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Valentin [98]
3 years ago
12

3. Give three examples of a pure substances

Chemistry
2 answers:
liq [111]3 years ago
7 0
Tin, sulfur, and water
Nadya [2.5K]3 years ago
5 0

Answer:

There will be weight, there will be volume, there will be height

Explanation:

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5. Write word equations for the following reactions
Andrews [41]

Answer:

a) Sulphur + Oxygen → Sulphur dioxide

b) Carbon + Oxygen → Carbon dioxide

c) Sulphur + Iron → Iron sulphide

6 0
3 years ago
What is the theory that can explain the model of Pangaea?
NISA [10]
The bones of the same animal found out continents far away from each other
4 0
3 years ago
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
When 3.00 g of sulfur are combined with 3.00 g of oxygen, 6.00 g of sulfur dioxide (SO2) are formed. What mass of oxygen would b
bazaltina [42]
Actually, we can answer the problem even without the first statement. All we have to do is write the reaction for the production of sulfur trioxide.

2 S + 3 O₂ → 2 SO₃

The stoichiometric calculations is as follows:

6 g S * 1 mol/32.06 g S = 0.187 mol S
Moles O₂ needed = 0.187 mol S * 3 mol O₂/2 mol S = 0.2805 mol O₂
Since the molar mas of O₂ is 32 g/mol,
Mass of O₂ needed = 0.2805 mol O₂ * 32 g/mol = 8.976 g O₂
4 0
3 years ago
How many grams of NaCl are needed to prepare 60 g of a 6.0% solution?
AlekseyPX

Answer:

3.6 grams of NaCl are needed

Explanation:

Percent solution are solutions whose concentrations are expressed in percentages. The amount(either weight or volume) of a solute is expressed as a percentage of the total weight or volume of solution. Percent solutions can either be expressed as  weight/volume % (wt/vol % or w/v %), weight/weight % (wt/wt % or w/w %), or volume/volume % (vol/vol % or v/v %).

A 6.0% wt/wt % solution contains 6 g of solute in 100 g of solution

Therefore, a 100 g solution contains 6.0 g of solute.

60 g of 6.0% solution will contain  60 g solution * 6.0 g solute/ 100 g solution

Mass of NaCl present =  3.6 g  of NaCl

3 0
3 years ago
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