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2 years ago

56. The electron configuration for phosphorous is [Ar]3s23p4. a. TRUE b. FALSE

Chemistry

2 answers:

melomori [17]2 years ago

7 0

Answer:

False

Explanation:

AlladinOne [14]2 years ago

3 0

Answer:

true

Explanation:

Ground state electron configurations are the foundation for understanding molecular bonding, properties, and structures. From the electrons in an atom, to the differing orbitals and hybridization, the ground state electron configuration sheds light on many different atomic properties.

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What are the mole fraction and the mass percent of a solution made by dissolving 0.21 g KBr in 0.355 L water? (d = 1.00 g/mL.) m

IRISSAK [1]

Answer:

Mol fraction H2O = 0.99991

Mol fraction KBr = 0.00009

mass % KBr = 0.059 %

mass % H2O = 99.941 %

Explanation:

Step 1: Data given

Mass of KBr = 0.21 grams

Molar mass KBr = 119 g/mol

Volume of water = 355 mL

Density of water = 1.00 g/mL

Molar mass water = 18.02 g/mol

Step 2: Calculate mass water

Mass water = 355 mL * 1g /mL

Mass water = 355 grams

Step 3: Calculate moles water

Moles water = mass water / molar mass water

Moles water = 355 grams / 18.02 g/mol

Moles water = 19.7 moles

Step 4: Calculate moles KBr

Moles KBr = 0.21 grams / 119 g/mol

Moles KBr = 0.00176 moles

Step 5: Calculate total moles

Total moles = 19.7 moles + 0.00176 moles

Total moles = 19.70176 moles

Step 6: Calculate mol fraction

Mol fraction H2O = 19.7 moles / 19.70176 moles

Mol fraction H2O = 0.99991

Step 7: Calculate mol fraction KBr

Mol fraction KBr = 0.00176 / 19.70176

Mol fraction KBr = 0.00009

Step 6: Calculate mass %

mass % KBR = (0.21 grams / (0.21 + 355) grams) *100%

mass % KBr = 0.059 %

mass % H2O = (355 grams / 355.21 grams) *100%

mass % H2O = 99.941 %

8 0

3 years ago

What part of the sun do we see from earth?

NISA [10]

Answer:

photosphere

Explanation:

photosphere

There are 3 main layers of the Sun that we can see. They are the photosphere, the chromosphere and the corona. Together they make up the "atmosphere" of the Sun. The part of the Sun that glows (and that we see with the naked eye) is called the photosphere

8 0

3 years ago

Read 2 more answers

Which describes the earth as an open system with respect to energy?

Naddik [55]

Answer:

Explanation:

Any system within the Earth system is considered an open system. Because energy flows freely into and out of systems, all systems respond to inputs and, as a result, have outputs.

6 0

2 years ago

Pls it’s urgent

Rainbow [258]

Answer:

0.02 moles.

Explanation:

volume of H₂ gas at R.T.P = 480 cm³

Where

R.T.P = room temperature and pressure

molar volume of gas at = 24000 cm³

no. of moles of hydrogen = ?

Solution:

formula Used

no. of moles = volume of gas / molar volume

put values in above equation

no. of moles = 480 cm³ / 24000 cm³/mol

no. of moles = 0.02 mol

So,

no. of moles of hydrogen in 480 cm³ is 0.02 moles.

4 0

2 years ago

Uranium-238 decays to lead-206 with a half-life of 4.5 x 109 yr. Determine how much uranium-238 decays in milligrams (to three s

Mamont248 [21]

Answer:

2.15 mg of uranium-238 decays

Explanation:

For decay of radioactive nuclide-

$N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where N is amount of radioactive nuclide after t time, $N_{0}$ is initial amount of radioactive nuclide and $t_{\frac{1}{2}}$ is half life of radioactive nuclide

Here $N_{0}=4.60 mg$ , $t=4.1\times 10^{9}yr$ and $t_{\frac{1}{2}}=4.5\times 10^{9}yr$

So, $N=(4.60mg)\times (\frac{1}{2})^{\frac{4.1\times 10^{9}}{4.5\times 10^{9}}}$

so, N = 2.446 mg

mass of uranium-238 decays = (4.60-2.446) mg = 2.15 mg

3 0

3 years ago