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3 years ago

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Soo Hwan found a flowering plant at school and used a classification key to help him classify the plant. The flowers of the plan

t are not clustered, and they have a deep brown center. What type of plant did Soo Hwan find?

1 answer:

KIM [24]3 years ago

8 0

Answer:

Black eyed susan

Explanation:

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Murljashka [212]

Answer:

2. The metal would lose one electrons and the non metal would gain one electrons

Explanation:

An atom of a certain element reacts with the atoms of other elements in order to fullfill its outermost shell (called valence shell).

We notice the following:

- The elements in Group 1 (which are metals) have only 1 electron in their valence shell

- The elements in Group 17 (which are non-metals) have 1 vacancy (lack of electron) in their valence shell

This means that in order for both an atom of group 1 and an atom of group 17 to fullfill the valence shell, they have to:

- The atom in group 1 has to give away its only electron of the valence shell

- The atom in group 17 has to gain one electron in order to fullfill the shell

Therefore, the correct option is

2. The metal would lose one electrons and the non metal would gain one electrons

7 0

3 years ago

What are the 4 different types of chemical reactions? And how to remember them

Tju [1.3M]

<span>Synthesis, decomposition, single replacement and double replacement.</span>

4 0

3 years ago

Read 2 more answers

How many subshells are there in the shell with n = 6?

andre [41]

The number of subshells within a certain shell can be identified using the orbital angular momentum quantum number "l"

"l" is given values from zero till (n-1)
So, for n=6
l is given the following values: 0,1,2,3,4,5
Counting the number of subshells, we will find that the shell with n=6 has 6 subshells

7 0

3 years ago

Describe the three values in a learner-centred curriculum that a teacher can use

ad-work [718]

Answer:

to talk in a way that the class can all understand . help them in thing they are falling . and have a good relashaship.

Explanation:

3 0

3 years ago

Questions 3-6 refer to rhe solutions below:

Yanka [14]

Under room temperature where $\text{pK}_w = 14$ :

3.) (A), (B), and (E).

4.) (D).

5.) (B).

<h3>Explanation</h3>

What makes a buffer solution? For a solution to be a buffer, it needs to contain large amounts of a weak acid and its conjugate base ion. Alternatively, the solution may contain large amounts of a weak base and its conjugate acid ion.

Not every one of the five solutions is a buffer solution.

<h3>(A)</h3>

Ethanoic acid CH₃COOH (a.k.a. acetic acid) is a weak acid. pKa = 4.756. CH₃COONa is a salt. It dissolves to produce CH₃COO⁻, which is the conjugate base ion of CH₃COOH. The solution in (A) contains equal number of CH₃COOH and CH₃COO⁻, both at 1.0 M.

Refer to the Henderson-Hasselbalch equation for buffers of weak acids.

$\displaystyle \text{pH} = \text{pK}_a + \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Acid}]}}$ .

$\displaystyle \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Acid}]}} =\ln{1} = 0$ .

The pH of the solution in (A) will be the same as the pKa of CH₃COOH. pH = 4.746.

<h3>(B)</h3>

Consider the hydrogen halides:

HF: weak acid.
HCl: strong acid.
HBr: strong acid.

The radius of halogen atoms increases down the group, and hydrogen-halogen bond becomes weaker. It becomes easier for water to break those bonds. As a result, the strength of hydrogen halides increases down the group. HF is the only weak acid among the common hydrogen halides.

Mixing HBr and KBr at equal ratio will be similar to mixing HCl and KCl at the same ratio. All HBr in the solution breaks down into H⁺ and Br⁻. The pH of the solution will depend only on the concentration of HBr.

$\displaystyle [\text{H}^{+}] = [\text{HBr}] \\\phantom{[\text{H}^{+}]}= \frac{n}{V} \\\phantom{[\text{H}^{+}]}= \frac{c(\text{HBr})\cdot V(\text{HBr})}{V(\text{HBr})+V(\text{KBr})}\\\phantom{[\text{H}^{+}]}=\frac{0.100\;\text{L}\times 1.0\;\text{mol}\cdot\text{L}^{-1}}{0.100\;\text{L}+0.100\;\text{L}} \\\phantom{[\text{H}^{+}]}= 0.50\;\text{mol}\cdot\text{L}^{-1}$ .

$\text{pH} = -\log{[\text{H}^{+}] = -\log{0.50} \approx {\bf 0.30}$ .

<h3>(C)</h3>

Similarly to HCl and HBr, HI is also a strong acid. Mixing HI and NaOH at equal ratio will produce a solution of NaI, which is similar to NaCl. The final solution will be neutral. pH = 7 if pKw = 14.

<h3>(D)</h3>

NH₃ is a weak base. NH₄Cl dissolves completely to produce NH₄⁺ and Cl⁻. NH₄⁺ is the conjugate acid of NH₃. The final solution will contain an equal number of NH₃ and NH₄⁺. pKb = 4.75 for ammonia NH₃.

Apply the Henderson-Hasselbalch equation for buffers of weak bases:

$\displaystyle \textbf{pOH} = \text{pK}_b + \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Base}]}}= 4.75 + \log{1} = 4.75$ .

Note that what this equation gives for buffers of weak bases is the pOH of the solution. pH = pKw - pOH. Assume that pKw = 14. pH = 14 - 4.75 = 9.25.

<h3>(E)</h3>

The solution in (E) will contain about 1.0 M of CH₃COOH. The volume of the solution will be 200 mL.

$n(\text{CH}_3\text{COO}^{-}) = n(\text{NaOH}] = c\cdot V = 0.10\;\text{mol}$ .

$\displaystyle [\text{CH}_3\text{COO}^{-}] = \frac{n}{V} = {0.10}{0.10 + 0.10} = 0.50 \;\text{mol}\cdot\text{L}^{-1}$ .

There's nearly no conjugate base of CH₃COOH. As a result, the solution will not be a buffer, and the Henderson-Hasselbalch Equation will not apply. Refer to an ICE table:

$\begin{array}{c|ccccccc}\text{R}&\text{CH}_3\text{COO}^{-} &+&\text{H}_2\text{O}&\rightleftharpoons &\text{CH}_3\text{COOH}&+&\text{OH}^{-}\\\text{I}&0.50\\\text{C}& -x &&&& +x &&+x\\\text{E} &0.50 - x &&&&x&&x\end{array}$

The value of pKa is large. Ka will be small. the value of $x$ will be much smaller than $0.50$ such that $0.50-x \approx 0.50$ .

The pKa of a weak acid is the same as pKw divided by the pKb of its conjugate base.

$\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]} = \text{K}_b(\text{CH}_3\text{COO}^{-}) \\\phantom{\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]} }= \frac{\text{K}_w}{\text{K}_a(\text{CH}_3\text{COOH})} \\\phantom{\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]}} = \frac{10^{-14}}{1.75\times 10^{-5}} = 5.71\times 10^{-10}$ .

$\displaystyle \frac{x^{2}}{0.50} =5.71\times 10^{-10}$ .

$[\text{OH}^{-}] = x \approx 1.69\times 10^{-5}\;\text{mol}\cdot\text{L}^{-1}$ .

$\text{pH} = \text{pK}_w + \log{[\text{OH}^{-}]} = 9.23$ .

7 0

3 years ago