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-Dominant- [34]
3 years ago
15

A 100g of sample of a compound is combusted in excess oxygen and the products are 2.492g of CO2 and 0.6495 of H2O. Determine the

empirical formula of the compound?
Chemistry
1 answer:
ruslelena [56]3 years ago
3 0

The empirical formula : C₁₁O₁₄O₃

<h3>Further explanation</h3>

The assumption of the compound consists of C, H, and O

mass of C in CO₂ =

\tt \dfrac{12}{44}\times 2.492=0.680~g

mass of H in H₂O =

\tt \dfrac{2.1}{18}\times 0.6495=0.072~g

mass of O :

mass sample-(mass C + mass H)

\tt 1-(0.68+0.072)=0.248`g

mol of  C :

\tt \dfrac{0.68}{12}=0.056

mol of H :

\tt \dfrac{0.072}{1}=0.072

mol of O :

\tt \dfrac{0.248}{16}=0.0155

divide by 0.0155(the lowest ratio)

C : H : O ⇒

\tt \dfrac{0.056}{0.0155}\div \dfrac{0.072}{0.0155}\div \dfrac{0.0155}{0.0155}=3.6\div 4.6\div  1\\\\\dfrac{11}{3}\div \dfrac{14}{3}\div \dfrac{3}{3}=11:14:3

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How many cu atoms are present in a piece of sterlingsilver jewelry weighing 33.24 g ? (sterling silver is a silver–copper alloy
emmasim [6.3K]

The mass of piece of sterling silver jewelry is 33.24 g. It contains 92.5% silver Ag by mass. Since, sterling silver is an alloy of Ag-Cu thus, percentage of Cu will be:

% Cu=100-92.5=7.5%

Thus, mass of copper will be:

m_{Cu}=\frac{7.5}{100}\times 33.24 g=2.493 g

Molar mass of Cu is 63.546 g/mol, thus, number of moles of Cu can be calculated as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass.

Putting the values,

n=\frac{2.493 g}{63.546 g/mol}=0.03923 mol

Now, in 1 mole of Cu there are 6.303\times 10^{23}atoms thus, in 0.03923 mol, number of Cu atoms will be:

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4 years ago
What characteristic of the light is affected by the amplitude of the light
Vadim26 [7]

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Suppose now that you wanted to determine the density of a small crystal to confirm that it is phosphorus. From the literature, y
denis23 [38]

Answer:

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

Explanation:

Given that:

the density of the mixture = 1.82 g/mL

From the density of the pure samples

The density of CHCl_3 = 1.492 g/mL

The density of CHBr_3 = 2.890 g/mL

The total volume of the liquid mixture = 20.0 mL

Suppose the volume of  CHCl_3 = P ml

and the volume of CHBr_3 = Q ml

the sum of their volumes should be equal to the total volume of the mixture

P \ ml + Q \ ml = 20 ml  ----- (1)

However, we know that Density = mass/volume

∴ mass = density × volume

The equation can now be expressed as:

\mathtt{(Density \ of  \ CHCl_3 \times Vol. \ of  \ CHCl_3 ) + (Density  \  of \  CHBr_3  \times \ volume \ of \ CHBr_3)} = \mathtt{ (Density  \ of \ mixture \times volume \ of \ the \ mixture)}

1.492 g/mL × P mL + 2.890 g/mL × Q mL = 1.82 g/mL × 20 mL  ---- (2)

From equation (1) ;

let Q = 20 - P

The replace the value of P into equation (2)

1.492 g/mL × P mL + 2.890g/mL × (20 - P) mL = 1.82 g/mL × 20 mL

1.492 P g + 57.8g - 2.890 P g =  36.4g

1.492 P g - 2.890 P g = 36.4g - 57.8g

-1.398 P g = -21.4g

P = -21.4g/-1.398g

P = 15.31 mL

Q = 20 - P

Q = (20 - 15.31) mL

Q = 4.69 mL

∴

Volume of CHCl_3   = 15.31 mL

Volume of CHBr_3 = 4.69 mL

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Answer:

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like nuculear power plants, create nuculear waste

you need to let it sit there for several years

unless if you want to get sued by the Enviromental protection agency

Explanation:

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