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xenn [34]
3 years ago
14

What are cells made of?

Chemistry
1 answer:
skad [1K]3 years ago
8 0

Answer: All cells are made from the same major classes of organic molecules: nucleic acids, proteins, carbohydrates, and lipids.

Explanation: please mark as brainliest

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KClO3 (s) KCl (s) + 02 (g)​<br><br><br>How do you balance this equation
Vitek1552 [10]

Answer:

2KCl(s) + 6O3O2

Explanation:

I think it's correct

7 0
2 years ago
Select the pair that consists of a base and its conjugate acid in that order. CO32−/CO22−
lana [24]

Answer: The pair that consists of a base and its conjugate acid in that order.NH_3/NH_4^+

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

H_3PO_4\rightarrow H_PO_4{2^-}+2H^+

H_2CO_3\rightarrow HCO_3^-+H^+

NH_3+H^+\rightarrow NH_4^+

HCO_3^-\rightarrow CO_3^{2-}+H^+

NH_3  is gaining a proton, thus it is considered as a brønsted-lowry base and after gaining a proton, it formsNH_4^+  which is a conjugate acid.

3 0
3 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
What do you understand by valency electron and valency shell?​
jasenka [17]

Answer:

<h2>total no. of electron present in Valency shell is called valency electron </h2><h2>___________________</h2>

<h2>valency shell is that in which last electron is present</h2>

7 0
2 years ago
PLEASE HELP!!!! What is the abbreviated electron configuration for nitrogen and for chlorine
tino4ka555 [31]

Answer:chlorine’s is (Ne) 3s2 3p5

Explanation:lol

5 0
3 years ago
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