If mixture didn't exist how would life be different PLEASEEE QUICKKKK

In Haber’s process, 30 moles of hydrogen and 30 moles of nitrogen react to make ammonia. If the yield of the product is 50%, wha

devlian [24]

Answer:

Therewill be produced 170.6 grams NH3, there will remain 25 moles of N2, this is 700 grams

Explanation:

Step 1: Data given

Number of moles hydrogen = 30 moles

Number of moles nitrogen = 30 moles

Yield = 50 %

Molar mass of N2 = 28 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate limiting reactant

For 1 mol of N2, we need 3 moles of H2 to produce 2 moles of NH3

Hydrogen is the limiting reactant.

The 30 moles will be completely be consumed.

N2 is in excess. There will react 30/3 =10 moles

There will remain 30 -10 = 20 moles (this in the case of a 100% yield)

In a 50 % yield, there will remain 20 + 0,5*10 = 25 moles. there will react 5 moles.

Step 4: Calculate moles of NH3

There will be produced, 30/ (3/2) = 20 moles of NH3 (In case of 100% yield)

For a 50% yield there will be produced, 10 moles of NH3

Step 5: Calculate the mass of NH3

Mass of NH3 = mol NH3 * Molar mass NH3

Mass of NH3 = 20 moles * 17.03

Mass of NH3 = 340.6 grams = theoretical yield ( 100% yield)

Step 6: Calculate actual mass

50% yield = actual mass / theoretical mass

actual mass = 0.5 * 340.6

actual mass = 170.3 grams

Step 7: The mass of nitrogen remaining

There remain 20 moles of nitrogen + 50% of 10 moles = 25 moles remain

Mass of nitrogen = 25 moles * 28 g/mol

Mass of nitrogen = 700 grams

6 0

4 years ago

Read 2 more answers

Adolfo drank some red punch and grimaced at the flavor. His little brother had made the pitcher of punch with too much sugar. Wh

drek231 [11]

I believe that the saliva interacting with sugar molecules is what allowed Adolfo’s brain to register the extra-sugary flavor of the punch

7 0

3 years ago

1) How many atoms are in 0.54 moles of Cu? show work

mario62 [17]

<h3>Answer:</h3>

3.3 × 10²³ atoms Cu

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

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[Given] 0.54 moles Cu

[Solve] atoms Cu

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

[DA] Set up: $\displaystyle 0.54 \ mol \ Cu(\frac{6.022 \cdot 10^{23} \ atoms \ Cu}{1 \ mol \ Cu})$
[DA] Multiply [Cancel out units]: $\displaystyle 3.25188 \cdot 10^{23} \ atoms \ Cu$

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

3.25188 × 10²³ atoms Cu ≈ 3.3 × 10²³ atoms Cu

7 0

3 years ago

Which one of the following compounds will NOT be soluble in water? Which one of the following compounds will NOT be soluble in w

Nadusha1986 [10]

Answer:

$BaSO_{4}$ will be not soluble in water

Explanation:

LiOH is a strong base. Hence it gets completely dissociated in aqueous solution.

$NaNO_{3}$ is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.

$MgCl_{2}$ is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.

$K_{2}S$ is a strong electrolyte. Hence it gets completely dissociated in aqueous solution.

$BaSO_{4}$ is a sparingly soluble salt. Hence it is not dissociated and hence dissolved in water. This is due to the fact that both $Ba^{2+}$ and $SO_{4}^{2-}$ ions are similar in size. Hence crystal structure of $BaSO_{4}$ is quite stable. Hence $BaSO_{4}$ is reluctant to undergo any dissociation in aqueous solution.

5 0

3 years ago

(a) Find the concentration of electrons and holes in a sample of germanium that has a concentration of donor atoms equal to 1015

matrenka [14]

Answer:

a) Germanium = 5.76 x 〖10〗^11 〖cm〗^(-3) , Semiconductor is n-type.

b) Silicon = 2.25 x 〖10〗^5 〖cm〗^(-3) , Semiconductor is n-type.

For clear view of the answers: Please refer to calculation 5 in the attachments section.

Explanation:

So, in order to find out the concentration of holes and electrons in a sample of germanium and silicon which have the concentration of donor atoms equals to 〖10〗^15 〖cm〗^(-3). We first need to find out the intrinsic carrier concentration of silicon and germanium at room temperature (T= 300K).

Here is the formula to calculate intrinsic carrier concentration: For calculation please refer to calculation 1:

So, till now we have calculated the intrinsic carrier concentration for germanium and silicon. Now, in this question we have been given donor concentration (N_d) (N subscript d), but if donor concentration is much greater than the intrinsic concentration then we can write: Please refer to calculation 2.

So, now we have got the concentration of electrons in both germanium and silicon. Now, we have to find out the concentration of holes in germanium and silicon (p_o). (p subscript o)

Equation to find out hole concentration: Please refer to calculation 3. and Calculation 4. in the attachment section.

Good Luck Everyone! Hope you will understand.

6 0

3 years ago

If mixture didn't exist how would life be different​PLEASEEE QUICKKKK

If mixture didn't exist how would life be differentPLEASEEE QUICKKKK