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Solnce55 [7]
3 years ago
10

What is the work required for the separation of air (21-mol-% oxygen and 79-mol-% nitrogen) at 25°C and 1 bar in a steady-flow p

rocess into product streams of pure oxygen and nitrogen, also at 25°C and 1 bar, of the thermodynamic efficiency of the process is 5% and if Tσ = 300 K
Chemistry
1 answer:
Ronch [10]3 years ago
4 0

Answer:

4oo

Explanation:

it

You might be interested in
Which of the following correctly lists the five atoms in order of increasing size (smallest to
ICE Princess25 [194]

Answer:

F - O - S - Mg - Ba

Explanation:

as you move left to right on the periodic table the number of electrons increase.

7 0
3 years ago
3. What do you notice about the temperature of the substance in the test tube after it is placed in the beaker of water?
34kurt

Answer:

C: The temperature of the substance increases as it sits in the beaker of water

Explanation:

This question was taken from a video where an attempt was made to investigate the changes in temperature when a substance undergoes change from it's solid phase to its liquid phase.

To do this, as seen in the video online, it shows a solid substance in a test tube being placed in a beaker of water.

From observation, the water in the beaker has a warmer temperature than the solid substance present in the test tube and this in turn makes the test tube gradually increase in temperature.

Thus, the solid substance will as well increase increase in temperature when it is placed in the beaker of water.

3 0
2 years ago
B) Quelle est la masse de tétraoxyde de trifer (Fe3O4) produite si 3,60 moles de trioxyde de
hram777 [196]

Answer:

626,4 g de Fe₃O₄

Explanation:

Nous commencerons par écrire l'équation équilibrée de la réaction entre le fer (Fe) et le trioxyde d'aluminium. Ceci est donné ci-dessous:

9Fe + 4Al₂O₃ -> 3Fe₃O₄ + 8Al

De l'équation équilibrée ci-dessus,

4 moles d'Al₂O₃ ont réagi pour produire 3 moles de Fe₃O₄.

Par conséquent, 3,6 moles d'Al₂O₃ réagiront pour produire = (3,6 × 3) / 4 = 2,7 moles de Fe₃O₄

Ainsi, 2,7 moles de Fe₃O₄ sont produites à partir de la réaction.

Enfin, nous déterminerons la masse massique de Fe₃O₄ produite par la réaction. Ceci peut être obtenu comme suit:

Mole de Fe₃O₄ = 2,7 moles

Masse molaire de Fe₃O₄ = (3 × 56) + (4 × 16)

= 168 + 64

= 232 g / mol

Masse de Fe₃O₄ =?

Mole = masse / masse molaire

2,7 = Masse de Fe₃O₄ / 232

Croiser multiplier

Masse de Fe₃O₄ = 2,7 × 232

Masse de Fe₃O₄ = 626,4 g

Par conséquent, 626,4 g de Fe₃O₄ sont produits à partir de la réaction

7 0
2 years ago
1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in
FrozenT [24]

Answer:

1)  <u>10.0 moles of NO</u>

<u>2) 25 moles of NaCl</u>

3) <u>1200 moles of CO2</u>

<u>4) 1.03 moles of MgO</u>

<u>5) 0.72 moles H2</u>

<u>6) 1041.15 grams BaCl2</u>

<u>7) </u>9.55 grams MgO

8) <u>45.5 grams Au</u>

<u>9 )14.93 grams AlCl3</u>

Explanation:

1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in the following composition reaction?

N2 + O2 → 2NO

For 1 mol N2 we need 1 mol O2 to produce 2 moles of NO

For 5.0 moles of N2 we need 5.0 moles of O2 to produce <u>10.0 moles of NO</u>

2. The neutralization of an acid with a base is a double replacement reaction in which a salt and water are formed. If you start with 25 moles of HCl and neutralize it with  NaOH how many moles of NaCl will be formed?

HCl + NaOH → NaCl + H2O

For 1 mol HCl we need 1 mol NaOH to produce 1 mol of NaCl and 1 mol H2O

For 25 moles of HCl we need 25 moles of NaOH to produce <u>25 moles of NaCl</u> and 25 moles of H2O

3. A car burns gasoline (octane – C8H18) with oxygen. If you drive to Salt Lake and  burn 150 moles of octane how many moles of carbon dioxide are you producing?

2C8H18 + 25O2 → 16CO2 + 18H2O

For 2 moles of octane we need 25 moles of O2 to produce 16 moles of CO2 and 18 moles of H2O

For 150 moles of octane we need 25*75 = 1875 moles of O2

To produce 16*75 = <u>1200 moles of CO2</u> and 18*75= 1350 moles

4. If 25 gram of magnesium combines with oxygen in a composition reaction, how  many moles of magnesium oxide will be formed?

2Mg + O2 → 2MgO

Moles of Mg = 25.0 g/24.3 g/mol = 1.03 moles

For 2 moles Mg we need 1 mol O2 to produce 2 moles MgO

For 1.03 moles Mg we'll have <u>1.03 moles of MgO</u>

<u />

<u />

<u />

5. . Lithium reacts with water in a single replacement reaction. How many moles of  hydrogen gas a produced by 10 grams of lithium?

2Li + 2H2O → 2LiOH + H2

Moles Li = 10.0 grams/ 6.94 g/mol = 1.44 moles

For 2 moles Li we need 2 mole H2O to produce 2 moles LiOH and 1 mol H2

For 1.44 moles Li we need 1.44 moles H2O to produce 1.44 moles H2O and <u>0.72 moles H2</u>

<u />

<u />

6. Barium chloride reacts with sodium sulfate in a double replacement reaction. How  many grams of barium chloride are required to react with 5 moles of sodium sulfate?

BaCl2 + Na2SO4 → BaSO4 + 2NaCl

For 1 mol of BaCl2 we need 1 mol of Na2SO4 to produce 1 mol of BaSO4 and 2 moles NaCl

For 5 moles Na2SO4 we need 5 moles BaCl2

mass BaCl2 = 5 moles * 208.23 g/mol = <u>1041.15 grams BaCl2</u>

7. Magnesium carbonate when heated decomposes to form magnesium oxide and carbon dioxide. How many grams of magnesium oxide will be formed if 20 grams of  magnesium carbonate are heated?

MgCO3 → MgO + CO2

Moles MgCO3 = 20.0 grams / 84.31 g/mol

Moles MgCO3 = 0.237 moles

For 1 mol MgCO3 we'll have 1 mol MgO and 1 mol CO2

For 0.237 moles MgCO3 we'll have 0.237 moles MgO and 0.237 moles CO2

Mass MgO = 0.237 moles * 40.30 g/mol = 9.55 grams MgO

8. If 70 grams of gold III chloride decomposes into its elements, how many grams of  gold will be produced?

2AuCl3 → 2Au + 3Cl2

Moles AuCl3 = 70 grams / 303.33 g/mol = 0.231 moles

For 2 moles AuCl3 we'll have 2 moles gold and 3 moles Cl2

For 0.231 moles AuCl3 we'll have 0.231 moles gold

Mass of gold =  0.231 moles * 196.97 g/mol = <u>45.5 grams Au</u>

<u />

<u />

9. Chlorine is more reactive element than bromine, thus chlorine will replace bromine in compound through a single replacement reaction. If 30 grams of aluminum bromide react with chlorine in this fashion how many grams of aluminum chloride will be formed?

2AlBr3 + 3Cl2 → 2AlCl3 + 3Br2

Moles AlBr3 = 30 g /266.69 g/mol = 0.112 moles

For 2 moles AlBr2 we need 3 moles Cl2 to produce 2 moles AlCl3 and 3 moles Br2

For 0.112 moles AlBr3 we need 3/2 * 0.112 = 0.168 moles of Cl2

To produce 0.112 moles of AlCl3 and 0.168 moles of Br2

Mass AlCl3 = 0.112 moles * 133.34 g/mol = <u>14.93 grams AlCl3</u>

8 0
3 years ago
HELPP! pls
pshichka [43]

Answer:

"2.48 mole" of H₂ are formed. A further explanation is provided below.

Explanation:

The given values are:

Mole of Al,

= 3.22 mole

Mole of HBr,

= 4.96 mole

Now,

(a)

The number of mole of H₂ are:

⇒  \frac{Mole \ of \ H_2}{3} =\frac{Mole \ of HBr}{6}

or,

⇒  Mole \ of \ H_2=\frac{1}{2}\times Mole \ of \ HBr

⇒                      =\frac{1}{2}\times 4.96

⇒                      =2.48 \ mole

(b)

The limiting reactant is:

= HBr

(c)

The excess reactant is:

= Al

6 0
2 years ago
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