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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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A kernel of popcorn contains a small amount of water and when heated the popcorn pops. How can you explain this to someone

Aliun [14]

When you put a popcorn kernel in a microwave, the microwave heats up the water. The water then evaporates, and the air wants to escape. There will be so much pressure that the skin/shell of the kernel will break, exposing the corn.

4 0

3 years ago

Given the reaction: solid sodium reacts with chlorine gas to form solid sodium chloride How many moles of the product result fro

lyudmila [28]

Answer:

number of moles of NaCl produce = 12 mol

Explanation:

Firstly, we need to write the chemical equation of the reaction and balance it .

Na(s) + Cl2(g) → NaCl(s)

The balanced equation is as follows:

2Na(s) + Cl2(g) → 2NaCl(s)

1 mole(71 g) of chlorine produces 2 moles(117 g) of sodium chloride

6 mole of chlorine gas will produce ? mole of sodium chloride

cross multiply

number of moles of NaCl produce = 6 × 2

number of moles of NaCl produce = 12 moles

number of moles of NaCl produce = 12 mol

3 0

3 years ago

Z forms chloride compounds with the formulae ZCl2 and ZC13.

nordsb [41]

Answer:

(B) II, IV.

hope this answer is helpful for u.

3 0

2 years ago

What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb

Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

$(\text{CH}_3)_3\text{N}$ in this question acts as a weak base. As seen in the equation in the question, $(\text{CH}_3)_3\text{N}$ produces $\text{OH}^{-}$ rather than $\text{H}^{+}$ when it dissolves in water. The concentration of $\text{OH}^{-}$ will likely be more useful than that of $\text{H}^{+}$ for the calculations here.

Finding the value of $[\text{OH}^{-}]$ from pH:

Assume that $\text{pK}_w = 14$ ,

$\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}$ .

$[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}$ .

Solve for $[(\text{CH}_3)_3\text{N}]_\text{initial}$ :

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}$

Note that water isn't part of this expression.

The value of Kb is quite small. The change in $(\text{CH}_3)_3\text{N}$ is nearly negligible once it dissolves. In other words,

$[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}$ .

Also, for each mole of $\text{OH}^{-}$ produced, one mole of $(\text{CH}_3)_3\text{NH}^{+}$ was also produced. The solution started with a small amount of either species. As a result,

$[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}$ .

$\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3}$ ,

$[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b}$ ,

$[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}$ .

8 0

3 years ago

In a titration, 25.0 mL of 0.500 M KHP is titrated to the equivalence point with NaOH. The final solution volume is 53.5 mL. Wha

Lena [83]

M1v1=m2v2
m2=(m1v1)/v2
Where m is the molarities and v is the volumes
<span>m2=(25.0*0.500)/53.5
m2=12.5/53.5
m2=0.2336
by rounding off:
m2=0.234 M
so the answer is C: 0.234 M</span>

3 0

3 years ago

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