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topjm [15]
2 years ago
12

Nicotine is a poisonous, addictive compound found in tobacco. A sample of nicotine contains 6.16 mmol of C, 8.56 mmol of H, and

1.23 mmol of N [1 mmol (1 millimole) = 10⁻³ mol]. What is the empirical formula of nicotine?
Chemistry
1 answer:
tiny-mole [99]2 years ago
8 0

C₁₀H₁₄N₂ is the empirical formula.

In chemistry, the empirical formula of a chemical compound is the simplest whole number ratio of atoms present in a compound.

<h3>Tell us about the empirical formula.</h3>

The empirical formula of a chemical compound in chemistry is the simplest whole number ratio of atoms in a compound. Two simple instances of this concept are the empirical formulas of sulfur monoxide (SO) and disulfur dioxide (S2O2).

Its empirical formula is the simplest whole number ratio of each type of atom in the compound. Data about the mass of each component in a compound or the composition's percentage can be used to calculate it.

To learn more about empirical formula visit:

brainly.com/question/14044066

#SPJ4

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Where would you have the least weight?​
Vikki [24]

In a place with no gravity

Explanation:

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A body will experience weightlessness in a place without no gravity.

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8 0
4 years ago
54.0g Al reacts with 64.0g O2 to form Al2O3 according to the equation.
pentagon [3]

Answer:

136 g Al₂O₃

Explanation:

Assuming you do not need to find the limiting reactant, to find the mass of Al₂O₃, you need to (1) convert grams O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles Al₂O₃ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles Al₂O₃ to grams Al₂O₃ (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value (64.0 g).

Molar Mass (O₂): 32 g/mol

Molar Mass (Al₂O₃): 102 g/mol

4 Al + 3 O₂ -----> 2 Al₂O₃

64.0 g O₂           1 mole           2 moles Al₂O₃            102 g
-----------------  x  --------------  x  ------------------------  x   -------------  =  136 g Al₂O₃
                             32 g               3 moles O₂             1 mole

6 0
2 years ago
Read 2 more answers
What is the ph of a solution made by mixing 100.00 ml of 0.20 m hcl with 50.00 ml of 0.10 m hcl? assume that the volumes are add
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I completely ignored sig figs though

3 0
3 years ago
Read 2 more answers
An ideal gas in a cylindrical container of radius r and height h is kept at constant pressure p. The bottom of the container is
Juli2301 [7.4K]

Answer:

m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}  

Explanation:

The gas ideal law is  

PV= nRT (equation 1)

Where:

P = pressure  

R = gas constant  

T = temperature  

n= moles of substance  

V = volume  

Working with equation 1 we can get  

n =\frac{PV}{RT}

The number of moles is mass (m) / molecular weight (mw). Replacing this value in the equation we get.

\frac{m}{mw} =\frac{PV}{RT}  or  

m =\frac{P*V*mw}{R*T}   (equation 2)

The cylindrical container has a constant pressure p  

The volume is the volume of a cylinder this is

V =(pi)*r^{2}*h

Where:

r = radius  

h = height  

(pi) = number pi (3.1415)

This cylinder has a radius, r and height, h so the volume is  V =(pi)*r^{2}*h

Since the temperatures has linear distribution, we can say that the temperature in the cylinder is the average between the temperature in the top and in the bottom of the cylinder. This is:  

T =\frac{T_{1} + T_{O}}{2}  

Replacing these values in the equation 2 we get:

m =\frac{P*V*mw}{R*T}   (equation 2)

m =\frac{p*(pi)*r^{2}*h*mw}{R*\frac{T_{1} + T_{O}}{2}}    

8 0
4 years ago
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