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Firlakuza [10]
3 years ago
13

The lithosphere and the asthenosphere are made up of solid,liquid,or gas?ASAP

Chemistry
1 answer:
REY [17]3 years ago
5 0
Gas, hope this helps
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An atom in the ground state contains 8 valence electrons this atom is classified as a
Alekssandra [29.7K]
Noble Gas. Metals have 1 or 2 Valence Electrons. Halogens have 7 Valence Electrons. Semi-Metals can have different amounts.
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3 years ago
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If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0
3 years ago
Consider the reaction between acetylene, c2h2, and oxygen in a welding torch: 2c2h2(g) + 5o2(g) → 4co2(g) + 2h2o(g) if 5.4 moles
fredd [130]
<span>                                      2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)
   from the reaction     2 mol                              4 mol
from the problem      5.4 mol                         10.8 mol

M(CO2) = 12.0 +2*16.0 = 44.0 g/mol
10.8 mol CO2 * 44.0 g CO2/1 mol CO2 = 475.2 g CO2 </span>≈480 = 4.8 * 10² g
Answer is  C. 4.8*10² g.
4 0
3 years ago
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PLEASE HELP :)) WILL GIVE BRAINLIEST
hjlf

Answer:

the answer is true

5 0
3 years ago
Calculate the volume in liters of a barium acetate solution that contains of barium acetate . Be sure your answer has the correc
Ilya [14]

Answer:

1.09 L

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the volume in liters of a 0.360 mol/L barium acetate solution that contains 100 g of barium acetate. Be sure your answer has the correct number of significant digits.</em>

<em />

The molar mass of barium acetate is 255.43 g/mol. The moles corresponding to 100 grams are:

100 g × (1 mol/255.43 g) = 0.391 mol

0.391 moles of barium acetate are contained in an unknown volume of a 0.360 mol/L barium acetate solution. The volume is:

0.391 mol × (1 L/0.360 mol) = 1.09 L

5 0
3 years ago
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