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vladimir2022 [97]
3 years ago
11

A nucleotide is composed of a(n) _____. phosphate group, a nitrogen-containing base, and a five-carbon sugar amino group, a nitr

ogen-containing base, and a five-carbon sugar sulfhydryl group, a nitrogen-containing base, and a five-carbon sugar glycerol, a nitrogen-containing base, and a five-carbon sugar phosphate group, a nitrogen-containing base, and a hydrocarbon
Chemistry
1 answer:
lapo4ka [179]3 years ago
5 0

A nucleotide is composed of a phosphate group, a nitrogen-containing base, and a five-carbon sugar amino group. A nucleotide is composed of a phosphate group, a nitrogen-containing base, and a five-carbon sugar amino group. A nucleotide is the building block or structural component of DNA and RNA. It consists of a base , that is one from adenine, thymine, guanine, and cytosine. and a molecule of sugar and one of phosphoric acid.


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Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4 What volume of 1.45 M Ca(OH)2 is needed to react with 25.0 moles of H2SO4?
Shkiper50 [21]

Answer:

We need 17.2 L of Ca(OH)2

Explanation:

Step 1: Data given

Concentration of Ca(OH)2 = 1.45 M

Moles of H2SO4 = 25.0 moles

Step 2: The balanced equation

Ca(OH)2 + H2SO4 ⟶2H2O + CaSO4

Step 3: Calculate moles Ca(OH)2

For 1 mol Ca(OH)2 we need 1 mol H2SO4 to produce 2 moles H2O and 1 mol CaSO4

For 25.0 moles H2SO4 we'll need 25.0 moles Ca(OH)2 to produce 50 moles H2O and 25.0 moles CaSO4

Step 4: Calculate volume of Ca(OH)2

Volume Ca(OH)2 = moles Ca(OH)2 / concentration Ca(OH)2

Volume Ca(OH)2 = 25.0 moles / 1.45 M

Volume Ca(OH)2 = 17.2 L

We need 17.2 L of Ca(OH)2

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3 years ago
A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas
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The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
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The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
     (800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
     V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
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Answer:

There are 14 cm in 0.14 m

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The two of them both made models of the current atom, which was very excellent because now we could imagine how an atom would look.  I would say that the best one was the electron orbital configuration because we needed to know what valence electrons are. 
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