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3 years ago

10

Wat do you pour 1st when diluting an acid? When diluting an acid, always pour 1. the water into the acid. 3. the acid into the w

ater.

2 answers:

Sauron [17]3 years ago

8 0

dilute acid into water

docker41 [41]3 years ago

3 0

The acid into the water

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It the mass of a material is 46 grams and the volume of the material is 8 cm ^3 What would the density of the material to be

Nat2105 [25]

5.75 Grams per cm^3

You do mass divided by volume

3 0

3 years ago

Hiii pls help me to write out the ionic equation

emmasim [6.3K]

Answer:

<u>STEP I</u>

This is the balanced equation for the given reaction:-

$2KOH_{(aq)} + H_2SO_4{}_{(aq)} \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)}$

<u>STEP II</u>

The compounds marked with (aq) are soluble ionic compounds. They must be

broken into their respective ions.

see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).

On breaking them into their respective ions :-

2KOH -> 2K+ + 2OH-
H2SO4 -> 2H+ + (SO4)2-
K2SO4 -> 2K+ + (SO4)2-

<u>STEP III</u>

Rewriting these in the form of equation

$\underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \: \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O$

<u>STEP </u><u>IV</u>

Canceling spectator ions, the ions that appear the same on either side of the equation

<em>(note: in the above step the ions in bold have gotten canceled.)</em>

$\boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}}$

This is the net ionic equation.

____________________________

$\\$

$\mathfrak{\underline{\green{ Why\: KOH \:has\: been\: taken\: as\: aqueous ?}}}$

KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.

[Alkali metal hydroxides are the only halides soluble in water ]

4 0

2 years ago

jok3333 [9.3K]

Answer:

1 mole represents 6.023×1023 particles.

1 mole of iodine atom= 6.023×1023

Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg

Given 48g of Mg = 2×6.023×1023

no. of Mg = 2 moles of Mg

1 mole of chlorine atom= 6.023× 1023

no. of chlorine atom = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 1023

no. of chlorine atom = 6.023×1023

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 1023

no. of atoms of hydrogen= 4 moles of hydrogen atom.

7 0

3 years ago

What method is used for chaff from grain<br>

Marta_Voda [28]

Answer:

Winnowing

Explanation:

Wind blows the lighter(in terms of mass) chaff from the whole grains,which are heavier(in terms of mass)

4 0

2 years ago

How many particles are present in 12.47 grams of NaCl

liq [111]

Answer:

1.26*10²³ particles are present in 12.47 grams of NaCl

Explanation:

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. The Avogadro number applies to any substance.

So, first of all you must know the amount of moles that represent 12.47 grams of NaCl. For that it is necessary to know the molar mass.

You know:

Na: 23 g/mole
Cl: 35.45 g/mole

So the molar mass of NaCl is: 23 g/mole + 35.45 g/mole= 58.45 g/mole

Now you apply a rule of three as follows: if 58.45 grams are present in 1 mole of NaCl, 12.47 grams in how many moles will they be?

$moles=\frac{12.47 grams*1 mole}{58.45 grams}$

moles= 0.21

You apply a rule of three again, knowing Avogadro's number: if in 1 mole of NaCl there are 6,023 * 10²³ particles, in 0.21 moles how many particles are there?

$number of particles=\frac{0.21 moles*6.023*10^{23} }{1 mole}$

number of particles= 1.26*10²³

<u><em>1.26*10²³ particles are present in 12.47 grams of NaCl</em></u>

<u><em></em></u>

5 0

3 years ago