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kirza4 [7]
3 years ago
11

Classify the following statements regarding the law of mass conservation as true or false.

Chemistry
1 answer:
Ahat [919]3 years ago
8 0
True true false true
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Explain what might happen to tundra animals, such as polar bears, as earth's climate warms. pls explain
SCORPION-xisa [38]

Answer:

They will die.

Explanation:

Polar Bears were made to live in the frigid temperatures of the tundras. If there is a temperature change it can affect their habitat. It can destroy it and they would have to move out to another place. They can die due to global warming.

8 0
3 years ago
If an atom has 34 electrons, 46 neutrons, and 34 protons, what is its mass number? what is the name of the element?
OleMash [197]
Mass number = protons + neutrons = 34+46 =80

the element with an atomic number of 34 and mass number of 80 is selenium
7 0
3 years ago
A certain orbital of the hydrogen atom has n = 4 and l = 2 what are the possible values of ml for this orbital?
Nadya [2.5K]

The ml is also called as the magnetic quantum number. The value of ml can range from –l to +l including zero. Hence all of the possible values for ml given that l = 2 are:

<span>-2, -1, 0, + 1, + 2</span>

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
100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temper
Olin [163]

Answer:

100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temperature rises from 21.80 °c to 23.20 °c. Assuming the density and specific heat of the resulting solution is 1.00 g/ml and 4.18 j/g ∙ °c respectfully, what is the ΔH°rxn?

39.013 kJ/mol.

Explanation:

AgNO3(aq) + HCl(aq) --------------> AgCl(s) + HNO3(aq)

We can calculate the amount of heat (Q) released from the solution using the relation:

Q = m.c.ΔT,

Where, Q is the amount of heat released from the solution (Q = ??? J).

m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.

c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).

ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).

∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.

∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of AgNO₃)/(Volume of the solution (L)).

∴ no. of moles of AgNO₃

               = (M)(Volume of the solution (L))

               = (0.3 M)(0.1 L) = 0.03 mol.

∴ ΔH°rxn

           = Qrxn/(no. of moles of AgNO₃)

            = (1170.4 J)/(0.03 mol)

            = 39013.33 J/mol

           = 39.013 kJ/mol.

7 0
3 years ago
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