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VLD [36.1K]
4 years ago
15

Why is it not a good idea to use a stopper when combining two unknown substances in  test?

Chemistry
1 answer:
Colt1911 [192]4 years ago
6 0
Hi!

You would not want to use a stopper when<span> combining two unknown substances because the gasses made from the combination could end up creating to much pressure and explode.</span>
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1. 17.0 grams of xenon hexafluoride is in a solid container. How many milliliters of that gas
BlackZzzverrR [31]

Answer: The volume of gas is 3020 ml

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 821.4 torr =  1.08 atm     (760 torr = 1atm)

V = Volume of gas in L = ?

n = number of moles = \frac{\text {given mass}}{\text {Molar mass}}=\frac{17.0g}{245.28g/mol}=0.069mol

R = gas constant =0.0821Latm/Kmol

T =temperature =302.7^0C=(302.7+273)K=575.7K

V=\frac{nRT}{P}

V=\frac{0.069mol\times 0.0821Latm/K mol\times 575.7K}{1.08atm}=3.02L=3020ml

Thus volume of gas is 3020 ml

4 0
3 years ago
Determine how many millilitres of a 4.25 M HCl solution are needed to react completely with 8.75 g CaCO3?
Helen [10]

Answer:

41 mL

Explanation:

Given data:

Milliliter of HCl required = ?

Molarity of HCl solution = 4.25 M

Mass of CaCO₃ = 8.75 g

Solution:

Chemical equation:

2HCl + CaCO₃      →    CaCl₂ + CO₂ + H₂O

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 8.75 g / 100.1 g/mol

Number of moles = 0.087 g /mol

Now we will compare the moles of  CaCO₃ with HCl.

                      CaCO₃         :          HCl

                          1               :            2

                      0.087           :         2/1×0.087 = 0.174 mol

Volume of HCl:

Molarity = number of moles / volume in L

4.25 M = 0.174 mol / volume in L

Volume in L = 0.174 mol /4.25 M

Volume in L = 0.041 L

Volume in mL:

0.041 L×1000 mL/ 1L

41 mL

8 0
3 years ago
A gas is collected in a 34.3L container at a temperature of 31.5°C. Later, the container has a volume of 29.2L, a temperature of
goldenfox [79]

Answer:

108 kPa  

Step-by-step explanation:

To solve this problem, we can use the <em>Combined Gas Laws</em>:

p₁V₁/T₁ = p₂V₂/T₂             Multiply each side by T₁

   p₁V₁ = p₂V₂ × T₁/T₂      Divide each side by V₁

      p₁ = p₂ × V₂/V₁ × T₁/T₂

Data:

p₁ = ?;                 V₁ = 34.3 L; T₁ = 31.5 °C

p₂ = 122.2 kPa; V₂ = 29.2 L; T₂ = 21.0 °C

Calculations:

(a) Convert temperatures to <em>kelvins </em>

T₁ = (31.5 + 273.15) K = 304.65 K

T₂ = (21.0 + 273.15) K = 294.15 K

(b) Calculate the <em>pressure </em>

p₁ = 122.2 kPa × (29.2/34.3) × (304.65/294.15)  

   = 122.2 kPa × 0.8542 × 1.0357

   = 108 kPa

4 0
3 years ago
A student determined that a 50mL sample of an unknown liquid had a density of 1 g/mL. A 100mL sample of the same liquid would ha
Lyrx [107]

Answer: False

Explanation: The density of the liquid will always be the same, no matter how much of the liquid you have.

For example, the density of the water in a pool is the same density as the water in a small cup.

4 0
3 years ago
Please help i got no clue how to do dis
mamaluj [8]

Answer:

Sulfur's Atomic #: 16

Fluorine's Atomic #: 9

Antimony's Atomic #: 51

Silver's Atomic #: 47

Rubidium's Atomic #: 37

Copper's Atomic #: 29

Tin's Atomic #: 50

Mercury's Atomic #: 80

Promethium's Atomic #: 61

3 0
3 years ago
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