters as half of hydrogens of the hydrocarbon and, in the last, balance oxygen with O₂, thus:

C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

10 oxygens, so you sholud add 5 O₂:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

b) The equation balanced says that you need 5 moles of O₂ per mole of propane.

c) To burn 100g of propane you need:

100 g C₃H₈×\frac{1 mol}{44,1g}×\frac{5 molO_2}{1mol C_{3}H_{8}}×\frac{16g}{1mol O_2}= 181 g of O₂

d) 181g of O₂ are 11,34 moles. The volume you require is:

V =nRT/P

where:

n are moles (11,34 moles)

R is gas constant (0,082 atmL/molK)

T is temperature (273 K at STP)

P is pressure (1 atm at STP)

V is 254 L of oxygen.

The liters of air are:

254L O₂ ₓ \frac{100 air}{21 O_2} = 1210 L of air

e) The volume of CO₂ produced is:

100 g C₃H₈×\frac{1 mol}{44,1g}×\frac{3 molCO_2}{1mol C_{3}H_{8}}= 6,80 moles of CO₂

V =nRT/P

where:

n are moles (6,80 moles)

R is gas constant (0,082 atmL/molK)

T is temperature (273 K at STP)

P is pressure (1 atm at STP)

V is 152 L of carbon dioxide.

f) C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O ΔH = -103,8 kJ/mol

1 kg of C₃H₈ are:

1000 g × \frac{1mol}{44,1 g} = 22,68 moles

Thus, the gross heat released is:

103,8 kJ/mol × 22,68 moles = 2354 kJ

I hope it helps!

4 0
3 years ago
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Which of the following statements is not true?
andreev551 [17]

Answer:

I think option (D)is not true

3 0
2 years ago
For chemistry please help pic is on link
chubhunter [2.5K]

Answer:

sheeesh That alot

Explanation:

6 0
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