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Pachacha [2.7K]
3 years ago
3

How many moles of CO2 are in 210 liters at STP of the compound​

Chemistry
1 answer:
Zina [86]3 years ago
8 0

Answer:

9.4 mol CO₂

General Formulas and Concepts:

<u>Chemistry - Gas Laws</u>

  • Using Dimensional Analysis
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Explanation:

<u>Step 1: Define</u>

210 L CO₂ at STP

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

22.4 L = 1 mol at STP

<u>Step 3: Convert</u>

<u />210 \ L \ CO_2(\frac{1 \ mol \ CO_2}{22.4 \ L \ CO_2} ) = 9.375 mol CO₂

<u>Step 4: Check</u>

<em>We are given 2 sig figs. Follow sig fig rules and round.</em>

9.375 mol CO₂ ≈ 9.4 mol CO₂

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How many moles of H2O2 are needed to react with 1.07 moles of N2H4?
I am Lyosha [343]

Answer:

2.14 moles of H₂O₂ are required

Explanation:

Given data:

Number of moles of H₂O₂ required = ?

Number of moles of N₂H₄ available = 1.07 mol

Solution:

Chemical equation:

N₂H₄  +   2H₂O₂       →   N₂ +  4H₂O

now we will compare the moles of H₂O₂ and N₂H₄

                          N₂H₄     :      H₂O₂  

                            1           :        2

                            1.07      :         2×1.07 = 2.14 mol

                   

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Name 2 separate technique in separating iodine crystal and iron filings​
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Read 2 more answers
If the sample has a total mass of 5.76 g and contains 1.79 g k, what are the percentages of kbr and ki in the sample by mass
Luden [163]

The percentages of KBr and KI  in the sample by mass is 80.68 and 19.32 % respectively.

<h3>What is Molar Mass ?</h3>

Molar mass is defined as the mass contained in 1 mole of sample.

It is given that

the sample has a total mass of 5.76 g contains KBr and KI

and contains 1.79 gm of K is present

what is the percentage of KBr and KI in the sample

Molecular weight of K = 39

Molecular weight of Br = 79.9

Molecular weight of I = 126.9

In KBr the mass percentage of K is 39/(39+79.9) = 32.89%

In KI the mass percentage of K is 39/(39+126.9) = 23.5%

Let the mass of KBr present in the sample is x

K will be 0.3289 x

and let the mass of KI present be y

K will be 0.235y

x +y =5.76

0.3289x+0.235y = 1.79

0.0939y = 0.1045

y = 1.1125 gm

x = 5.76-1.1125

x = 4.6475 gm

% of KBr = (4.6475/5.76  )*100 = 80.68 %

% of KI = (1.1125/5.76) *100 = 19.32%

To know more about Molar Mass

brainly.com/question/12127540

#SPJ1

8 0
2 years ago
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