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noname [10]
2 years ago
11

A sample of a compound is determined to have 1.17 g of carbon and 0.287 g of hydrogen. what is the correct representation of the

empirical formula for the compound?
Chemistry
1 answer:
yarga [219]2 years ago
5 0

CH3 is the empirical formula for the compound.

A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.

The number of atom or moles in the compound is

1.17 g C X  1 mol of C / 12.011 g C = 0.097411 mol of C.

0.287 g H x 1 mol of  H / 1 g H = 0.28474 mol H.

This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.

So we can represent the compound with the formula C0.974H0.284.

Subscripts in formulas can be made into whole numbers by multiplying the smaller subscript by the larger subscript.

we can divide 0.284 by 0.0974.

0.284 / 0.0974 = 3.

So here, Carbon is one and hydrogen is 3.

We can write the above formula as a CH3.

Hence the empirical formula for the sample compound is CH3.

For a detailed study of the empirical formula refer given link brainly.com/question/13058832.

#SPJ1.

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diamong [38]

Answer:

3.6 × 10²⁴ molecules

Explanation:

Step 1: Given data

Moles of methane (n): 6.0 moles

Step 2: Calculate the number of molecules of methane in 6.0 moles of methane

In order to convert moles to molecules, we need a conversion factor. In this case, we will use Avogadro's number: there are 6.02 × 10²³ molecules of methane in 1 mole of molecules of methane.

6.0 mol × 6.02 × 10²³ molecules/1 mol = 3.6 × 10²⁴ molecules

6 0
3 years ago
If gas is initially at 350L and 500k then changes to 295K what is the new volume
Dimas [21]
<h2>Hello!</h2>

The answer is:

The new volume is equal to 206.5 L.

<h2>Why?</h2>

To solve this problem, we need to assume that the pressure is constant, and use the Charle's Law equation, so, solving we have:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

We are given:

V_1=350L\\T_1=500K\\T_2=295K

Then, using the Charle's Law equation, we have:

\frac{350L}{500K}=\frac{V_2}{295K}

\frac{350L}{500K}=\frac{V_2}{295K}\\\\V_2=\frac{350L}{500K}*295K=206.5L

Hence, we have that the new volume is equal to 206.5 L.

Have a nice day!

5 0
3 years ago
The fish died after living in the aquarium for many years
miv72 [106K]

In an aquarium the water quantity is limited and fish excrete ammonia through their gills and body, this dissolves in water and creates toxins.

<u>Explanation</u>:

  • In an aquarium the water quantity is limited and fish excrete ammonia through their gills and body, this dissolves in water and creates toxins. Over some time some bacteria develop in water which converts this ammonia into nitrites and then into nitrates. Till this process is complete, the aquarium remains a death trap.

There are some reasons for a fish to die early,

  • This one is common for beginners. They don’t have any idea about the nitrogen cycle. And they simply buy a fish tank and fish on the same day, go home and set it up.
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4 0
3 years ago
Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

[H^+]=x=0.01346 M

The pH of the solution is ;

pH=-\log[H^+]=-\log[0.01346 M]=1.87

The pH of an 0.280 M HF solution is 1.87.

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