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2 years ago

22.4 liters is the same as: cm3 and mL

Chemistry

2 answers:

beks73 [17]2 years ago

8 0

Answer;

22400 cm³ or 22400 mL

Explanation;

1000 mL are equivalent to 1 Liter

But; cubic centimeters is the same as milliliters

Thus; 1000 cubic centimeters are equivalent to 1000 mL and also equivalent to 1 Liter.

1000 cm³ = 1000 mL = 1 L

Therefore; 22.4 Liters will be equivalent to;

22.4 × 1000 = 22400 cm³ or 22400 mL

harina [27]2 years ago

8 0

To find mL you multiply by 1000 So 22.4 L*1000= 22400mL
mL is the same as cm3 so you have 22400cm3 .

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Convert 2.1 mole of Al2(SO4)3 ionic units to a number of particles.

Semenov [28]

1.26 x 10²⁴particles

Explanation:

Given parameters:

Number of moles of the compound = 2.1 moles

Unknown:

Number of particles contained in the compound = ?

Solution:

A mole can be defined as the amount of a substance that contains Avogadro's number of particles 6.02 x 10²³.

The particles mentioned here can be atoms, molecules, formula units, ions, electrons, protons, neutrons and so forth.

To solve for the number of moles:

Number of particles = number of moles x 6.02 x 10²³

Number of particles = 2.1 x 6.02 x 10²³ = 12.64 x 10²³ = 1.26 x 10²⁴particles

Learn more:

mole calculation brainly.com/question/1841136

#learnwithBrainly

8 0

2 years ago

How much maleic anhydride would you need to react 178 mg of anthracene? Assume 1:1 ratio from maleic anhydride to anthracene.

Licemer1 [7]

Answer:

(1) 0.10 (2) 17.8 g

Explanation:

Since the reaction ratio is 1:1 what we need is to convert the given masses to moles and you will have the answer:

MW anthracene = 178.23 g/mol

MW maleic anhydride = 98.06 g/mol

a) mass anthracene = 178 mg x 1 g/ 1000 mg = 0.178 g anthracene

Moles anthracene = 0.178 g anthracene/ 178.23 g/mol

= 0.001 mol anthracene

0.001 mol anthracene x 1 mol maleic acid/mol anthracene

= 0.001 mol maleic anhydride

mass maleic anhydride = 0.001 mol x 98.06 g/mol = 0.10 g

b) moles maleic anhydride = 9.8 g/ 98.06 g/mol = 0.099 moles

0.099 moles maleic anhydride x 1 mol anthracene/mol maleic anhydride =

0.099 mol anthracene

g anthracene = 0.10mol x 178 g/mol = 17.8 g

8 0

2 years ago

The volume of a gas is 36.0 ml at 10.0°c and 4.50 atm. at what temperature (°c) will the gas have a pressure of 3.50 atm and a v

galben [10]

Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.

5 0

2 years ago

Be sure to answer all parts. For each reaction, find the value of ΔSo. Report the value with the appropriate sign. (a) 3 NO2(g)

Liono4ka [1.6K]

Answer:

a. -268.13 J/K

b. -279.95 J/K

c. + 972.59 J/K

Explanation:

The value of the change in entropy (ΔS°) can be calculated by:

ΔS° = ∑n*S° products - ∑n*S° reactants, where n is the stoichiometric number of moles.

The values of S° for each substance can be found on a thermodynamic table.

a. 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

S°, NO2(g) = 240.06 J/mol.K

S°, H2O(l) = 69.91 J/mol.K

S°, HNO3(l) = 155.60 J/mol.K

S°, NO(g) = 210.76 J/mol.K

ΔS° = (210.76 + 2*155.60) - (3*240.06 + 69.91)

ΔS° = -268.13 J/K

b. N2(g) + 3F2(g) → 2NF3(g)

S°, N2(g) = 191.61 J/mol.K

S°, F2(g) = 202.78 J/mol.K

S°, NF3(g) = 260.0 J/mol.K

ΔS° = (2*260.0 ) - (191.61 + 3*202.78)

ΔS° = -279.95 J/K

c. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(g)

S°, C6H12O6(s) = 212 J/mol.K

S°, O2(g) = 205.138 J/mol.K

S°, CO2(g) = 213.74 J/mol.K

S°, H2O(g) = 188.83 J/mol.K

ΔS° = (6*213.74 + 6*188.83) - (212 + 6*205.138)

ΔS° = +972.59 J/K

3 0

2 years ago

Express the quantity 556.2 x 10^-12 in each units A.) ms B.) ns C.) ps D.) fs

zavuch27 [327]

The answer:
we should know the meaning of each abbreviation:
ms means millisecond, its value is 10^-3 s
ns means means nanosecond, its value is 10^-9 s
ps means picosecond, its value is 10^-12 s
fs means femtosecond, its value is 1x 10^15 s

Expressions of the quantity 556.2 x 10^-12 are
556.2 x 10^-12 =556.2 ps
556.2 x 10^-12 =556.2 x 10^-9 x 10^-3= 556.2 x 10^-9 ms
556.2 x 10^-12 = 556.2 x 10^-3 x 10^-9 = 556.2 x 10^-3 ns
556.2 x 10^-12 = 556.2 x 10^- 27 x 10^15 = 556.2 x 10^- 27 fs

6 0

2 years ago