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2 years ago

Which is a starting material for cellular respiration?

Chemistry

2 answers:

Ratling [72]2 years ago

8 0

Answer:

glucose is the starting material

lina2011 [118]2 years ago

3 0

Answer:

glucose

Explanation:

Oxygen and glucose are both reactants in the process of cellular respiration. The main product of cellular respiration is ATP; waste products include carbon dioxide and water.

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Which plant characteristic is the least important with regard to attracting bats? (1 point)

Sav [38]

Answer:

O flower shape

Explanation:

Bats are just as important for pollination as insects and just like insects, plants use artifices to attract bats to their flowers, allowing them to have access to the pollen that will be taken elsewhere.

Bats are mainly attracted by the color and smell of flowers, the size and shape also play a role in attracting these animals, but the shape is less influential.

7 0

3 years ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

How many moles of sodium hydroxide are there in 1.0mL of 2.0M NaOH

kvasek [131]

To find the moles, you can use the following formula

moles= Molarity x Liters

Molarity= 2.0 M
Liters= 0.0010 Liters ---------------->>>>>>>>>> 1.0 mL= 0.0010 Liters

moles= 2.0 M x 0.0010 Liters= 0.0020 moles

4 0

2 years ago

What is the freezing point (in °C) of a 0.743 m

LUCKY_DIMON [66]

Answer:

man i sure wish u read the explanation and all of it

Explanation:

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Your heart's been aching but you're too shy to say it

Inside we both know what's been going on

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And if you ask me how I'm feeling

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Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

(Give you up)

(Ooh) Never gonna give, never gonna give

(Give you up)

We've known each other for so long

Your heart's been aching but you're too shy to say it

Inside we both know what's been going on

We know the game and we're gonna play it

I just wanna tell you how I'm feeling

Gotta make you understand

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

4 0

3 years ago

Consider the half reactions below for a chemical reaction.

ladessa [460]

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

7 0

2 years ago