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Gnoma [55]
2 years ago
6

A book sits motionless on a table. Which statement is true? Group of answer choices There are no forces acting on the book. The

forces acting on the book are balanced by each other. The upward force is greater than the downward force on the book. The downward force is greater than the upward force on the book
Chemistry
1 answer:
Katena32 [7]2 years ago
6 0

Answer:

The forces acting on the book are balanced by each other.

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Check the approach to obtain the correct conversion equation. Use the temperature in °F to find the temperature in °C.
Maru [420]

Answer:

( °F − 32) × 5/9 =  °C

Explanation:

Also there is a mental calculation to convert from Fahrenheit to Celsius. The ratio 5/9 is approximately equal 0.55555….

Subtract 32º to adapt the equivalent in the Fahrenheit scale.

Divide the degrees Celsius by 2 (multiply by 0.5).

Take 1/10 of this number (0.5 * 1/10 = 0.05) and add it to the number obtained previously.

Example: Convert 98.6º F to Centigrade.

98.6 - 32 = 66.6

66.6 * 1/2 = 33.3

33.3 * 1/10 = 3.3

33.3 + 3.3 = 36.6 which is an approximation in degrees Centigrade

8 0
3 years ago
The following data were collected for the rate of disappearance of NO in the reaction 2NO(g)+O2(g)→2NO2(g)::
Anit [1.1K]

Answer:

a) The rate law is: v = k[NO]² [O₂]

b) The units are: M⁻² s⁻¹

c) The average value of the constant is: 7.11 x 10³ M⁻² s⁻¹

d) The rate of disappearance of NO is 0.8 M/s

e) The rate of disappearance of O₂ is 0.4 M/s

Explanation:

The experimental rates obtained can be expressed as follows:

v1 = k ([NO]₁)ᵃ ([O₂]₁)ᵇ = 1.41 x 10⁻² M/s

v2 = k ([NO]₂)ᵃ ([O₂]₂)ᵇ = 5.64 x 10⁻² M/s

v3 = k ([NO]₃)ᵃ ([O₂]₃)ᵇ = 1.13 x 10⁻¹ M/s

where:

k = rate constant

[NO]₁ = concentration of NO in experiment 1

[NO]₂ = concentration of NO in experiment 2

[NO]₃ = concentration of NO in experiment 3

[O₂]₁ = concentration of O₂ in experiment 1

[O₂]₂ = concentration of O₂ in experiment 2

[O₂]₃ = concentration of O₂ in experiment 3

a and b = order of the reaction for each reactive respectively.

We can see these equivalences:

[NO]₂ = 2[NO]₁

[O₂]₂ = [O₂]₁

[NO]₃ = [NO]₂

[O₂]₃ = 2[O₂]₂

So, v2 can be written in terms of the concentrations used in experiment 1 replacing [NO]₂ for 2[NO]₁ and [O₂]₂ by [O₂]₁ :

v2 = k (2 [NO]₁)ᵃ ([O₂]₁)ᵇ

If we rationalize v2/v1, we will have:

v2/v1 = k *2ᵃ * ([NO]₁)ᵃ * ([O₂]₁)ᵇ / k * ([NO]₁)ᵃ * ([O₂]₁)ᵇ (the exponent "a" has been distributed)

v2/v1 = 2ᵃ

ln(v2/v1) = a ln2

ln(v2/v1) / ln 2 = a

a = 2

(Please review the logarithmic properties if neccesary)

In the same way, we can find b using the data from experiment 2 and 3 and writting v3 in terms of the concentrations used in experiment 2:

v3/v2 = k ([NO]₂)² * 2ᵇ * ([O₂]₁)ᵇ / k * ([NO]₂)² * ([O₂]₂)ᵇ

v3/v2 = 2ᵇ

ln(v3/v2) = b ln 2

ln(v3/v2) / ln 2 = b

b = 1

Then, the rate law for the reaction is:

<u>v = k[NO]² [O₂]</u>

Since the unit of v is M/s and the product of the concentrations will give a unit of M³, the units of k are:

M/s = k * M³

M/s * M⁻³ = k

<u>M⁻² s⁻¹ = k </u>

To obtain the value of k, we can solve this equation for every experiment:

k = v / [NO]² [O₂]

for experiment 1:

k = 1.41 x 10⁻² M/s / (0.0126 M)² * 0.0125 M = 7.11 x 10³ M⁻² s⁻¹

for experiment 2:

k = 7.11 x 10³ M⁻² s⁻¹

for experiment 3:

k = 7.12 x 10³ M⁻² s⁻¹

The average value of k is then:

(7.11 + 7.11 + 7.12) x 10³ M⁻² s⁻¹ / 3 = <u>7.11 x 10³ M⁻² s⁻¹ </u>

The rate of the reaction when [NO] = 0.0750 M and [O2] =0.0100 M is:

v = k [NO]² [O₂]

The rate of the reaction in terms of the disappearance of NO can be written this way:

v = 1/2(Δ [NO] / Δt) (it is divided by 2 because of the stoichiometric coefficient of NO)

where (Δ [NO] / Δt) is the rate of disappearance of NO.

Then, calculating v with the data provided by the problem:

v = 7.11 x 10³ M⁻² s⁻¹ * (0.0750M)² * 0.0100M = 0.4 M/s

Then, the rate of disappearance of NO will be:

2v = Δ [NO] / Δt = <u>0.8 M/s</u>

The rate of disappearance of O₂ has to be half the rate of disappearance of NO because two moles of NO react with one of O₂. Then Δ [O₂] / Δt = <u>0.4 M/s</u>

With calculations:

v = Δ [O₂] / Δt = 0.4 M/s (since the stoichiometric coefficient is 1, the rate of disappearance of O₂ equals the rate of the reaction).

3 0
3 years ago
Which is a factor in determining the average atomic mass of an element?
kvasek [131]
The factor in determining the average atomic mass of an element is: 
B or 2 relative abundance of each isotope because the by looking at how many protons , electrons and neutrons the most isotope is of the element has relative abundance. 
7 0
3 years ago
Read 2 more answers
2. The pressure of the oxygen gas inside a
Flauer [41]

Answer: 4.41 atm

Explanation:

Given that,

Original pressure of oxygen gas (P1) = 5.00 atm

Original temperature of oxygen gas (T1) = 25°C

[Convert 25°C to Kelvin by adding 273

25°C + 273 = 298K

New pressure of oxygen gas (P2) = ?

New temperature of oxygen gas (T2) = -10°C

[Convert -10°C to Kelvin by adding 273

-10°C + 273 = 263K

Since pressure and temperature are given while volume is held constant, apply the formula for Charle's law

P1/T1 = P2/T2

5.00 atm /298K = P2/263K

To get the value of P2, cross multiply

5.00 atm x 263K = 298K x V2

1315 atm•K = 298K•V2

V2 = 1315 atm•K / 298K

V2 = 4.41 atm

Thus, the new pressure inside the canister is 4.41 atmosphere

4 0
3 years ago
The Moon shines white light on Earth. Is the Moon a good blackbody
Hitman42 [59]

Answer:

C.) No, because the Moon reflects most of the Sun's light rather than

absorbing it

Explanation:

:p

8 0
3 years ago
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