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2 years ago

What is a nuclear acid

Chemistry

2 answers:

Ronch [10]2 years ago

8 0

Answer:

Nucleic acid is an important class of macromolecules found in all cells and viruses. Deoxyribonucleic acid (DNA) encodes the information the cell needs to make proteins.

A related type of nucleic acid, called ribonucleic acid (RNA), comes in different molecular forms that participate in protein synthesis.

grin007 [14]2 years ago

3 0

A complex organic substance present in living cells, especially DNA or RNA, whose molecules consist of many nucleotides linked in a long chain.

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Arturiano [62]

Answer:

5SiO2 + 2CaC2 = 5Si + 2CaO + 4CO2

Explanation:

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3 years ago

Sailin' Steve likes to test the buoyancy of different objects. He threw a spherical object into a fluid that has a greater densi

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3 years ago

Nana76 [90]

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3 years ago

Read 2 more answers

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the balanced equation 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

3 0

3 years ago

What mass of calcium carbonate is produced when 250 mL of 6.0 M sodium carbonate is added to 750 mL of 1.0 M calcium fluoride

Savatey [412]

Given:

Volume of Na2CO3 = 250 ml = 0.250 L

Molarity of Na2CO3 = 6.0 M

Volume of CaF2 = 750 ml = 0.750 L

Molarity of CaF2 = 1.0 M

To determine:

The mass of CaCO3 produced

Explanation:

Na2CO3 + CaF2 → CaCO3 + 2NaF

Based on the reaction stoichiometry:

1 mole of Na2CO3 reacts with 1 moles of Caf2 to produce 1 mole of caco3

Moles of Na2CO3 present = V * M = 0.250 L * 6.0 moles/L = 1.5 moles

Moles of CaF2 present = V* M = 0.750 * 1 = 0.750 moles

CaF2 is the limiting reagent

Thus, # moles of CaCO3 produced = 0.750 moles

Molar mass of CaCO3 = 100 g/mol

Mass of CaCO3 produced = 0.750 moles * 100 g/mol = 75 g

Ans: Mass of CaCO3 produced = 75 g

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3 years ago

What is a nuclear acid ​

What is a nuclear acid