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3 years ago

How are the isotopes of a particular element alike?

1 answer:

6 0

The isotopes of a particular elements means that every element has there own isotopes like H has three isotopes protium, dutrium,tritium. like that cl has also thier own isotopes.

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vanillin is a flavoring agent in vanilla it has a mass percent composition of 61.15% c,5.3%H,31.55% o determine the empirical fo

Nutka1998 [239]

The empirical and molecular formulas will be $C_5H_5O_2$ and $C_1_0H_1_0O_4$ respectively.

<h3>Empirical and molecular formula</h3>

The compound contains C, H, and O.

C = 61.15/12 = 5.0958

H = 5.3/1 = 5.3

O = 31.55/16 = 1.9719

Divide by the smallest

C = 2.6

H = 2.7

O = 1

Thus, the empirical formula is $C_5H_5O_2$

Empirical formula mass = (12x5) + (1x5) + 16x2 = 97

n = 152.15/97 = 2

The molecular formula is $C_1_0H_1_0O_4$

More on molecular and empirical formulas can be found here: brainly.com/question/14425592

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6 0

2 years ago

If m = 45g and V = 15ml, what is D in g/ml?

ch4aika [34]

Answer:

3 g/mL

Explanation:

We know that the density of an object can be measured by dividing its mass (g) to its volume (mL).

Formula

D=m/v

Given data:

Mass= 45 g

Volume= 15 mL

Now we will put the values in formula:

D=45 g/ 15 mL= 3 g/mL

6 0

3 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

3 0

4 years ago

What is the Answer of the picture

Neko [114]

Answer: d

Explanation: I got it right

8 0

3 years ago

Read 2 more answers

IUPAC NAME FOR:<br>CH2(OH)-CH2-CH(C2H5)-OH

fgiga [73]

The IUPAC name for the organic compound CH2(OH)-CH2-CH(C2H5)-OH is Pentan-1,3-diol

This compound above belong to class of organic compound called the alkanols

<h3>What are organic compounds?</h3>

Organic compounds are compounds which contains carbon and hydrogen only.

Some few classes of organic compounds includes the following:

Alkanes
Alkenes
Alkynes
Alkanols
Alkanals
Alkanones
Esters
Ketones and other

So therefore, the IUPAC name for the organic compound is CH2(OH)-CH2-CH(C2H5)-OH is Pentan-1,3-diol

Learn more about organic compounds:

brainly.com/question/704297

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6 0

2 years ago