In a physical change the make up of matter is changed. O True O False

Nitrogen gas reacts with hydrogen gas to produce ammonia. How many liters of hydrogen gas at 95kPa and 15∘C are required to prod

viktelen [127]

Answer:

222.30 L

Explanation:

We'll begin by calculating the number of mole in 100 g of ammonia (NH₃). This can be obtained as follow:

Mass of NH₃ = 100 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 100 / 17

Mole of NH₃ = 5.88 moles

Next, we shall determine the number of mole of Hydrogen needed to produce 5.88 moles of NH₃. This can be obtained as follow:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, Xmol of H₂ is required to p 5.88 moles of NH₃ i.e

Xmol of H₂ = (3 × 5.88)/2

Xmol of H₂ = 8.82 moles

Finally, we shall determine the volume (in litre) of Hydrogen needed to produce 100 g (i.e 5.88 moles) of NH₃. This can be obtained as follow:

Pressure (P) = 95 KPa

Temperature (T) = 15 °C = 15 + 273 = 288 K

Number of mole of H₂ (n) = 8.82 moles

Gas constant (R) = 8.314 KPa.L/Kmol

Volume (V) =?

PV = nRT

95 × V = 8.82 × 8.314 × 288

95 × V = 21118.89024

Divide both side by 95

V = 21118.89024 / 95

V = 222.30 L

Thus the volume of Hydrogen needed for the reaction is 222.30 L

8 0

3 years ago

Which of the following is an acid? Be(OH)2 HCl LiBr NH3

Lyrx [107]

Answer:

HCl is the correct answer

4 0

3 years ago

Read 2 more answers

Why calcium oxide and magnesium oxide used as soil treatment

BartSMP [9]

Answer:

Chemically speaking, lime refers only to calcium oxide

(CaO); however, in common usage the term includes the calcination products of calcitic and dolomitic limestones. Calcitic (high-calcium) limes are produced by calcination of

calcareous materials (e.g., calcitic limestone, calcite,

oyster shells, and chalk) containing from 95 to 99 percent

calcium carbonate (CaCO^). Dolomitic limes are produced from

dolomitic limestone or dolomite which contains from 30 to ^+0

percent magnesium carbonate (MgCO^), the rest being calcium

carbonate.

At atmospheric pressure, calcite in limestone decomposes

at approximately 900°C to form CaO and COg. The decomposition of dolomite, CaMg is a two-stage process. At

temperatures between 650°C to 750°C dolomite decomposes to

form MgO, CO^ and CaCO^. It is necessary to raise the temperature to 900°C to decompose the CaCO^ (15, 35)» This

phenomenon is extremely important, as is shown later.

Various investigators have studied the effects of stone

size, temperature, and time of calcination of commercial

Explanation:

4 0

3 years ago

Will mark the brainiest later for correct answers! Please show work.

Elodia [21]

Answer:

According to avogadro's law, 1 mole of every substance contains avogadro's number $6.023\times 10^{23}$ of particles and weighs equal to its molecular mass.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

$\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadros number}}$

a. moles in 14.08 g of $C_{12}H_{22}O_{11}$ = $\frac{14.08g}{342.3g/mol}=0.04113moles$

molecules in 14.08 g of $C_{12}H_{22}O_{11}$ = $0.04113\times 6.023\times 10^{23}=0.2477\times 10^{23}$

b. moles in 17.75 g of NaCl = $\frac{17.75g}{58.5g/mol}=0.3034moles$

molecules in 17.75 g of $NaCl$ = $0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}$

formula units 17.75 g of $NaCl$ = $0.3034\times 6.023\times 10^{23}=1.827\times 10^{23}$

c. moles in 20.06 g of $CuSO_4.5H_2O$ = $\frac{20.06g}{249.68g/mol}=0.08034moles$

formula units in 20.06 g of $CuSO_4.5H_2O$ = $0.08034\times 6.023\times 10^{23}=0.4839\times 10^{23}$

7 0

3 years ago

How many milligrams of product can be produced from the complete Diels-Alder reaction of 180. mg of anthracene and 100. mg of ma

ZanzabumX [31]

Answer: It will be produced 276,3 mg of product

Explanation: The reaction of anthracene (C14H10) and maleic anhydride (C4H2O3) produce a compound named 9,10-dihydroanthracene-9,10-α,β-succinic anhydride (C18H12O3), as described below:

C14H10 + C4H2O3 → C18H12O3

The reaction is already balanced, which means to produce 1 mol of C18H12O3 is necessary 1 mol of anthracene and 1 mol of maleic anhydride.

1 mol of C14H10 equals 178,23 g. As it is used 180 mg of that reagent, we have 0,001 mol of anthracene. With it, the reaction produces 0,001 mol of C18H12O3.

As 1 mol of C18H12O3 equals 276,3 g, the mass produced is 276,3 mg.

3 0

3 years ago