13. Find the total number of atoms present in the following molecules. a. 5 H₂O b. Zn Cl₂

Please help!! I don’t understand

laiz [17]

<h3>Answer:</h3>

150 g Si

<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

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Use Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[Given] 3.2 × 10²⁴ atoms Si

[Solve] grams Si

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Si - 28.09 g/mol

Step 3: Convert

[DA] Set up: $\displaystyle 3.2 \cdot 10^{24} \ atoms \ Si(\frac{1 \ mol \ Si}{6.022 \cdot 10^{23} \ atoms \ Si})(\frac{28.09 \ g \ Si}{1 \ mol \ Si})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 149.266 \ g \ Si$

Step 4: Check

Follow sig fig rules and round. Instructed to round to 2 sig figs.

149.266 g Si ≈ 150 g Si

4 0

3 years ago

Which of the following contains only one

kipiarov [429]

Answer:

A. elements

I hope it's helps you

7 0

2 years ago

Read 2 more answers

Which of the following reactions would you find in a radioisotope thermal generator

slavikrds [6]

The answer to your question is option 1. I hope this has helped.

3 0

3 years ago

Read 2 more answers

1) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data

earnstyle [38]

Answer:

1) The absolute zero temperature is -272.74 °C

2) The absolute zero temperature is -269.91°C

Explanation:

1) The specific volume of air at the given temperatures are;

At -85°C, v = 1/(1.877) = 0.533 dm³/g, the pressure =

At 0°C, v = 1/1.294 ≈ 0.773 dm³/g

At 100°C, v = 1/0.946 ≈ 1.057 dm³/g

We therefore have;

v₁ = 0.533 T₁ = 188.15 K

v₂ = 0.773 T₂ = 273.15 K

v₃ = 1.057 T₃ = 373.15 K

The slope of the graph formed by the above data is therefore given as follows;

m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K

The equation is therefor;

v - 0.533 = 0.00284×(T - 188.15)

v = 0.00284×T - 0.534 + 0.533 = 0.00284×T - 0.001167

v = 0.00284×T - 0.001167

Therefore, when the temperature, at absolute 0, we have;

v = 0

Which gives;

0 = 0.00284×T - 0.001167

0.00284×T = 0.001167

T = 0.001167/0.00284 ≈ 0.411 K

Which is 0.411 -273.15 ≈ -272.74 °C

The absolute zero temperature is -272.74 °C

2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm

The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)

The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;

0°C = 0 + 273.15 K = 273.15 K

Therefore, the equation of the graph can be presented as follows;

v - 20.00 = 0.0741 × (T - 273.15)

Which gives;

v = 0.0741·T - 0.0741 ×(273.15) + 20

v = 0.0741·T - 20.2404125 + 20

v = 0.0741·T - 0.240415

Therefore at absolute 0, v = 0, we have;

0 = 0.0741·T - 0.240415

0.0741·T = 0.240415

T = 0.240415/0.0741 = 3.2445 K

The temperature in degrees is therefore;

3.2445 K - 273.15 ≈ -269.91°C

The absolute zero temperature is therefore, -269.91°C

3 0

3 years ago

A gas is at a pressure of 3.70 atm. What is this pressure in kilopascals?

Yanka [14]

Answer:
= 374.90 kPa

Calculation:
As we know atm and kiloPascal are related to each other as,

1 atm = 101.325 kPa
So,
3.70 atm = X
Solving for X,
X = (3.70 atm × 101.325 kPa) ÷ 1 atm

X = 374.90 kPa

7 0

3 years ago

13. Find the total number of atoms present in the following molecules. a. 5 H₂O b. Zn Cl₂​

13. Find the total number of atoms present in the following molecules. a. 5 H₂O b. Zn Cl₂