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3 years ago

Detritivores feed on organic material produced from dead plants and animals called

Chemistry

1 answer:

densk [106]3 years ago

6 0

The answer is b. Detritus

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How do you draw 5 arrows showing uniformity

makkiz [27]

Answer:

I don't really know what that is so here is a picture of it

Explanation:

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3 years ago

Which object traveled at a greater speed?

agasfer [191]

Object B I know it lol

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3 years ago

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The The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k=0.029 W/ m measured temperature differ

vfiekz [6]

Answer:

Heat flux = 13.92 W/m2

Rate of heat transfer throug the 3m x 3m sheet = 125.28 W

The thermal resistance of the 3x3m sheet is 0.0958 K/W

Explanation:

The rate of heat transfer through a 3m x 3m sheet of insulation can be calculated as:

$q=-k*A*\frac{\Delta T}{\Delta X}\\\\q=-0.029\frac{W}{m*K}*(3m*3m)*\frac{12K}{0.025m} =125.28W$

The heat flux can be defined as the amount of heat flow by unit of area.

Using the previous calculation, we can estimate the heat flux:

$heat \, flux=\frac{q}{A}=\frac{125.28 W}{9 m^{2} } =13.92 W/m^{2}$

It can also be calculated as:

$q/A=-k*\frac{\Delta T}{\Delta X}$

The thermal resistance can be expressed as

$\Delta T=R_t*Q\\R_t=\Delta T/Q=\frac{\Delta X}{k*A}$

For the 3m x 3m sheet, the thermal resistance is

$R_t = \frac{\Delta X}{k*A}=\frac{0.025m}{0.029W/mK*9m^{2}}=0.0958 \, K/W$

4 0

3 years ago

9. The “power plant” of the cell is the

Daniel [21]

Answer:

D. Mitochondria

Explanation:

The mitochondria, labeled the powerhouse of the cell, are the organelle responsible for energy production within the cell

8 0

2 years ago

Read 2 more answers

Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g)

Fynjy0 [20]

Answer:

Rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

Explanation:

According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

⇒ $-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}$ -----------------------------(1)

Given that the rate of disappearance of oxygen = $-\frac{d[O_{2} ]}{dt}$ = 3.64 x 10⁻³ M/s

So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = ?

from equation (1) we can write

$\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]$

⇒ $\frac{d[SO_{3}] }{dt}$ = 2 x 3.64 x 10⁻³ M/s

⇒ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ $[\frac{d[SO_{3}] }{dt}]$ = 7.28 x 10⁻³ M/s

7 0

4 years ago