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Arada [10]
3 years ago
11

A Venus Flytrap is a carnivorous plant, meaning it undergoes photosynthesis for food but also consumes insects for nutrients. In

to which kingdom should the Venus Flytrap be classified?
can some write the answer
Chemistry
2 answers:
nasty-shy [4]3 years ago
5 0

Answer:

the food kingdom

Explanation:

look kid i just need some points socratic will help with

ur answer

Natasha_Volkova [10]3 years ago
4 0

Answer:

Venus Flytrap:

The Plant Kingdom, also termed Plantae, is the broadest classification under which the Venus flytrap fits.

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__Ca(OH)2 + __ (NH4)2SO4 ----- &gt; ___CaSO4 + ___NH3 + __H2O<br><br> A)11<br> B)5<br> C)7<br> D)9
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Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold
Mama L [17]

<u>Answer:</u> The number of gold atoms per cubic centimeters in the given alloy is 1.83\times 10^{22}

<u>Explanation:</u>

To calculate the number of gold atoms per cubic centimeters for te given silver-gold alloy, we use the equation:

N_{Au}=\frac{N_AC_{Au}}{(\frac{C_{Au}M_{Au}}{\rho_{Au}})+(\frac{M_{Au}(100-C_{Au})}{\rho_{Ag}})}

where,

N_{Au} = number of gold atoms per cubic centimeters

N_A = Avogadro's number = 6.022\times 10^{23}atoms/mol

C_{Au} = Mass percent of gold in the alloy = 42 %

\rho_{Au} = Density of pure gold = 19.32g/cm^3

\rho_{Ag} = Density of pure silver = 10.49g/cm^3

M_{Au = molar mass of gold = 196.97 g/mol

Putting values in above equation, we get:

N_{Au}=\frac{(6.022\times 10^{23}atoms/mol)\times 48\%}{(\frac{48\%\times 196.97g/mol}{19.32g/cm^3})+(\frac{196.97g/mol\times 58\%}{10.49g/cm^3})}\\\\N_{Au}=1.83\times 10^{22}atoms/cm^3

Hence, the number of gold atoms per cubic centimeters in the given alloy is 1.83\times 10^{22}

5 0
3 years ago
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Bumek [7]

Answer:

I believe it is B

Explanation:

hope it helps. please let me know if it's wrong

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3 years ago
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