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2 years ago

What kind of changes cause iron to rust ?

Chemistry

1 answer:

loris [4]2 years ago

8 0

Answer:

ways it rust are

1.you put water on it

2.let it sit out side for about a month or 2

you put the iron in water still wait a month or 2 then

you got the full or sum rusty parts

Explanation:

ether way you need water, air ,and the main thing... IRON

You might be interested in

State the total number of electrons occupying all the p- orbitals in one atom of chlorine .

V125BC [204]

Answer:A chlorine atom in its ground state has a total of seven electrons in orbitals related to the atoms third energy level.

Explanation:

8 0

3 years ago

Epsom salts, a strong laxative used in veterinary medicine, is a hydrate, which means that a certain number of water molecules a

alexira [117]

The values of x represents that number of moles of water molecules that is present per mole of the salt magnesium sulfate. To determine the value for this, we need to know how much is the water that is lost after heating the sample assuming that all of the water molecules are evaporated leaving only the unhydrated form of the salt. We calculate as follows:

Mass of hydrated salt = 3.484 g
Mass after heating = 1.701 g
Mass lost = 3.484 g - 1.701 g = 1.783 g

The mass lost is equal to the mass of water lost.

Moles water lost = 1.783 g ( 1 mol / 18.02 g ) = 0.0989 mol H2O
Moles of unhydrated salt = 1.701 g ( 1 mol / 120.37 g ) = 0.0141 mol MgSO4

moles water / moles MgSO4 = 0.0989 mol H2O / 0.0141 mol MgSO4 = 7

Therefore, the value of x is 7.

8 0

3 years ago

HELP! ASAP! Iron (Fe) and copper (II) chloride (CuCl2) combine to form iron (III) chloride (FeCl3) and copper (Cu). If you start

Helga [31]

Answer: 30.978

Explanation:

From the equation 2 moles of Fe will result in 3 moles copper

so .325 moles Fe will result in .4875 moles Cu

Cu weights 63.546 gm per mole

.4875 moles * 63.546 gm / mole = 30.978 gm of Cu

7 0

2 years ago

What occurs when energy is removed from a liquid-vapor system in equilibrium

yulyashka [42]

The answer is that the number of water molecules increases and that of vapor decreases. When energy is removed from the system it would do so mostly as heat. Therefore the energy in the system would decrease and the molecules would vibrate at lower energies. Therefore, several water vapor molecules would turn back to water.

4 0

3 years ago

How many grams of sodium phosphate monobasic would we add to a liter and how many grams of sodium phosphate dibasic would we add

Nostrana [21]

Answer :

The correct answer for Mass of Na₂HPO₄ = 4.457 g and mass of NaH₂PO₄ = 8.23 g

Given : pH = 6.86

Total concentration of Phosphate buffer = 0.1 M

Asked : Mass of Sodium phosphate monobasic (NaH₂PO₄) = ?

Mass of Sodium phosphate dibasic(Na₂HPO₄)= ?

Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :

(In phosphate buffer , Na+ ion from NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )

Step 1 : To find pka

H₂PO₄⁻ <=> HPO₄²⁻

The above reaction has pka = 7.2 ( from image shown )

Step 2 : Plug values in Hasselbalch- Henderson equation .

Hasselbalch -Henderson equation is to find pH for buffer solution which is as follows :

$pH = pka + log\frac{[A^-]}{[HA]}$

pH = 6.86 pKa = 7.2

$6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

Subtracting both side by 7.2

$6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

$-0.34 = log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

Removing log

$10^-^0^.^3^4 = \frac{[HPO_4^2^-]}{[H_2PO_4^-]}$

$\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457$ ---------------- equation (1)

Step 3 : To find molarity of H₂PO₄⁻ and HPO₄²⁻

Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M

Hence, [H₂PO₄⁻ ] + [ HPO₄²⁻ ] = 0.1 M

Assume [H₂PO₄⁻ ] = x

So , [x ] + [ HPO₄²⁻ ] = 0.1 M

[ HPO₄²⁻ ] = 0.1 - x

Step 4 : Plugging value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ]

[H₂PO₄⁻ ] = x

[ HPO₄²⁻ ] = 0.1 - x

Equation (1) = > $\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457$

Plug value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ] ( from step 3 ) into equation (1) as :

$\frac{[0.1 - x ]}{[ x]} = 0.457$

Cross multiplying

$0.1 - x = 0.457 x$

Adding x on both side

$0.1 -x + x = 0.457 x + x$

$0.1 = 1.457 x$

Dividing both side by 1.457

$\frac{0.1}{1.457} = \frac{1.457 x }{1.457}$

x = 0.0686 M

Hence , [H₂PO₄⁻ ] = x = 0.0686 M

[ HPO₄²⁻ ] = 0.1 - x

[ HPO₄²⁻ ] = 0.1 - 0.0686

[ HPO₄²⁻ ] = 0.0314 M

Step 5 : To find moles of H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .

Molarity is defined as mole of solute per 1 L volume of solution .

Molarity of NaH₂PO₄ = 0.0686 M or 0.0686 mole per 1 L

Molarity of Na₂HPO₄ = 0.0314 M or 0.0314 mole per 1 L

Since that volume of buffer solution is 1 L , so Molarity = mole

Hence Mole of NaH₂PO₄ = 0.0686 mol

Mole of Na₂HPO₄ = 0.0314 mol

Step 6 : To find mass of Na₂HPO₄ and NaH₂PO₄

Moles of Na₂HPO₄ and NaH₂PO₄ can be converted to their masses using molar mass as follows :

Molar mass of Na₂HPO₄ = $141.96 \frac{g}{mol}$

Molar mass of NaH₂PO₄ = $119.98 \frac{g}{mol}$

$Mass (g) = mole (mol)* molar mass(\frac{g}{mol})$

$Mass of Na_2HPO_4 = 0.0314 mol * 141.96 \frac{g}{mol}$

Mass of Na₂HPO₄ = 4.457 g

$Mass of NaH_2PO_4 = 0.0686 mol * 119.98 \frac{g}{mol}$

Mass of NaH₂PO₄ = 8.23 g

5 0

2 years ago