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3 years ago

What type of bond would occur between lithium (Li) and oxygen (O)?

Chemistry

1 answer:

Ghella [55]3 years ago

5 0

Answer:

D. Ionic bond

Explanation:

First some definitions: ionic bonds always have metals that transfer electrons. Covalent bonds are between nonmetals that share electrons. (Sharing is caring!)

Li is a metal, so right off the bat we know this is not a covalent bond. However, it's not a metal, so we can cross off C as well. The phrase "paired bond" isn't common, so we know the answer must be D. ionic bond.

And in this bond, Li's electrons will transfer to O to make both have a full valence shell. (As opposed to covalent bonds, which share electrons, not transfer completely.)

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In this experiment the crest of one wave combined with the crest of the second wave to produce _______________.

timurjin [86]

Answer:

B a large upward displacement

Explanation:

3 0

2 years ago

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40.0L of N₂ gas are in a sealed container at STP.How many moles of N₂ are present?9 mol

Vinvika [58]

Explanation:

We have to find the number of moles of N₂ that are present in a sample that has a volume of 40.0 L at STP.

STP means Standard Conditions of Temperature and Pressure. These conditions are 273.15 K and 1 atm. We know that 1 mol of N₂ will occupy 22.4 L. We can use that ratio to find the answer to our problem.

1 mol of N₂ = 22.4 L

moles of N₂ = 40.0 L * 1 mol/(22.4 L)

moles of N₂ = 1.79 mol

Answer: 1.79 moles of nitrogen are present.

8 0

1 year ago

Be sure to answer all parts. Dimercaprol (HSCH2CHSHCH2OH) was developed during World War I as an antidote to arsenic-based poiso

Sauron [17]

Answer:

For A: The number of arsenic atoms are $3.4\times 10^{21}$

For B: The percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

Explanation:

For A:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of dimercaprol = 696 mg = 0.696 g (Conversion factor: 1 g = 1000 mg)

Molar mass of dimercaprol = 124.21 g/mol

Putting values in above equation, we get:

$\text{Moles of dimercaprol}=\frac{0.696g}{124.21g/mol}=0.0056mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.0056 moles of dimercaprol will contain $0.0056\times 6.022\times 10^{23}=3.4\times 10^{21}$ number of molecules.

As, 1 molecule of dimercaprol binds with 1 atom of Arsenic

So, $3.4\times 10^{21}$ number of dimercaprol molecules will bind with = $1\times 3.4\times 10^{21}=3.4\times 10^{21}$ number of arsenic atoms

Hence, the number of arsenic atoms are $3.4\times 10^{21}$

For B:

We know that:

Molar mass of dimercaprol = 124.21 g/mol

Molar mass of mercury = 200.59 g/mol

Molar mass of thallium = 204.38 g/mol

Molar mass of chromium = 51.99 g/mol

Also, 1 molecule of dimercaprol binds with 1 metal atom.

To calculate the percentage composition of metal in a complex, we use the equation:

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

For mercury:

Mass of Hg-complex = (200.59 + 124.21) = 324.8 g

Mass of mercury = 200.59 g

Putting values in equation 1, we get:

$\%\text{ composition of mercury}=\frac{200.59g}{324.8g}\times 100=61.76\%$

For thallium:

Mass of Tl-complex = (204.38 + 124.21) = 328.59 g

Mass of thallium = 204.38 g

Putting values in equation 1, we get:

$\%\text{ composition of thallium}=\frac{204.38g}{328.59g}\times 100=62.2\%$

For chromium:

Mass of Cr-complex = (51.99 + 124.21) = 176.2 g

Mass of chromium = 51.99 g

Putting values in equation 1, we get:

$\%\text{ composition of chromium}=\frac{51.99g}{176.2g}\times 100=29.51\%$

Hence, the percent composition of mercury, thallium and chromium in their complexes are 61.76 %, 62.2 % and 29.51 % respectively.

8 0

3 years ago

The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a

mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

3 0

3 years ago

What number minus 6 equals 58?

Novay_Z [31]

Answer: 64

Explanation:

3 0

3 years ago

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