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Nutka1998 [239]

2 years ago

7

a classroom has an aquarium. In the aquarium are a goldfish and a water plant. the teacher explains that both organisms use cell

ular respiration to transform food into energy. identify how the plant and the goldfish get the materials needed for cellular respiration.

1 answer:

Anna71 [15]2 years ago

4 0

the thing that helps is always gonna help that’s why u should always help

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What is true of greenhouse gases

marysya [2.9K]

<h2>Let us explain about green house gases</h2>

Explanation:

Green house effect

It can be explained as :

The sun emit radiations that consist of infra red rays , ultra violet rays and visible light .
The infra red rays (responsible for heating ) when come from sun , it travels through atmosphere and reaches earth surface .
On striking the earth surface , it gets reflected and reflected infra red ray has greater wavelength .
Then this reflected rays moves back through atmosphere .
On doing so , they encounter certain gas molecules like carbon dioxide , methane etc .
This gases absorb these rays and do not let them to escape through them .
As a result, they get trapped .
As we have already mention that they cause heating effect , due to this property they raise the temperature of atmosphere leading global warming .

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3 years ago

HCIO4 is identified as what acid

ivann1987 [24]

It is Perchloric acid

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3 years ago

"What can you tell about the rate of the two reactions shown in the graph?

Juliette [100K]

I believe the correct response is C. They occurred at different rates.

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3 years ago

Anyone know what the answer is?

malfutka [58]

2 Ag2O = 4 Ag + 1 O2

4 0

3 years ago

Read 2 more answers

What would be the voltage (Ecell) of a voltaic cell comprised of Cr (s)/Cr3+(aq) and Fe (s)/Fe2+(aq) if the concentrations of th

zaharov [31]

Answer:

0.35 V

Explanation:

(a) Standard reduction potentials

<u> E°/V</u>

Fe²⁺ + 2e- ⇌ Fe; -0.41

Cr³⁺ + 3e⁻ ⇌ Cr; -0.74

(b) Standard cell potential

<u> E°/V</u>

2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74

<u>3Fe ⇌ 3Fe²⁺ + 6e-; </u> <u>-0.41 </u>

2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33

3. Cell potential

2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr

<u>3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e- </u>

2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)

The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation

$E = E^{\circ} - \dfrac{RT}{zF}\ln Q$

(a) Data

E° = 0.33 V

R = 8.314 J·K⁻¹mol⁻¹

T = 298 K

z = 6

F = 96 485 C/mol

(b) Calculations:

$Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}$

7 0

3 years ago