All categories

3 years ago

Under which conditions would you expect freezing rain

1 answer:

3 0

There has to be a layer of cold enough air that is shallow enough so that the rain falling through it doesn't have time to freeze before it reaches the surface.

Hope this helps!

You might be interested in

Give 10 chemical properties of common polymers

algol [13]

Answer:

Some of the useful properties of various engineering polymers are high strength or modulus to weight ratios (light weight but comparatively stiff and strong), toughness, resilience, resistance to corrosion, lack of conductivity (heat and electrical), color, transparency, processing, and low cost

Explanation:

3 0

3 years ago

Read 2 more answers

What volume of 2.50 M Na2SO4 solution contains 71.0 grams of Na2SO4?

statuscvo [17]

Answer

0.02

Explanation:

7 0

4 years ago

2AlF3 + 3K2O → 6KF + Al2O3 How many grams of AlF3would it take to make 15.524 g of KF?

mr_godi [17]

<h3>Answer:</h3>

7.4797 g AlF₃

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Stoichiometry

Reading a Periodic Table

<h3>Explanation:</h3>

Step 1: Define

[RxN] 2AlF₃ + 3K₂O → 6KF + Al₂O₃

[Given] 15.524 g KF

Step 2: Identify Conversions

[RxN] 2 mol AlF₃ = 6 mol KF

Molar Mass of Al - 26.98 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of K - 39.10 g/mol

Molar Mass of AlF₃ - 26.98 + 3(19.00) = 83.98 g/mol

Molar Mass of KF - 39.10 + 19.00 = 58.10 g/mol

Step 3: Stoichiometry

Set up: $\displaystyle 15.524 \ g \ KF(\frac{1 \ mol \ KF}{58.10 \ g \ KF})(\frac{2 \ mol \ AlF_3}{6 \ mol \ KF})(\frac{83.98 \ g \ AlF_3}{1 \ mol \ AlF_3})$
Multiply/Divide: $\displaystyle 7.47966 \ g \ AlF_3$

Step 4: Check

Follow sig fig rules and round. We are given 5 sig figs.

7.47966 g AlF₃ ≈ 7.4797 g AlF₃

3 0

3 years ago

Read 2 more answers

Calculate the vapor pressure (in torr) of a 35°C solution of ethanol and propanol where the mole fraction of propanol is .75. As

ASHA 777 [7]

The vapor pressure (in torr) of a 35°C solution : 52 torr

<h3>Further explanation</h3>

Given

x propanol = 0.75

P°ethanol=100 torr

P°propanol=36 torr

Required

the vapor pressure

Solution

Raoult's Law

P solution = xA.P°A +xB.P°B

x ethanol = 1- 0.75 = 0.25

Input the value :

P solution = 0.25 x 100 + 0.75 x 36

P solution = 25 + 27

P solution = 52 torr

7 0

3 years ago

A 0.025-g sample of a compound composed of Boron and hydrogen with the molecular mass of 28 AMU burns spontaneously when exposed

nalin [4]

Answer:

The empirical formula is BH₃ and molecular formula is B₂H₆

Explanation:

To know the number of borons you must obtain the moles number of the initial and B₂O₃ compound. So:

Moles of initial compound:

0,025 g of X × ( 1 mol / 28 g) = 8,9 × 10⁻⁴ moles

Moles of B₂O₃ compound:

AMU of B₂O₃ = (2 × 10,8 g/mol + 3 × 16 g/mol) = 69,6 g/mol

0,063 g ( 1 mol B₂O₃ / 69,6 g) = 9,1 × 10⁻⁴ moles

As moles number of initial and final compounds are the same, the number of borons must be equals. So, our compound has two borons.

These two borons weight: 2 × 10,8 g/mol = 21,6 g/mol

If UMA number of our compound is 28 g/mol we need, yet,

28 g/mol - 21,6 g/mol = 6,4 g/mol

These 6,4 g/mol comes from hydrogen that weights 1 g/mol. So, we have 6 hydrogens.

Thus, the molecular formula is B₂H₆

The empirical formula is the simplest way to represent the atoms of a chemical compound. If we divide the molecular formula in two, we will obtain the empirical formula: BH₃

I hope it helps!

4 0

3 years ago