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2 years ago

What is a monomer of nucleic acids called?

Chemistry

1 answer:

Viefleur [7K]2 years ago

4 0

Monomer of nucleic acids is called nucleotide.

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How many moles of salt are in 13.8 g of sodium chloride?

Hoochie [10]

Answer: 0.24 moles

Explanation:

Molecular Mass of NaCl (23 + 35.5) = 58.5g

58.5g of Sodium Chloride -------> 1 mole of NaCl

∴ 13.8g of Sodium Chloride ------> 1 ÷58.5 x 13.8 = 0.2358974 ≈ 0.24moles

3 0

3 years ago

Read 2 more answers

Is volume and mass size-dependent

Otrada [13]

A size dependent property is a physical property that changes when the size of an object changes.

6 0

3 years ago

If you have 8.943 L of carbon monoxide gas at SATP, how many moles would it contain?

viktelen [127]

1mole contains 22.4Lmol^-1
xmole contains 8.943
cross-multiply
x=1×8.943/22.4
x=0.40mole
there it contains 0.40moles.

8 0

3 years ago

The molecular formula of a compound is C7H14O7. what is the empirical formula

nata0808 [166]

Empirical formula is the simplest way the molecular formula can be wrote so here 7 goes into all of these so it would be CH2O

6 0

2 years ago

1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this

LenKa [72]

<h3>Answer:</h3>

0.6 g NaCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

<u>Step 2: Identify Conversions</u>

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})$
Multiply/Divide: $\displaystyle 0.551373 \ g \ NaCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

5 0

2 years ago