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Hoochie [10]
2 years ago
8

What do blizzards consist of? (Select all that apply)

Chemistry
1 answer:
Maru [420]2 years ago
7 0

Answer:

Heavy Snowfall

High Winds

Extremely Low Temperatures

Reasoning:

Lots of snow, kinda self explanatory

high winds are needed to carry the snow that fast

its has to be below freezing for snow to even form

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True or False: If you have 2 moles of water and 2 moles of carbon, you have the same number of molecules.
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Calculate the pH of the following <br> A.0.02mol/L HCI<br>B. 0.1 mol/L NaOH​
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If 125g of KClO3 is heated, what is the total mass of the products?​
Andrej [43]

Given parameters:

Mass of KClO₃  = 125g

Unknown:

Total mass of the products = ?

When  KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;

               2KClO₃  →  2KCl + 3O₂

All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.

The total mass of the products must give us 125g according to this law of conservation of matter.

Now to find the masses of each product,

  1. Find the number of moles of the given reactant:

     Number of moles  = \frac{mass}{molar mass}

  molar mass of  KClO₃  = 39 + 35.5 + 3(16)  = 122.5g/mol

    So number of moles of KClO₃ = \frac{125}{122.5}  = 1.02moles

    2. Now, using this number of moles, find the number of moles of the products using this value;

   2 moles of KClO₃ produced 2 moles of KCl

  1.02 moles of KClO₃ will also produce 1.02moles of KCl

   2 moles of KClO₃ produced 3 moles of O₂

   1.02 moles of KClO₃ will produce   \frac{1.02 x 3} {2} mole = 1.53 moles of O₂

   3. Now find the masses of each product;

Mass  = number of moles x molar mass

  molar mass of KCl  = 39 + 35.5 = 74.5g/mol

  molar mass of O₂  = 16 x 2  = 32g/mol

  Mass of KCl  = 74.5 x 1.02  = 75.99g

  Mass of O₂  = 32 x 1.53 = 48.96g

Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g

                                        = 124.95g

This value is approximately the same as that of mass of  KClO₃

 

7 0
3 years ago
What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)
lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

\Delta T = i \times Kf \times m =   1 \times 6.90 \°C/m  \times 1.324m = 9.14  \°C

where,

  • i: van 't Hoff factor (1 for non-electrolytes)
  • Kf: cryoscopic constant
  • m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

T = 80.26 \° C - 9.14 \° C = 71.12 \° C

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

3 0
2 years ago
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