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gayaneshka [121]
3 years ago
9

The fuel used in many disposable lighters is liquid butane, C4H10

Chemistry
1 answer:
Charra [1.4K]3 years ago
6 0

Answer:

1.45 *10^23 atoms C

Explanation:

3.50 g butane * 1 mol butane/58.1 g butane =0.06024 mol butane

in 1 mol C4H10           -------- 4 mol C

in 0.06024 mol C4H10 -------- 4*0.6024 = 0.241 mol C

0.241 mol C * 6.02*10^23 atoms C/1 mol C = 1.45 *10^23 atoms C

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The answer is london dispersion forces. It is one of two intermolecular forces that is present in all molecules! 


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Only oppositely charged objects can attract each other. true false
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Which element could provide one atom to make an ionic bond with fluorine
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A cylinder of compressed gas has a volume of 350 ml and a pressure of 931 torr. What volume in Liters would the gas occupy if al
hodyreva [135]

Answer:

0.384\ \text{L}

Explanation:

P_1 = Initial pressure = 931 torr = 931\times \dfrac{101.325}{760}=124.12\ \text{kPa}

P_2 = Final pressure = 113 kPa

V_1 = Initial volume = 350 mL

V_2 = Final volume

From the Boyle's law we have

P_1V_1=P_2V_2\\\Rightarrow V_2=\dfrac{P_1V_1}{P_2}\\\Rightarrow V_2=\dfrac{124.12\times 350}{113}\\\Rightarrow V_2=384.44\ \text{mL}=0.384\ \text{L}

The volume the gas would occupy is 0.384\ \text{L}.

3 0
3 years ago
A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
Lostsunrise [7]

Answer:

THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

                     mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

5 0
3 years ago
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