This is Bohr's Model of: A. Oxygen B. Magnesium C. Sulfur

What is the name of the compound with the formula B2C14?

agasfer [191]

Answer:

Diboron tetrachloride ---> B2Cl4

6 0

3 years ago

Just as one dozen eggs always has 12 eggs in it, one mole of a

zzz [600]

Answer:

$6.022x10^{23}atoms \ Al$

Explanation:

Hello,

In this case, given the described concept regarding the Avogadro's number, we can easily notice that 27.0 g of aluminium foil has 6.022x10²³ atoms as shown below based on the mass-mole-particles relationship:

$27.0gAl*\frac{1molAl}{27.0gAl} *\frac{6.022x10^{23}atoms \ Al}{1molAl} \\\\=6.022x10^{23}atoms \ Al$

Notice this is backed up by the fact that aluminium molar mass if 27.0 g/mol.

Best regards.

8 0

3 years ago

1. 7.000 mmhg = ______ KPa? 2. 10.00 kPa = ________atm? 3. 15.00 kPa______mmHg?

Minchanka [31]

1 kPa = 7.5 mmHg so 7.0 mmHg / 7.5 mmHg x 1 kPa = .93 kPa

101.3 kPa = 1 atm so 10 kPa / 101.3 kPa x 1 atm = .0987 atm

1 kPa = 7.5 mmHg so 15 kPa x 7.5 mmHg / 1 kPa = 112.5 mmHg

7 0

3 years ago

List some of applications with chemistry in daily life

Nuetrik [128]

Answer:

while cooking or washing something lol

6 0

3 years ago

Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2N

Gnesinka [82]

Answer:

NH3 is the limiting reactant

The theoretical yield is 216.0 kg urea

The % for this reaction is 78.8 %

Explanation:

Step 1: Data given

Mass of ammonia = 122.5 kg

Mass of carbon dioxide = 211.4 kg

Mass of urea produced = 170.3 kg

Molar mass of ammnoia = 17.031 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Moalr mass of urea = 60.06 g/mol

Step 2: The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

Step 3: Calculate moles of NH3

Number of moles = mass / Molar mass

Moles NH3 = 122500 grams / 17.031 g/mol

Moles NH3 = 7192.77 moles

Step 4: Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

Step 5: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).

CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles

There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2

Step 6: Calculate moles of urea produced:

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

Step 7: Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

Step 8: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

The % for this reaction is 78.8 %

3 0

3 years ago

This is Bohr's Model of:A. OxygenB. MagnesiumC. Sulfur​

This is Bohr's Model of:A. OxygenB. MagnesiumC. Sulfur