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3 years ago

Attempt 2

1 answer:

4 0

Lets let our mass equal 3 on alletals and solve using d=m/v equation

Aluminum
V=3/2.70=1.11
Silver
V=3/10.5=.286
Rhenium
V=3/20.8=.144
Nickel
V=3/8.90=.337
This gives us the following list from largest to smallest Aluminum, Nickel, Silver, and Rhenium

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Karolina [17]

Can you reword it im confused

6 0

2 years ago

If two identical atoms are bonded together, what kind of molecule is formed?

AlekseyPX

Answer:

C. A linear, nonpolar molecule

Explanation:

Molecules which are alike usually have the same degree of pull which results in them sharing electrons. This sharing of electrons is known as the molecules exhibiting Covalent bonding between them.

The equal pull also results in the cancelling out of electrons and favoring non polar bonds due to the absence of free electrons which would have been able to interact with H2O in a polar binding system.

6 0

3 years ago

Read 2 more answers

How many moles of tungsten carbide will be produced if plenty of tungsten is reacted with 1.25 moles of carbon?

Tanya [424]

Answer:

1.25 moles of tungsten carbide

Explanation:

Tungsten carbide, WC, is a substance that is produced prom carbon and tungsten as follows:

W + C → WC

Based on the reaction, 1 mole of Carbon produce 1 mole of tungsten carbide. That means if 1.25 moles of carbon are added in excess of tungsten, the moles of tungsten carbide produced are:

<h3>1.25 moles of tungsten carbide</h3>

8 0

3 years ago

A compound is 7.74% hydrogen and 92.26% carbon by mass. At 100°C a 0.6883 g sample of the gas occupies 250 mL when the pressure

ycow [4]

Answer: The molecular formula for the compound is $C_6H_6$

Explanation:

We are given:

Percentage of C = 92.26 %

Percentage of H = 7.74 %

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

Mass of C = 92.26 g

Mass of H = 7.74 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{92.26g}{12g/mole}=7.68moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles$

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 7.68 moles.

For Carbon = $\frac{7.68}{7.68}=1$

For Hydrogen = $\frac{7.74}{7.68}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

The empirical formula for the given compound is $CH$

Calculating the molar mass of the compound:

To calculate the molecular mass, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{m}{M}RT$

where,

P = pressure of the gas = 820 torr

V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )

m = mass of gas = 0.6883 g

M = Molar mass of gas = ?

R = Gas constant = $62.3637\text{ L. torr }mol^{-1}K^{-1}$

T = temperature of the gas = $100^oC=(100+273)K=373K$

Putting values in above equation, we get:

$820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol$

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 = 78.10 g/mol

Mass of empirical formula = 13 g/mol

Putting values in above equation, we get:

$n=\frac{78.10g/mol}{13g/mol}=6$

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

$C_{(1\times 6)}H_{(1\times 6)}=C_6H_6$

Hence, the molecular formula for the compound is $C_6H_6$

8 0

3 years ago

What is the empirical formula for a compound that contains 79.86 % iodine and 20.14 % oxygen by mass?

serg [7]

Answer:

IO₂

Explanation:

We have been given the mass percentages of the elements that makes up the compound:

Mass percentage given are:

Iodine = 79.86%

Oxygen = 20.14%

To calculate the empirical formula which is the simplest formula of the compound, we follow these steps:

> Express the mass percentages as the mass of the elements of the compound.

> Find the number of moles by dividing through by the atomic masses

> Divide by the smallest and either approximate to nearest whole number or multiply through by a factor.

> The ratio is the empirical formula of the compound.

Solution:

I O

% of elements 79.86 20.14

Mass (in g) 79.86 20.14

Moles(divide by

Atomic mass) 79.86/127 20.14/16

Moles 0.634 1.259

Dividing by

Smallest 0.634/0.634 1.259/0.634

1 2

The empirical formula is IO₂

7 0

3 years ago