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3 years ago

The air exhaled from the lungs of a smoker has a concentration of 24 ppm CO. Express the concentration as a percent.

Chemistry

1 answer:

Nana76 [90]3 years ago

7 0

<h3><u>Answer;</u></h3>

0.0024%

<h3><u>Explanation</u>;</h3>

Parts per million or ppm means out of a million. Usually describes the concentration of something in water or soil.
One ppm is equivalent to 1 milligram of something per liter of water (mg/l) or 1 milligram of something per kilogram soil (mg/kg).

Therefore to convert 24 ppm of CO as a percentage.

24 ppm × 100%

(24/10⁶)× 100%

<u>= 0.0024%</u>

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Naddik [55]

Answer:

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Explanation:

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3 years ago

The opposite of an endothermic reaction is a(n) reaction.

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Answer:

The opposite of an endothermic process is an exothermic process, one that releases or "gives out" energy, usually in the form of heat and sometimes as electrical energy.

Explanation:

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3 years ago

Read 2 more answers

How many moles would you need to have 7.85 x 10^22 molecules of ammonia (NH3).

RSB [31]

<h3>Answer:</h3>

0.130 moles NH₃

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

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

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 7.85 × 10²² molecules NH₃

[Solve] moles NH₃

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 7.85 \cdot 10^{22} \ molecules \ NH_3(\frac{1 \ mol \ NH_3}{6.022 \cdot 10^{23} \ molecules \ NH_3})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 0.130355 \ moles \ NH_3$

<u>Step 4: Check</u>

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

0.130355 moles NH₃ ≈ 0.130 moles NH₃

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3 years ago

In the first 15.0 s of the reaction, 1.7×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.440 L . What is the

Step2247 [10]

Answer:

$Rate=2.57x10^{-3}\frac{M}{s}$

Explanation:

Hello,

In this case, for the reaction:

$2N_2O(g) \rightarrow 2N_2(g)+O_2(g)$

We can easily compute the average rate by firstly computing the final concentration of oxygen:

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Then, we compute it by using the given interval of time: from 0 seconds to 15.0 seconds and concentration: from 0 M to 0.0386M as oxygen is being formed:

$Rate=\frac{0.0386M-0M}{15.0s-0s}\\ \\Rate=2.57x10^{-3}\frac{M}{s}$

Regards.

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4 years ago

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VARVARA [1.3K]

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