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3 years ago

Equations can be balanced by using the half-reaction method. Which step should be completed first when using this method?

Chemistry

2 answers:

Tamiku [17]3 years ago

8 0

Answer:

Separate the reaction into half equations

Explanation:

When we want to balance redox equations, we must take cognizance of the fact that a specie was oxidized and another specie was reduced.

Hence we must identify the specie that was oxidized and the one that was reduced and then break up the whole redox reaction into oxidation and reduction half equations.

nikitadnepr [17]3 years ago

4 0

Answer:

D. finding the oxidation states of atoms

Explanation:

i did the review on edge2020

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What four factors influence the rate of a homogeneous reaction.

Anika [276]

1) concentration or partial pressure of species
involved. 2) temperature • 3) presence of catalyst
4) nature of reactants.

5 0

2 years ago

You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye

dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

$V$ = 10 mL solution

$V_{added}$ = 12.20 mL

$V_{total}$ = 22.20 mL

then,

M 0.103 M of NaOH,

$V_{rinsed}$ = experiment will not be affected

$V_{total \ base}$ = 10.38 mL

Now,

⇒ mol of NAOH = MV

= $0.103\times 10.38$

= $1.06914 \ m$

Whether Sulfuric acid, then

⇒ $H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O$

⇒ $mol \ of \ acid =\frac{1}{2}\times \ mol \ of \ base$

⇒ $1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid$

Before any dilution:

$V_{sample} = 10 \ mL$

⇒ $M \ acid = \frac{m \ mol}{V}$

$=\frac{ 0.53457 }{10}$

$=0.053457 \ M$ (Sulfuric acid)

6 0

3 years ago

1.20 x 10^22 molecules NaOH to gram

Trava [24]

Answer:

$\boxed {\boxed {\sf 0.797 \ g \ NaOH}}$

Explanation:

<u>1. Convert Molecules to Moles</u>

First, we must convert molecules to moles using Avogadro's Number: 6.022*10²³. This tells us the number of particles in 1 mole of a substance. In this case, the particles are molecules of sodium hydroxide.

$\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Multiply by the given number of molecules.

$1.20*10^{22} \ molecules \ NaOH *\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Flip the fraction so the molecules cancel out.

$1.20*10^{22} \ molecules \ NaOH *\frac {1 \ mol \ NaOH} {6.022*10^{23} \ molecules \ NaOH}}$

$1.20*10^{22} *\frac {1 \ mol \ NaOH} {6.022*10^{23}}}$

$\frac {1.20*10^{22} \ mol \ NaOH} {6.022*10^{23}}}$

$0.0199269345732 \ mol \ NaOH$

<u>2. Convert Moles to Grams</u>

Next, we convert moles to grams using the molar mass.

We must calculate the molar mass using the values on the Periodic Table. Look up each individual element.

Na: 22.9897693 g/mol
O: 15.999 g/mol
H: 1.008 g/mol

Since the formula has no subscripts, we can simply add the molar masses.

NaOH: 22.9897693+15.999+1.008=39.9967693 g/mol

Use this as a ratio.

$\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

Multiply by the number of moles we calculated.

$0.0199269345732 \ mol \ NaOH*\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

The moles of sodium hydroxide cancel.

$0.0199269345732 *\frac {39.9967693 \ g \ NaOH }{1}$

$0.0199269345732 *39.9967693 \ g \ NaOH$

$0.79701300498 \ g \ NaOH$

The original measurement of molecules has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 0 tells us to leave the 7 in the hundredth place.

$0.797 \ g \ NaOH$