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2 years ago

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A reaction produces a gas, which is collected in a glass cylinder. When a glowing splint is placed inside the container, it burn

s brightly, as shown. Which gas was produced in the reaction?

1 answer:

Gnesinka [82]2 years ago

5 0

Answer:

true

Explanation:

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Which statement about the electron-cloud model is true? It is the currently accepted atomic model. It can easily be replaced by

yan [13]

Answer:

3 choice

Explanation:

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1 year ago

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When a complex ion forms, ______________ arrange themselves around _________, creating a new ion with a charge equal to the sum

alisha [4.7K]

Answer:

lignands, the central atom/metal ion

Explanation:

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1 year ago

The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

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1 year ago

During this reaction, p4 + 5o2 → p4o10, 0.800 moles of product was made in 15.0 seconds. what is the rate of reaction? 56.8 g/mi

Eduardwww [97]

The solution for this problem would be:
The mass of P4O10 is computed by: 0.800 mol x 284 g/mol = 227g t = 15.0 s ( 1 min / 60 s) = 0.25 min
So solving for the rate will be mass over t = m/t = 227/0.25 = 908 g/min would be the answer for this problem.

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1 year ago

Uranium–232 has a half–life of 68.9 years. A sample from 206.7 years ago contains 1.40 g of uranium–232. How much uranium was or

givi [52]

<u>Answer:</u> The initial amount of Uranium-232 present is 11.3 grams.

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

We are given:

$t_{1/2}=68.9yrs$

Putting values in above equation, we get:

$k=\frac{0.693}{68.9}=0.0101yr^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $0.0101yr^{-1}$

t = time taken for decay process = 206.7 yrs

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 1.40 g

Putting values in above equation, we get:

$0.0101yr^{-1}=\frac{2.303}{206.7yrs}\log\frac{[A_o]}{1.40}$

$[A_o]=11.3g$

Hence, the initial amount of Uranium-232 present is 11.3 grams.

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1 year ago

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