Lagta ha bhaijaan sa sab bahenjaan naraz haan koi bat ni kr raha

Nitrogen (N2) and hydrogen (H2) react to make NH3 according to the following

Schach [20]

The number of moles of NH3 that could be made would be 0.5 moles

<h3>Stoichiometric reactions</h3>

From the balanced equation of the reaction:

N2 (g) + 3 H2(g) ----> 2NH3 (g)

The mole ratio of N2 to H2 is 1:3

Thus, for 0.50 moles of N2, 1.5 moles of H2 should be present. But 0.75 moles of H2 was allowed to react. Meaning that H2 is limiting in this case.

Mole ratio of H2 and NH3 = 3:2

Thus for 0.75 moles H2, the mole of NH3 that would be produced will be:

2 x 0.75/3 = 0.5 moles

More on stoichiometric calculations can be found here: brainly.com/question/8062886

3 0

2 years ago

"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"

Salsk061 [2.6K]

Answer:- The Ka for the acid is $1.53*10^-^5$ .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

$Ka = [H^+][A^-]\frac{1}{HA}$

Where, Ka is the acid ionization constant. Let's plug in the values.

$Ka = \frac{X^2}{0.25-X}$

Let's calculate the value of X first using the equation:

$pH = -log[H^+]$ [/tex]

on taking antilog ob above equation we get:

$[H^+]=10^-^p^H$

$[H^+]=10^-^2^.^7^1$

$[H^+]$ = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

$Ka=\frac{(0.00195)^2}{0.25-0.00195}$

$Ka=1.53*10^-^5$

So, the value of Ka for butyric acid is $1.53*10^-^5$ .

8 0

3 years ago

Read 2 more answers

Given the following unbalanced equation:

antiseptic1488 [7]

<h3>Answer:</h3>

11.84 mol CoF₂

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[RxN - Balanced] CoCl₂ + F₂ → CoF₂ + Cl₂

[Given] 11.84 moles CoCl₂

[Solve] moles CoF₂

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol CoCl₂ → 1 mol CoF₂

<u>Step 3: Stoich</u>

[DA] Set up: $\displaystyle 11.84 \ mol \ CoCl_2(\frac{1 \ mol \ CoF_2}{1 \ mol \ CoCl_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 11.84 \ mol \ CoF_2$

7 0

3 years ago

Drinking water may contain a low concentration of lead ion (Pb2 ) due to corrosion of old lead pipes. The EPA has determined tha

Lynna [10]

Answer:

3.861x10⁻⁹ mol Pb⁺²

Explanation:

We can <u>define ppm as mg of Pb²⁺ per liter of water</u>.

We<u> calculate the mass of lead ion in 100 mL of water</u>:

100.0 mL ⇒ 100.0 / 1000 = 0.100 L

0.100 L * 0.0080 ppm = 8x10⁻⁴ mg Pb⁺²

Now we <u>convert mass of lead to moles</u>, using its molar mass:

8x10⁻⁴ mg ⇒ 8x10⁻⁴ / 1000 = 8x10⁻⁷ g

8x10⁻⁷ g Pb²⁺ ÷ 207.2 g/mol = 3.861x10⁻⁹ mol Pb⁺²

4 0

3 years ago

QUESTION 7

Temka [501]

The student's test average is : 0.92

Total score obtained = 369

Number of tests = 4

Since Each test is exactly 100 points ;

The total score obtainable is (100 × 4) = 400

Average = score obtained / total score obtainable

Average = 369 / 400

Average = 0.9225

Average = 0.92

Learn more : brainly.com/question/9416487?referrer=searchResults

5 0

3 years ago

Lagta ha bhaijaan sa sab bahenjaan naraz haan koi bat ni kr raha​

Lagta ha bhaijaan sa sab bahenjaan naraz haan koi bat ni kr raha