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s344n2d4d5 [400]

2 years ago

5

Calculate the ratio of the effusion ratesbetween chlorine gas (Cl2, molar mass = 71 g/mol) and iodine gas(I2, molar mass = 254 g

/mol).
a)0.280
b)1.89
c)3.58
d)8.43

1 answer:

inysia [295]2 years ago

5 0

Answer:

d is the correct answer

Explanation:

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Explain how synthesis and decomposition reactions can be reverse of one another

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Decompostion reaction have the general chemical reaction of: AB = A + B while synthesis reactions are of the general equation A + B = AB. These reaction can be the reverse of each other because ones building a compound (synthesis) and ones breaking down a compound (decomposition). For example, when reacting carbon and oxygen for a synthesis reaction we have: C + O2 = CO2 and for a decomposition reaction we have: CO2 = C + O2. Thus, these two reactions are the reverse of each other.

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A certain first-order reaction (a→products) has a rate constant of 9.60×10−3 s−1 at 45 ∘c. how many minutes does it take for the

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The integrated rate law expression for a first order reaction is

$ln\frac{[A_{0}]}{[A_{t}]}=kt$

where

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[At]=6.25

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k = 9.60X10⁻³s⁻¹

Putting values

$ln\frac{100}{6.25}=9.6X10^{-3}t$

taking log of 100/6.25

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Which of the following could be one reason why the earth is getting warmer?

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2 years ago

A 37.2 g sample of copper at 99.8 °C is carefully placed into an insulated container containing 188 g of water at 18.5 °C. Calcu

klasskru [66]

Answer:

T₂ = 19.95°C

Explanation:

From the law of conservation of energy:

$Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w$

where,

mc = mass of copper = 37.2 g

Cc = specific heat of copper = 0.385 J/g.°C

mw = mass of water = 188 g

Cw = specific heat of water = 4.184 J/g.°C

ΔTc = Change in temperature of copper = 99.8°C - T₂

ΔTw = Change in temperature of water = T₂ - 18.5°C

T₂ = Final Temperature at Equilibrium = ?

Therefore,

$(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}$

<u>T₂ = 19.95°C</u>

6 0

2 years ago