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3 years ago

When do plants use the process of cellular respiration

Chemistry

1 answer:

d1i1m1o1n [39]3 years ago

8 0

Answer:

So the answer is every day

Explanation:

Plants respire at all times of the day and night because their cells need a constant energy source to stay alive.

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PLZ HELP

poizon [28]

Answer:

Explanation:

Convection is single or multiphase fluid flow that occurs spontaneously due to the combined effects of material property heterogeneity and body forces on a fluid, most commonly density and gravity (see buoyancy). When the cause of the convection is unspecified, convection due to the effects of thermal expansion and buoyancy can be assumed. Convection may also take place in soft solids or mixtures where particles can flow.

4 0

2 years ago

The cell potential of the following electrochemical cell depends on the pH of the solution in the anode half-cell:Pt(s)|H2(g, 1a

meriva

Answer:

0.51

Explanation:

Given the Nernst equation;

E= E° - 0.0592/n logQ

E= 355 mV or 0.355 V

E° = 0.34 - 0= 0.34 V

n= 2(two electrons were transferred in the process)

Equation of the reaction;

H2(g) + Cu^2+(aq) -----> 2H^+(aq) + Cu(s)

Substituting values;

0.355 = 0.34 - 0.0592/2 log([H^+]/1)

0.355 - 0.34 = - 0.0296 log [H^+]

0.015/-0.0296 = log [H^+]

Antilog (-0.5068) = [H^+]

[H^+] = 0.311 M

pH = -log[H^+]

pH= - log(0.311 M)

pH = 0.51

6 0

2 years ago

Propane (C3H8) can be burned to produce heat for homes. The products of the reaction are CO2 and H2O. For complete combustion to

Natalija [7]

Answer:

1- 3 Moles of CO2

2- 132 g of CO2

3- 105,6 g of CO2

4- Limiting Reagent O2

Products form based on limiting reagent (384g O2) :

CO2: 316,8 g

H2O: 172,8 g

Products form based on C3H8 (132,33 g):

CO2: 396,99 g

H2O: 216,54 g

Explanation:

Atomic Masses:

C: 12

H: 1

O: 16

Molecular weights:

C3H8: 44 g

O2: 32 g

H2O: 18 g

CO2: 44 g

C3H8 + 5 O2⇒ 3 CO2 + 4 H2O

C3H8 (44g)+ O2 (160 g) ⇒ CO2 (132 g) + H2O (72 g)

In 5 moles of O2 are produced 3 moles of CO2, equivalent to 132 g

For 160g of O2 are produced 132 g of CO2, so 128 g of O2

160 g O2 ⇒ 132 g CO2

128 g O2⇒ × = 105,6 g CO2

(128×132÷160= 105,6)

The limiting reagent is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, because the reaction cannot continue without it.

If I have 132,33 g of C3H8 and 384 g O2 we can calculate:

For 44 g of CH3H8 ⇒160 g of O2

With 132,33 g of CH3H8 ⇒ ×= 481,2 g of O2

(132,33×160÷44=481,2)

As this amount exceeds the quantity of O2 that we have, we can assume that the 384 g O2 will be totally consumed.

Calculations of the products formed in base of quantity of O2 (limiting reagent):

160 g O2 ⇒ 132 g CO2

384 g O2⇒ × = 316,8 g CO2

(384×132÷160= 316,8)

160 g O2 ⇒ 72 g H2O

384 g O2⇒ × = 172,8 g H2O

(384×72÷160= 172,8)

Calculations of the products formed in base of quantity of C3H8 (excess reagent):

44 g C3H8 ⇒ 132 g CO2

132,33 g C3H8 ⇒ × = 396,99 g CO2

(132,33×132÷44=396,99)

44 g C3H8 ⇒ 72 g H2O

132,33 g C3H8⇒ × = 216,54 g H2O

(132,33×72÷44= 216,54)

3 0

2 years ago

The CN– ion is widely used in the synthesis of organic compounds. What is the pattern of electron pairs in this ion? How many bo

OleMash [197]

For the Lewis diagram of the cyanide ion, a figure is shown.

There are 3 pairs of bonding electrons. There is a one lone pair each for the carbon and the nitrogen atoms.

I hope I was able to answer your question. Thank you!

8 0

2 years ago

For doping silicon with boron, silicon specimen was kept in gaseous atmosphere containing B2O3 that maintained the B concentrati

Kay [80]

Solution :

From Fick's law:

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=N_a$

Mass balance: Exits = Accumulation

$-N_A A = \frac{dm}{dt}$

$-N_A A = \frac{dVp}{dt}$

$-N_A A = \frac{dV}{dt}p$

$-N_A A = \frac{dhA}{dt}$

$-N_A A = \frac{dh}{dt} \times Ap$

From the last step, area cancels out and thus leaves :

$-N_A = \frac{dh}{dt} \times p$

So now we can substitute the $N_A$ by the Fick's law

$\frac{D_{AB}}{\Delta z } \times (C_{A1}-C_{A2})=\frac{dh}{dt} p$

Substituting the values we get

$=\frac{-4 \times 10^{-13}}{0.0001} \times (3 \times 10^{26} - C_{A2}) = \frac{dh}{dt} \times 2.46$

$=-4 \times 10^{-9} \times( 3 \times 10^{26} - C_{A2}) = 0.0001 \times 2.46$

$= -7800 \times 4 \times 10^{-9} \times (3 \times 10^{26}-C_{A2})=0.000246$ $=-(3 \times 10^{26}-C_{A2}) = 7.8846 \times 100 \times \frac{1}{69.62} \times 6.022 \times 10^{23}$

$C_{A2} = 6.85 \times 10^{28} \ \text{ boron atoms} /m^3$

3 0

2 years ago