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3 years ago

Homeostasis is the balance between your internal and external environments. We

Chemistry

1 answer:

topjm [15]3 years ago

8 0

Answer:

They are already matched for you. It goes

in the order of the questions.

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One question for chem?? Image attached? please help ! :)

lana [24]

The only bond that involves the complete transfer of electrons is ionic bonding.

A metal that does not have much care for its electrons, such as Sodium, Na, is willing to give up its electron more freely. And an atom that really cares about having electrons might be a gas such as chlorine, Cl. The Chlorine atom takes the electron from Sodium, and then the newly formed ions of Na+ and Cl- bond because they have equal and opposite charges and opposites attract. NaCl is formed.

Hydrogen bonds aren’t actually real bonds (I know, the name is deceptive)

Covalent bonds involve the sharing of electrons.

I’m pretty sure metallic bonds is a made-up term.

7 0

3 years ago

Period 4 of this periodic table is highlighted. What happens as you move from right to left across the period?

ivanzaharov [21]

The elements change from metals to nonmetals.

4 0

2 years ago

Read 2 more answers

B. The following reaction takes place in a basic solution. (7 points)

Natali5045456 [20]

<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

<u>Explanation:</u>

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

$MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)$

To balance the given redox reaction in basic medium, there are few steps to be followed:

Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electrons

Oxidation half-reaction: $NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-$

Reduction half-reaction: $MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-$

Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions

Oxidation half-reaction: $NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-$ ( × 3)

Reduction half-reaction: $MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-$ ( × 2)

The half-reactions now become:

Oxidation half-reaction: $3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-$

Reduction half-reaction: $2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-$

Add the equations and simplify to get a balanced equation

Overall redox reaction: $3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-$

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

3 0

2 years ago

3. Which is not true about natural<br> selection?

tatuchka [14]

Answer:

A:There must be diversity in the population.

Explanation:

Hopefully this helps!

5 0

3 years ago

A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What

NeTakaya

Answer:

Approximately $1.854\; \rm mol\cdot L^{-1}$ .

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

<h3>Formula mass of strontium hydroxide</h3>

Look up the relative atomic mass of $\rm Sr$ , $\rm O$ , and $\rm H$ on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

$\rm Sr$ : $87.62$ .
$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Calculate the formula mass of $\rm Sr(OH)_2$ :

$M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}$ .

<h3>Number of moles of strontium hydroxide in the solution</h3>

$M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}$ means that each mole of $\rm Sr(OH)_2$ formula units have a mass of $121.634\; \rm g$ .

The question states that there are $10.60\; \rm g$ of $\rm Sr(OH)_2$ in this solution.

How many moles of $\rm Sr(OH)_2$ formula units would that be?

$\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}$ .

<h3>Molarity of this strontium hydroxide solution</h3>

There are $8.71467\times 10^{-2}\; \rm mol$ of $\rm Sr(OH)_2$ formula units in this $47\; \rm mL$ solution. Convert the unit of volume to liter:

$V = 47\; \rm mL = 0.047\; \rm L$ .

The molarity of a solution measures its molar concentration. For this solution:

$\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}$ .

(Rounded to four significant figures.)

5 0

3 years ago