Does H form cation, anion both or neither ions?

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Answer 1

Answer:

Answer:

hydrogen have electronic configuration of 1s¹ to acquire a stable state it can either lose electron and form H+cation or gain an electron( to compete its 1s subshell ) to form H- anion. As it has ±1 valency it is placed neither in group 1(alkaline metals ) nor in group 17 (halogens) .

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What are the starting substances (molecules) in a chemical equation called?

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Answer:

A chemical reaction is the process in which atoms present in the starting substances rearrange to give new chemical combinations present in the substances formed by the reaction. These starting substances of a chemical reaction are called the reactants, and the new substances that result are called the products.

You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol,MW=60.05 g/mol, pKa=4.76), pKa=4.76), 3.00 M NaOH,3.00 M NaOH, and water. Answer the questions regarding the preparation of the buffer. 1. How many grams of acetic acid will be needed to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution.

Answers

The quantity of acetic acid that is needed to prepare the 500 mL buffer is 9.0075 grams.

Given the following data:

Volume of acetate buffer = 500 mL to L = 0.5 L.

Molarity of acetate buffer = 0.300 M.

pH = 4.90.

MW = 60.05 g/mol.

pKa = 4.76.

How to calculate the mass of acetic acid.

First of all, we would write the equilibrium chemical reaction for acetate-acetic acid as follows:

$CH_3COOH \rightleftharpoons CH_3COO^(-)+H^+$

Next, we would calculate HA by applying Henderson-Hasselbalch equation:

$pH =pka+ log_(10) (A^-)/(HA)$

Where:

HA is acetic acid.

$A^-$ is acetate ion.

Substituting the given parameters into the formula, we have;

$4.90 =4.76+ log_(10) (A^-)/(HA)\n\n4.90 -4.76+ log_(10) (A^-)/(HA)\n\n(A^-)/(HA)=1.38\n\nA^- = 1.38[HA]$

For the concentration of both acids, we have:

$[HA]+[A^-]=0.300M\n\n[HA]+1.38[HA]=0.300M\n\n2.38[HA]=0.300M\n\nHA = 0.126$

For acetate ion:

$A^- = 1.38[HA] = 1.38 * 0.126\n\nA^- =0.174$

At a volume of 0.5 liters, we have:

$HA = 0.5 * 0.126\n\nHA = 0.063 \;moles$

$A^- = 0.5 * 0.174\n\nA^- =0.087 \;moles$

By stoichiometry:

Total moles = $0.063 + 0.087$ = 0.15 moles.

$Mass = number \;of \;moles * molar\;mass\n\nMass =0.15 * 60.05$

Mass = 9.0075 grams.

Read more on moles here: brainly.com/question/3173452

Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀ $([A^(-)])/([HA])$

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀ $([A^(-)])/([HA])$

1,38 = $([A^(-)])/([HA])$ (1)

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] (2)

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid ≡ 0,15 moles of acetic acid × $(60,05 g)/(1mol)$ = 9,0 g of acetic acid

I hope it helps!

How many moles of MgS2O3 are in 223 g of the compound

Answers

Answer: 1.63 moles

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023* 10^(23)$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}=$

Given mass = 223 g

Molar mass = 136.4

$\text{Number of moles}=(223g)/(136.4g/mol)=1.63moles$

Thus there are 1.63 moles in 223 g of the compound.

Moles of MgS2O3 = 223/molar mass of MgS2O3

= 223/136.42
= 1.634 moles.

Hope this helps!

After 56.0 min, 40.0% of a compound has decomposed. What is the half‑life of this reaction assuming first‑order kinetics?

Answers

Answer:

Go ahead and plug in the percentages and time to find the answer.

Explanation:

The amount of a substance with half-life h, that remains after time t is 0.5t/h

Since 26% has decomposed, 74% remains.

So .74 = 0.580/h

ln .74 = (80/h) ln 0.5

h/80 = ln 0.5 / ln .74

h = 80 ln 0.5 / ln .74

h = 184.16 minutes

A carbon dioxide sample weighing 44.0 g occupies 32.68 l at 65°c and 645 torr. what is its volume at stp?

Answers

Combined gas law,

$(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))$

$((645torr)(32.68 L))/(338 K) = ((760 torr)V_(2))/(273 K)$

$v_(2)=22.4 L$

So, the gas will occupy 22.4 L at STP

For each of the following redox reactions in the, identify the oxidizing agent and the reducing agent.2Na(aq)+2H2O(l)→2NaOH(aq)+H2(g)
C(s)+O2(g)→CO2(g)
2MnO−4(aq)+5SO2(g)+2H2O(l)→2Mn2+(aq)+5SO2−4(aq)+4H+(aq)

Answers

anywhere between 5sogt2 and 776sogt2

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