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2 years ago

What is the best definition for vehicle contamination

Chemistry

1 answer:

Serjik [45]2 years ago

6 0

Answer:

What is the best definition of "vehicle of contamination"?

-A preventive vaccine

-Any source of biological -contamination

An ambulance

-A food truck

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A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th

kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = $\frac{38.8}{61.2^{2} }$

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = $\frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }$

0.0104 = $\frac{200 - P(NO_{2}) }{[P(NO_{2} )]^{2}}$

0.0104 $[P(NO_{2} )]^{2}$ + $P(NO_{2} )$ - 200 = 0

Resolving the second degree equation:

$P(NO_{2} )$ = $\frac{-1+\sqrt{9.32} }{0.0208}$

$P(NO_{2} )$ = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are $P(NO_{2} )$ = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0

3 years ago

The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10

Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0 125 kPa

At t = teq 125 - 2x 4x x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $2.8\times 10^{-3}$ min⁻¹

Initial concentration $[A_0]$ = 125 kPa

Final concentration $[A_t]$ = 91 kPa

Time = ?

Applying in the above equation, we get that:-

$91=125e^{-2.8\times 10^{-3}\times t}$

$125e^{-2.8\times \:10^{-3}t}=91$

$-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)$

$t=113\ min$

3 0

3 years ago

The reaction between iron(II) oxide and carbon monoxide produces iron and carbon dioxide. How many moles of iron can be obtained

aleksandrvk [35]

Answer:

1.5 moles of Fe produced.

Explanation:

Given data:

Moles of FeO react = 1.50 mol

Moles of iron produced = ?

Solution:

Chemical equation:

FeO + CO → Fe + CO₂

Now we will compare the moles of ironoxide with iron.

FeO : Fe

1 : 1

1.5 : 1.5

Thus from 1.5 moles of FeO 1.5 moles of Fe are produced.

5 0

3 years ago

Sn + 2 H2SO4 → SnSO4 + SO2 + 2 H2O If 219.65 grams of SnSO4 are produced, how many moles of H2SO4 were reacted?

anygoal [31]

Answer:

2.05moles

Explanation:

The balanced chemical equation in this question is as follows;

Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O

Based on the above equation, 2 moles of H2SO4 reacted to produce 1 mole of SnSO4

However, the mass of SnSO4 produced is 219.65 grams. Using mole = mass/molar mass, we can find the number of moles of SnSO4 produced.

Molar mass of SnSO4 where Sn = 118.7, S = 32, O = 16

= 118.7 + 32 + 16(4)

= 150.7 + 64

= 214.7g/mol

mole = 219.65/214.7

mole = 1.023mol

Therefore, if 2 moles of H2SO4 reacted to produce 1 mole of SnSO4

1.023 mol of SnSO4 produced will cause: 1.023 × 2/1

= 2.046moles of H2SO4 to react.

8 0

3 years ago

Chemical Name for Rb2S

Westkost [7]

Answer:

Rubidium Sulfide

Explanation:

The chemical name for Rb2S is Rubidium Sulfide

6 0

3 years ago

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