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Veronika [31]
3 years ago
8

Could a mixture be made up of only one element and no compounds​

Chemistry
1 answer:
Firdavs [7]3 years ago
5 0

Yes. A mixture can be made up of just elements without the use of a compound, you can also just use compounds in a mixture without any elements

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In preparation for a demonstration, your professor brings a 1.50−L bottle of sulfur dioxide into the lecture hall before class t
solmaris [256]

Answer:

4.81 moles

Explanation:

The total pressure of the gas = Pressure at which gauge reads zero + pressure read by it.

Pressure at which gauge reads zero = 14.7 psi

Pressure read by the gauge = 988 psi

Total pressure = 14.7 + 988 psi = 1002.7 psi

Also, P (psi) = P (atm) / 14.696

Pressure = 1002.7 / 14.696  = 68.2297 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 1.50 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

68.2297 atm × 1.5 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 4.81 moles

4 0
3 years ago
HELP ASAP !!!!!
Anna35 [415]

Answer:

Explanation:

Take a random sample of nuts from the jar.  Let's take two handfuls, after shaking the jar and mixing the nuts thoroughly.  Separate the nuts into almonds and cashews.  Count each pile, then do the following calculation (these numbers are random, for example only).

                <u> Count</u>   <u>Percentage %</u>

Almonds       38        (38)/(87)x100

Cashews      <u> 49</u>         49/87x100

                      87          87/87 = 100%

Ratio of Almonds to Cashews:  <u>38/49</u>

3 0
3 years ago
Non chemical industry examples​
Luba_88 [7]

Answer:

Light , heat , and sound are examples.

4 0
3 years ago
Read 2 more answers
4. How many atom are in <br> C3H5(NO3)3
shepuryov [24]

Answer:

carbon = 3

hydrogen=5

nitrogen=3

Oxygen=9

8 0
2 years ago
Calculate the vapor pressure at 35ºC of a solution made by dissolving 20.2 g of sucrose (C12H22O11)in 60.5 g of water. The vapor
sukhopar [10]

Answer:

P' = 41.4 mmHg → Vapor pressure of solution

Explanation:

ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')

Xm = Mole fraction for solute (Moles of solvent /Total moles)

Firstly we determine the mole fraction of solute.

Moles of solute → Mass . 1 mol / molar mass

20.2 g . 1 mol / 342 g = 0.0590 mol

Moles of solvent → Mass . 1mol / molar mass

60.5 g . 1 mol/ 18 g = 3.36 mol

Total moles = 3.36 mol + 0.0590 mol = 3.419 moles

Xm = 0.0590 mol / 3.419 moles → 0.0172

Let's replace the data in the formula

42.2 mmHg - P' = 42.2 mmHg . 0.0172

P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)

P' = 41.4 mmHg

5 0
3 years ago
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