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Juli2301 [7.4K]

3 years ago

8

A person's heartbeat is 89 beats per minute. If his/her heart beats 3.1e9 times in a lifetime, how long (in whole years) does th

e person live? Disregard leap years.

1 answer:

andrezito [222]3 years ago

5 0

I think the answer is 82.5

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How hard is Stoichiometry?<br> Just want to know since I'm about to do a lesson on it..

DanielleElmas [232]

Stoichiometry is not really hard if you are good with your numbers. All you have to do is pay close attention and follow you teacher because if you don't you will end up like me not knowing what you are doing. Listen and pay real close attention and I can guarantee you it will be just fine!

6 0

3 years ago

What is the chemical formula for the binary compound nitrogen dioxide?

-Dominant- [34]

NO2 Hope this helped!

4 0

3 years ago

Read 2 more answers

2C3H7OH + 9O2 --> 6CO2 + 8H2O

Ne4ueva [31]

<h3>Answer:</h3>

733 g CO₂

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

[Given] 5.55 mol C₃H₇OH

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol C₃H₇OH → 6 CO₂

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Stoichiometry</u>

Set up conversion: $\displaystyle 5.55 \ mol \ C_3H_7OH(\frac{6 \ mol \ CO_2}{2 \ mol \ C_3H_7OH})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})$
Multiply/Divide: $\displaystyle 732.767 \ g \ CO_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

732.767 g CO₂ ≈ 733 g CO₂

8 0

3 years ago

What are three applications of the<br>coal

labwork [276]

Three uses of coal:

1. Electricity generation

2. Steel production

3. Cement manufacturing and as a liquid fuel

Hope it helps!

3 0

2 years ago

A colorless liquid has a molar mass of 60.01 g/mol. When the liquid was analyzed, it was 46.7% nitrogen and 53.3% oxygen. What i

MAXImum [283]

first we have to find the empirical formula of the compound

empirical formula is the simplest ratio of whole numbers of components making up a compound

for 100 g of the compound

N O

mass 46.7 g 53.3 g

number of 46.7 g/ 14 g/mol 53.3 g/ 16 g/mol

moles = 3.34 mol = 3.33 mol

divide by the least number of moles

3.34/3.33 = 1.00 3.33/ 3,33 = 1.00

therefore number of atoms are

N - 1

O - 1

empirical formula is - NO

mass of empirical unit - 14 g/mol + 16 g/mol = 30 g

molecular formula is actual composition of elements in the compound

molecular mass - 60.01 g/mol

number of empirical units = molecular mass / empirical unit mass

= 60.01 g/mol / 30 g = 2

there are 2 empirical units

2(NO)

molecular formula = N₂O₂

6 0

3 years ago