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Oksi-84 [34.3K]
3 years ago
9

A balanced chemical equation used to prepare ammonium carbonate, (nh4)2co3 , is: 2 n h 3 ( g ) + c o 2 ( g ) + h 2 o ( l ) ⟶ ( n

h 4 ) 2 c o 3 which choice of reactant quantities shown below would result in the greatest amount of ammonium carbonate being formed?
Chemistry
1 answer:
Tems11 [23]3 years ago
7 0

Answer:

This a simple stoichiometry problem using the ideal gas law.  

First take the grams of ammonium carbonate and convert it to moles using its molar mass and dividing. 11.9 g/96.0932 g/mol= .12384 mol  

Now use a molar conversion using the balanced equation,  

1 mol (NH4)2CO3 ---> 4 mol gas formed (2 mol NH3 + 1 mol CO2 + 1 mol H2O) = .12384 x 4 = .49535 mol gas  

PV=nRT  

V=nRT/P= .49535mol (.08206 Lxatm/molxK) (296K)/ (1.03 atm)=11.682 L

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tino4ka555 [31]
Displaced volume:

final volume - initial volume

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38.5 mL - 35.0 mL = 3.5 cm³

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3 years ago
A solution is made by dissolving 4.8 moles of salt in 1.6L of solution. What is the molarity
goldenfox [79]

Answer:

3M

Explanation:

moles ÷ liters = molarity

4.8 ÷ 1.6 = 3M

3 0
3 years ago
Determine if each statement is True or False. [ Select ] Central atoms with four electron groups will be sp3 hybridized. [ Selec
meriva

Answer:

Central atoms with four electron groups will be sp3 hybridized. True

Hybrid orbitals are delocalized over the entire molecule. False

The number of hybrid orbitals is equal to the number of atomic orbitals that are blended together. True

Atoms with a single pi bond and an octet are sp2 hybridized. True

Sometimes oxygen atoms will be sp3d hybridized in organic molecules. False

All resonance structures must be considered when assigning hybridization. False

Explanation:

When a central atom has four electron groups attached to it, then it must be sp3 hybridized. This is the case in ammonia, water, hydrogen sulphide, methane etc.

Hybridization is a valence bond concept while delocalization is a molecular orbital theory concept. Hybridized orbitals are localized on central atoms in a molecule while delocalized orbitals spread across the entire molecule.

According to valence bond theory, the number of atomic orbitals that combined to give hybrid orbitals must be equal to the number of hybrid orbitals formed.

When an atom has a single pi bond and on octet of electrons, then it must be sp2 hybridized. Remember that in ethene for instance, carbon has one pi bond and an octet of electrons.

Oxygen has an empty n=3 level hence it can not have d-orbitals involved in hybridization.

The electron domain geometry of the molecule is considered when assigning hybridization and not the resonance structures. All the resonance structures must have the central atom in the same hybridization state.

8 0
3 years ago
Non metal are usually poor conductor of hart and electricity . The are non -lustourous,non-sonours,non-malleable and are coloure
vekshin1
Check the solution in the attachment.

7 0
3 years ago
A sample of gas (1.9 mol) is in a flask at 21 °C and 697 mm Hg. The flask is opened and more gas is added to the flask. The new
Artyom0805 [142]

Answer: There are now 2.07 moles of gas in the flask.

Explanation:

PV=nRT

P= Pressure of the gas = 697 mmHg = 0.92 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = ?

n = number of moles = 1.9

T = Temperature of the gas = 21°C=(21+273)K= 294 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

V=\frac{nRT}{P}=\frac{1.9\times 0.0821 \times 294}{0.92}=49.8L

When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.

PV=nRT

P= Pressure of the gas = 775 mmHg = 1.02 atm  (760 mmHg= 1 atm)

V= Volume of gas = volume of container = 49.8 L

n = number of moles = ?

T = Temperature of the gas = 26°C=(26+273)K= 299 K   (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm\K mol

n=\frac{PV}{RT}=\frac{1.02\times 49.8}{0.0821\times 299}=2.07moles

Thus the now the container contains 2.07 moles.

6 0
3 years ago
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