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Lelechka [254]
3 years ago
12

Perform the following conversions:

Chemistry
1 answer:
jek_recluse [69]3 years ago
8 0

Answer :

The conversion used for the temperature from Fahrenheit to degree Celsius is:

^oC=\frac{5}{9}\times (^oF-32)

The conversion used for the temperature from degree Celsius to Kelvin is:

K=^oC+273

(a) 1.06^oF

^oC=\frac{5}{9}\times (^oF-32)

^oC=\frac{5}{9}\times (1.06^oF-32)

^oC=-17.188

The temperature in degree Celsius is, -17.188^oC

K=^oC+273

K=-17.188+273

K=255.812

The temperature in Kelvin is, 255.812 K

(b) 3410^oC

^oC=\frac{5}{9}\times (^oF-32)

3410^oC=\frac{5}{9}\times (^oF-32)

^oF=6106

The temperature in degree Fahrenheit is, 6106^oF

K=^oC+273

K=3410+273

K=3683

The temperature in Kelvin is, 3683 K

(c) 6.1\times 10^3K

K=^oC+273

6.1\times 10^3K=^oC+273

^oC=5827

The temperature in degree Celsius is, 5827^oC

^oC=\frac{5}{9}\times (^oF-32)

5827^oC=\frac{5}{9}\times (^oF-32)

^oF=10456.6

The temperature in degree Fahrenheit is, 10456.6^oF

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6.8 × 10ⁿ

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The graph given below shows the motion of four runners P, Q, R, and S in a 5 km marathon.
wel

Answer:-

As we can see from the graphical data,

The distance covered by all the four runners is the same 5 km.

Among the four athletes, Athlete P covers the distance in under three hours.

It is the minimum time taken among the four athletes.

Thus Athlete P covers the 5 km distance in the minimum amount of time.

We know that speed = \frac{Distance}{Time}

Since time taken for P is minimum, his speed is the maximum. P ran the fastest.

Time taken by Q = 4.5 hours.

Speed of Q = \frac{Distance}{Time taken by Q}

                     = \frac{5km}{4.5hour}

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Time taken by R = 6 hours.

Speed of R = \frac{Distance}{Time taken by R}

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3 0
3 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
4 years ago
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