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3 years ago

Calculate the atomic mass of copper if copper-63 is 69.17% abundant and copper-65 is 30% abundant

Chemistry

1 answer:

astraxan [27]3 years ago

6 0

Answer:List of isotopes

Nuclide Z Isotopic mass (Da)

Excitation energy

63Cu 29 62.9295975(6)

64Cu 29 63.9297642(6)

65Cu 29 64.9277895(7)

Explanation:

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Which of the following substances would you predict to have the highest DHvap? Group of answer choices CH3CH2CH2CH3 CH3CH2OH HF

Katyanochek1 [597]

Answer:

a) $CH_3CH_2CH_2CH_3$

Explanation:

In this question we have the following answer choices:

a) $CH_3CH_2CH_2CH_3$

b) $CH_3CH_2OH$

c) $HF$

d) $CH_3Cl$

e) $HOCH_2CH_2OH$

We have to remember the relationship between intermolecular forces and vapor pressure. If we have stronger intermolecular forces we will have less vapor pressure because the molecules have more interactions between them, so, the molecules will prefer to stay in a liquid state rather than a gaseous state. Now, we have to check each molecule:

a) $CH_3CH_2CH_2CH_3$ (Van der waals interactions)

b) $CH_3CH_2OH$ (Hydrogen bonding)

c) $HF$ (Hydrogen bonding)

d) $CH_3Cl$ (Dipole-dipole interaction)

e) $HOCH_2CH_2OH$ (Hydrogen bonding)

For molecules b, c and e we have hydrogen bond to a heteroatom (O, N, S, or P). In this case oxygen, therefore we will have hydrogen bonding interactions (a very strong interaction). So, we can discard these ones.

In molecule e, we have "Cl" bond to a "C" therefore we will have the presence of a dipole (due to the electronegativity difference). If we have a dipole, we will have a dipole-dipole interaction (a strong interaction, less than hydrogen bonding but still is a strong interaction).

In molecule a, we have only Van der Waals interactions because in this molecule we have only carbon and hydrogen atoms bonded by single bonds. So, we will have a non-polar molecule. These interactions are the weakest interactions of all the molecules given. So, if we have weaker interactions the molecules can be converted to a gas state more easily and we have more vapor pressure. 

7 0

3 years ago

compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of

arlik [135]

Answer: The empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

Explanation:

We are given:

Mass of $CO_2=3.926g$

For calculating the mass of carbon:

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

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$ ......(1)

For Carbon:

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

For Fluorine:

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

Step 2: 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 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

Step 3: Taking the mole ratio as their subscripts.

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

The empirical formula for the given compound is $CH_2F$

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 = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

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

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

3 0

3 years ago

What is the name of the instrument used to measure the pressure of a contained gas? *

Natasha_Volkova [10]

Answer:Barometer :

Barometer is a scientific device. It is used to measure air pressure in environment. Air pressure changes with distance upper and bottom sea level.

Barometer is also used for measure altitude.

Barometric pressure has main effects on water and atmosphere conditions.

Hence, Barometer is used to measure air pressure.

Explanation:

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Explanation:

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What is water supply ?

Tom [10]

Answer:

a water suppley is water that pepople buy to dreak and ues

Explanation:

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