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3 years ago

Container A holds 5.0 L of nitrogen gas. Container B holds 5.0 L of carbon dioxide gas. Both containers are at STP.

Chemistry

1 answer:

jeyben [28]3 years ago

3 0

Answer:7.5 gram of a gas occupies 5.6 litres as STP. The gas is (a) CO B)NO C)CO2 D)N,O

Explanation:

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What kind of simple machine is a shovel

Minchanka [31]

I think a shovel is a wedge simple machine.

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3 years ago

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1.40 m3 is how many mL

Zarrin [17]

$\LARGE{ \boxed{ \rm{ \pink{Solution:}}}}$

We know, 1 m³ of space can hold 1000 l of the substance.

⇛ 1 m³ = 1000 l----(1)

And, 1 l is 1000 times more than 1 ml

⇛ 1 l = 1000 ml------(2)

So, From (1) and (2),

⇛ 1 m³ = 1000 × 1000 ml

⇛ 1m³ = 1000000 ml

We had to find,

⇛ 1.40 m³ = 1.40 × 1000000 ml

⇛ 1.40 m³ = 140/100 × 1000000 ml

⇛ 1.40 m³ = 1400000 ml

⇛ 1.40 m³ = 14,00,000 ml / 14 × 10⁵ ml / 1.4 × 10⁶ ml

☃️ So, 1.40 m³ = 14 × 10⁵ ml / 1.4 × 10⁶ ml.

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3 years ago

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The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mo

aleksandrvk [35]

Explanation:

Relation between temperature and activation energy according to Arrhenius equation is as follows.

k = $A exp^{\frac{-E_{a}}{RT}}$

where, k = rate constant

A = pre-exponential factor

$E_{a}$ = activation energy

R = gas constant

T = temperature in kelvin

Also,

$ln (\frac{k_{2}}{k_{1}}) = (\frac{-E_{a}}{R}) \times (\frac{1}{T_{2}} - \frac{1}{T_{2}})$

$T_{1}$ = $244^{o}C$ = (244 + 273) K = 517.15 K

$T_{2}$ = $324^{o}C$ = (597.15 + 273) K = 597.15 K

$k_{1}$ = 6.7 $M^{-1} s^{-1}$ , $k_{2}$ = ?

R = 8.314 J/mol K

$E_{a}$ = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows.

$ln (\frac{k_{2}}{6.7}) = (\frac{-71000}{8.314} \times (\frac{1}{597.15} - \frac{1}{517.15})$

= 2.2123

$\frac{k_{2}}{6.7} = exp(2.2123)$

= 9.1364

$k_{2} = 61 M^{-1}s^{-1}$

Thus, we can conclude that rate constant of this reaction is $61 M^{-1}s^{-1}$ .

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3 years ago

Consider the mechanism. Step 1: A+B↽−−⇀CA+B↽−−⇀C equilibrium Step 2: C+A⟶DC+A⟶D slow Overall: 2A+B⟶D2A+B⟶D Determine the rate la

tatuchka [14]

Answer:

rate = k[A][B] where k = k₂K

Explanation:

Your mechanism is a slow step with a prior equilibrium:

$\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}$

(The arrow in Step 1 should be equilibrium arrows).

1. Write the rate equations:

$-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]$

2. Derive the rate law

Assume k₋₁ ≫ k₂.

Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.

In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is

rate = k[A][B] where k = k₂K

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4 years ago

Match each vocabulary word with the correct definition.

djyliett [7]

Answer:

Match each vocabulary word with the correct definition.

1. measure of how quickly velocity is changing

Friction

2. speed in a given direction

magnitude

3. force that resists moving one object against another

inertia

4. measure of the pull of gravity on an object

weight

5. tendency of an object to resist a change in motion

velocity

6. size

acceleration

Hope I Helped

8 0

3 years ago