Which of the following is true of a balanced equation?

A horizontal pipe 15 cm in diameter and 4 m long is buried in the earth at a depth of 20 cm. The pipe-wall temperature is 75 deg

Basile [38]

Explanation:

The given data for case (1) is as follows.

h = 20 cm = 0.2 m

Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.

Q = $kA \frac{\Delta T}{L}$

where, A = area

L = length

k = thermal conductivity = 0.8 W/m

$\Delta T$ = change in temperature.

Therefore, putting the given values into the above formula as follows.

Q = $kA \frac{\Delta T}{L}$

= $0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{0.2}$

= 168 W

For case (2), h = 180 cm = 1.8 m

Therefore, heat lost will be calculated as follows.

Q = $kA \frac{\Delta T}{L}$

= $0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{1.8}$

= 18.67 W

Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.

8 0

3 years ago

PLZ HELP ITS A TEST AND GIVE ME THE RIGHT ANSWER

Rzqust [24]

There’s no question?

6 0

3 years ago

Explain the energy flowing as you hold an ice cream cone. (Be sure to mention "the system," "the surroundings," and how energy i

Serggg [28]

Answer:

This is what I put, it hasn't been graded though...

Within the system, the energy is flowing from your hand to the ice cream cone. Energy is also flowing from the surroundings to the ice cream cone unless it's reeeeaally cold in the surroundings in which case why are you even holding an ice cream cone when it's that cold?

4 0

3 years ago

What is the sequence of the complementary strand for the DNA ,gcaa

Vinil7 [7]

Answer:

1. GCAA= CGTT

2. TCAG =AGTC

3. CGTT= GCAA

4. ATAT = TATA

5. ATGA = TACT

Explanation:

A goes to T

C goes to G

T goes to A

G goes to C

4 0

3 years ago

Read 2 more answers

Bromine (63 g) and fluorine (60 g) are mixed to give bromine trifluoride. a) Write a balanced chemical reaction. b) What is the

fiasKO [112]

Answer:

a) Br2 + 3F2 → 2BrF3

b) Br2 is the limiting reactant.

c) There will be formed 86.3 grams of BrF3

d) There will remain 0.326 moles of F2 = 12.97 grams

Explanation:

Step 1: The balanced equation

Br2 + 3F2 → 2BrF3

Step 2: Given data

Mass of Bromine = 63grams

Mass of fluorine = 60 grams

percent yield = 80%

Molar mass of bromine = 79.9 g/mol =

Molar mass of fluorine = 19 g/mol

Molar mass of bromine trifluoride = 136.9 g/mol

Step 3: Calculating moles

Moles Br2 = 63 grams / (2*79.9)

Moles Br2 = 0.394 moles

Moles F2 = 60 grams / (2*19.9)

Moles F2 = 1.508

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

Br2 is the limiting reactant. It will completely be consumed (0394 moles).

There will react 3*0.394 = 1.182 moles of F2

There will remain 1.508 - 1.182 = 0.326 moles of F2 = 12.97 grams

Step 5: Calculate moles of BrF3

For 1 mole Br2 consumed, we need 3 moles of F2 to produce 2 moles of BrF3

So there is 2*0.394 moles = 0.788 moles of BrF3 moles produced

Step 6: Calculate mass of BrF3

mass = Moles * Molar mass

mass of BrF3 = 0.788 moles * 136.9 g/mol = 107.88 grams = Theoretical yield

Step 7: Calculate actual yield

% yield = 0.80 = actual yield / theoretical yield

actual yield = 0.80 * 107.88 grams = 86.3 grams

actual yield = 86.3 grams

7 0

2 years ago