Please help me I'm so lost

Nitrogen monoxide, NO, is formed in automobile exhaust by the reaction of nitrogen and oxygen in the air: N2 (g) + O2 (g) ↔ NO (

Solnce55 [7]

Answer:

The equilibrium concentration of NO is 0.001335 M

Explanation:

Step 1: Data given

The equilibrium constant Kc is 0.0025 at 2127 °C

An equilibrium mixture contains 0.023M N2 and 0.031 M O2,

Step 2: The balanced equation

N2(g) + O2(g) ↔ 2NO(g)

Step 3: Concentration at the equilibrium

[N2] = 0.023 M

[O2] = 0.031 M

Kc = 0.0025 = [NO]² / [N2][O2]

Kc = 0.0025 = [NO]² / (0.023)(0.031)

[NO] = 0.001335 M

The equilibrium concentration of NO is 0.001335 M

6 0

3 years ago

What is 8.3 x 10^4 in standard form

seraphim [82]

Explanation:

83,000 is 8.3x10^4 in standard form

when it's a positive move decimal to the right as many as exponent say

if it's negative move decimal to left and and zeros till you move what exponent says

8 0

3 years ago

Read the following chemical equation

Naddika [18.5K]

Answer:

For eacht 4 moles Fe consumed, we will produce 2 moles Fe2O3

The mole ration is 4:2 (option 1)

Explanation:

Step 1: The unbalanced equation

Fe + O2 → Fe2O3

Step 2: Balancing the equation

Fe + O2 → Fe2O3

On the left side we have 2x O (in O2) and on the right side we have 3x O (in Fe2O3) . To balance the amount of O on both sides, we have to multiply O2 by 3 and Fe2O3 by 2.

Fe + 3O2 → 2Fe2O3

On the left side we have 1x Fe, On the right side we have 4x (in 2Fe2O3). To balance the amount of Fe we have to multiply Fe (on the left side) by 4.

Now the equation is balanced.

4Fe + 3O2 → 2Fe2O3

For eacht 4 moles Fe consumed, we will produce 2 moles Fe2O3

The mole ration is 4:2 (option 1)

8 0

3 years ago

wheat has been bred by farmers for thousands of years to improve its ability to be ground into flour. this is an example of what

Vlada [557]

The correct answer is letter B. Genetic engineering has shown a great leap in Science because we can now harness the ability of mutations of cells in crops and in other aspects. Flour is a commodity that is greatly needed by the masses and learning how to genetically modify its components to help it breed better is a great advantage for any society.

8 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