Unsorted sands have a lower porosity. True O False

6.) 50.0 mol H2O ? molecules

8_murik_8 [283]

<h3>Answer:</h3>

3.01 × 10²⁵ molecules H₂O

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

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

Step 1: Define

50.0 mol H₂O

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

$\displaystyle 50.0 \ mol \ H_2O(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O} )$ = 3.011 × 10²⁵ molecules H₂O

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

3.011 × 10²⁵ molecules H₂O ≈ 3.01 × 10²⁵ molecules H₂O

5 0

3 years ago

Help please first person to get it right gets my old Netflix account

spin [16.1K]

Butterflies are cold-blooded and need the light from the sun to warm the muscles they use to fly. Not only do butterflies like the sun, the plants the they thrive on need full direct sun. Most plants need at least 8 hours of sunlight to bloom properly and provide enough nectar.

7 0

3 years ago

Read 2 more answers

What do these substances have in common?

Akimi4 [234]

They are compounds Good Luck!

4 0

3 years ago

Read 2 more answers

Group the following electron configurations in pairs that would represent similar chemical properties at their atoms;

marishachu [46]

Answer: Pairs are: (a) and (d), (b) and (f), (c) and (e)

Explanation:

In a periodic table, elements are arranged in 18 vertical columns known as groups and 7 horizontal rows known as periods.

Elements arranged in a group show similar chemical properties because of the presence of same number of valence electrons.

Valence electrons are defined as the electrons which are present in the outermost shell of an atom. Outermost shell has the highest value of 'n' that is principal quantum number.

For the given options:

For a:

The given electronic configuration is: $1s^22s^22p^63s^2$

The number of valence electrons in the given configuration are 2

For b:

The given electronic configuration is: $1s^22s^22p^63s^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

For c:

The given electronic configuration is: $1s^22s^22p^63s^23p^64s^23d^{10}4p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For d:

The given electronic configuration is: $1s^22s^2$

The number of valence electrons in the given configuration are 2

For e:

The given electronic configuration is: $1s^22s^22p^6$

The number of valence electrons in the given configuration are [2 + 6] = 8

For f:

The given electronic configuration is: $1s^22s^22p^63s^23p^3$

The number of valence electrons in the given configuration are [2 + 3] = 5

Electronic configuration of (a) and (d) will form a pair, (b) and (f) will form a pair, (c) and (e) will form a pair and will have similar chemical properties.

Hence, the pairs are: (a) and (d), (b) and (f), (c) and (e)

4 0

3 years ago

Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907

melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent. $\frac{^0 C}{m}$

m = molality of solute (m or $\frac{mol}{Kg}$ )

Given : kf = 1.86 $\frac{^0 C*Kg}{mol}$

m = 0.907 $\frac{mol}{Kg}$ )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 $\frac{^0 C*Kg}{mol}$ * 0.907 $\frac{mol}{Kg}$ )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0

3 years ago