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REY [17]
2 years ago
13

Help bc i need it lol

Chemistry
1 answer:
nexus9112 [7]2 years ago
6 0

Answer:

<u>2</u> XE + <u>1</u> AB. --> <u>1</u> AE2 + <u>2</u> XB

Above equation is now balanced

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The reaction (CH3)3CBr + OH- (CH3)3COH + Br- in a certain solvent is first order with respect to (CH3)3CBr and zero order with r
son4ous [18]

Answer and Explanation:

The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation

R = molar gas constant

K = A(e^(-Ea/RT))

Taking natural log of both sides

In K = In A - (Ea/RT)

In K = (-Ea/R)(1/T) + In A

Comparing this to the equation of a straight line; y = mx + c

y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A

a) From the question, m = (-Ea/R) = -1.10 × (10^4) K

(-Ea/R) = -1.10 × (10^4) = -11000

R = 8.314 J/K.mol

Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol

b) c = In A = 33.5

A = e^33.5 = (3.54 × (10^14))/s

c) K = A(e^(-Ea/RT))

A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol

K = (3.54 × (10^14))(e^(-91454/(8.314×298.15))) = 0.0336/s

QED!

5 0
3 years ago
Ik i already asked this but i need diff point of views
salantis [7]

Answer:

Is it prescribe to you?If so than yes if not then no need to

Explanation:

5 0
2 years ago
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Determine how many carbon dioxide (CO2) molecules are produced if 8.45 x 1023 molecules of water (H2O) are produced 2 C2H6 (g) +
Andreas93 [3]
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The temperature of a substance increases when the ____________________ energy of the substance increases.
kykrilka [37]
Molecular

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3 years ago
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A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids.
Sergio039 [100]

Explanation:

Let us take the volume of block is x.

Since, the block is floating this means that it is in equilibrium. Formula to calculate net force will be as follows.

                F_{net} = Buoyancy force(F_{b}) - weight force(w)

Also, buoyancy force (F_{b}) = (volume submerged in water × density of water) + (volume in oil × density of oil)

          (F_{b}) = (0.592 V \times \rho) + (1 - 0.592)V \times 1000 g          

                      = (0.592 V \times \rho + 408 V) g

As,   W = V × density of graphite × g

It is given that density of graphite is 2.16 g/cm^{3} or 2160 kg/m^{3}.

So, W = 2160 V g

F_{net} = (0.592 V \rho + 408 V) g - 2160 V g = 0

            0.592 \rho = 1752

     \rho = 2959.46 kg/m^{3} or 2.959 g/cm^{3} is the density of oil.

It is given that mass of flask is 124.8 g.

Mass of 35.3 cm^{3} oil = 35.3 \times 2.959 104.7 g

Hence, in second weighing total mass will be calculated as follows.

                       (124.8 + 104.7) g

                       = 229.27 g

Thus, we can conclude that in the second weighing mass is 229.27 g.

5 0
2 years ago
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