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3 years ago

How many elements are in Li2SO4?how many elements are in li2s 04 how many elements are in li2s 04

Chemistry

1 answer:

Free_Kalibri [48]3 years ago

6 0

Answer:

Explanation:

the three elements involved in this compound are Li, S, O.

lithium, sulfur, and oxygen. Which create the ionic compound, "Lithium Sulfate."

7 atoms total, since there are two lithium, four oxygen, and one sulfate atom. this is a white inorganic salt.

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What are the chemical properties of water?

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It's a compound of hydrogen and oxygen

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Your blood contains 2 L of Nitrogen at 1 atm. When you go scuba diving at the Great Barrier Reef, the pressure increases to 8.7

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Explanation:

I hope this helped,i m sorry if its not the exact answer you need.

6 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

3 years ago

The gas arsine, AsH3, decomposes as follows: In an experiment at a certain temperature, pure AsH3(g) was placed in an empty, rig

bazaltina [42]

The equilibrium pressure of $H_{2}$ is 288 torr. and $K_{p}$ for this reaction is 0.786 atm.

from the reaction:- $2AsH_{3}$ ⇄ $2As + 3H_{2}$

initial concentration 392 torr. 0

at equilibrium. 392 - $2x$ . $3x$

and the final pressure in the flask = 488 torr.

Hence,

$( 392 - 2x ) + 3x = 488\\ x = 488-392\\x = 96 torr.$

The partial pressure of $H_{2}$ is 3 × 96 = 288 torr.

and $AsH_{3}$ is 392 - ( 2 × 96 ) = 392 - 192 = 200 torr.

Now, to find $K_{p}$ for this reaction, we will use $K_{p} = \frac{(P_{H_{2} })^{2} }{( P_{AsH_{3} } )^{2}}$

putting all the values, we get,

$K_{p} = \frac{(288)^{3} }{(200)^{2} }$

= 597.1968 torr.

= 0.786 atm. ( 1 atm = 760 torr. )

what do you mean by equilibrium?

Equilibrium in chemistry is the phase that exists when a chemical reaction and its opposite reaction happen at the same rates. This word's Latin origin dates back to the prefix aequi-, which means equal, and lbra, that indicates scale or balance.

Learn more about equilibrium reaction here:-

brainly.com/question/15118952

#SPJ4

3 0

2 years ago

How many elements are in Li2SO4?how many elements are in li2s 04 how many elements are in li2s 04 ​

How many elements are in Li2SO4?how many elements are in li2s 04 how many elements are in li2s 04