1.20 L of oxygen gas is held in a flexible vessel at STP. If the

Use the conversion factor 1amu=6.66054 x 10^-24 to answer the following questions.. . a) 1.674 x 10^-24 g of neutrons is how man

Wewaii [24]

A.) The conversion factor is 1 amu = $1.66054 ^{-24}g$

To know how many amu in $1.674x10^{-24}g$ grams of neutrons:
$1.674x10^{-24}g ( \frac{x (amu) }{ 1.66054x10^{-24} g)}$

=1.00811 amu

b.) The mass in grams of one lithium ion which has an atomic weight of 6.94 amu.

$6.94 amu (\frac{6.66054 ^{-24} g}{1 amu})$ = $4.62241x10 ^{23} g$

c.) How many amu in 6.492x10^-23g potassium?

$6.492x10^{-23}g ( \frac{x (amu) }{ 1.66054x10^{-24}g })$ = 39.0957 amu

6 0

2 years ago

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Bonds between two atoms that are equally electronegative are _____. bonds between two atoms that are equally electronegative are

Anna [14]

Bonds of two atoms of equal electronegativity are nonpolar covalent bonds.

Your second sentence is identical to the first sentence; I'll bet the second sentence is "Bonds between two atoms that are unequally electronegative are polar covalent bonds."

4 0

2 years ago

ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of

Luden [163]

Answer: The empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

Explanation:

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 $CO_2=9.06g$

Mass of $H_2O=5.58g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 9.06 g of carbon dioxide, $\frac{12}{44}\times 9.06=2.47g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 5.58 g of water, $\frac{2}{18}\times 5.58=0.62g$ of hydrogen will be contained.

Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles$

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 0.206 moles.

For Carbon = $\frac{0.206}{0.206}=1$

For Hydrogen = $\frac{0.62}{0.206}=3$

For Oxygen = $\frac{0.206}{0.206}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is $C_1H_{3}O_1=CH_3O$

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 = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

$n=\frac{124g/mol}{31g/mol}=4$

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

$C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4$

Thus, the empirical and molecular formula for the given organic compound is $CH_3O$ and $C_4H_{12}O_4$

3 0

2 years ago

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl),

choli [55]

Answer:

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?

A. 101.53° C

B. 100.00° C

C. 101.02° C

D. 100.51° C

Which of the following solutions will have the lowest freezing point?

A. 1.0 mol/kg sucrose (C12H22O11)

B. 1.0 mol/kg lithium chloride (LiCl)

C. 1.0 mol/kg sodium phosphide (Na3P)

D. 1.0 mol/kg magnesium fluoride (MgF2)

Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?

A. sodium iodide (NaI)

B. magnesium sulfate (MgSO4)

C. potassium bromide (KBr)

D. All will be equally effective.

Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?

A. 98.7 kPa

B. 96.3 kPa

C. 101.3 kPa

D. 100.2 kPa

Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?

A. 101.2°C

B. 105.9°C

C. 102.7°C

D. 108.1°C

Explanation:

The following are the answers to the different questions:

The four rows of data below show the boiling points for a solution with no solute, sucrose (C12H22O11), sodium chloride (NaCl), and calcium chloride (CaCl2) (not in that order). Which boiling point corresponds to calcium chloride?

A. 101.53° C

Which of the following solutions will have the lowest freezing point?

D. 1.0 mol/kg magnesium fluoride (MgF2)

Which of the following compounds will be most effective in melting the ice on the roads when the air temperature is below zero?

A. sodium iodide (NaI)

Four different solutions have the following vapor pressures at 100°C. Which solution will have the greatest boiling point?

B. 96.3 kPa

Four different solutions have the following boiling points. Which boiling point corresponds to a solution with the lowest freezing point?

D. 108.1°C

5 0

2 years ago

Which of these elements are you least likely to find in nature as a pure element (that is, not combined with any other elements

Delvig [45]

The answer is sodium (Na)

Let me know if u have anymore questions ☺

7 0

2 years ago

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