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3 years ago

How do moss leaves and fish differ? How are they the same?

1 answer:

4 0

Moss leaves and fish are different in that, the moss leave is a producer, that is, it produces its own food through photosynthesis while the fish is a consumer, it feeds on foods that are not produced by it.
Both moss and fish are the same in the sense that both have cell as their basic unit of life, that is, they both possess cells.

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For each compound determine whether it is organic or inorganic. C10h16kno9s2 naaso2 hsicl3 (ch)4as2 (bio)2co3 h2p2o7 h2o co2 c6h

Sindrei [870]

a. Organic: C₁₀H₁₆KNO₉S₂; (CH₃)₄As₂; C₆H₁₂O₆

b. Inorganic: NaAsO₂; HSiCl₃; (BiO)₂CO₃; H₂P₂O₇; H₂O; CO₂

Compounds containing both C and H are organic.

Compounds that are not organic are inorganic.

8 0

3 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at 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:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

What do we call genes

qwelly [4]

Hi there so i need one more brainliest mark so if i get it right can i get brainliest please :)

A gene is the basic physical and functional unit of heredity. Genes are made up of DNA. Some genes act as instructions to make molecules called proteins. However, many genes do not code for proteins. In humans, genes vary in size from a few hundred DNA bases to more than 2 million bases

4 0

2 years ago

Manganese, Mn, forms two ions, one with a 2+ charge and one with a 3+ charge. What is the formula for manganese (II) sulfide?

maria [59]

It would be MnSO4

The (II) lets you know it’s the form with a 2+ charge and Sulfate has a 2- charge

These will cancel out making it plain MnSO4

If it was manganese (iii) sulfide the answer would be Mn2(SO4)3

4 0

3 years ago

Calculate the mass, in grams, of each sample. a. 1.1 * 1023 gold atoms b. 2.82 * 1022 helium atoms c. 1.8 * 1023 lead atoms d. 7

Vlada [557]

Answer:blablablablablabla

Explanation:

Blablabla

5 0

3 years ago