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Citrus2011 [14]
3 years ago
15

Why do electrons enter the 4s orbital before entering the 3d orbital?

Chemistry
2 answers:
LUCKY_DIMON [66]3 years ago
8 0
C. Because as it get farther from the nucleus the amount of energy is decreasing
salantis [7]3 years ago
7 0

Answer:

C. because the 4s orbital is at a lower energy level

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For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a
Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁   -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁   ----> P cancels out  

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0
3 years ago
What decides if an atom will be a cation or an anion?
JulijaS [17]

The loss or gain of electrons

Explanation:

The loss or gain of electrons determines if an atom will become a cation or anion.

A cation is a positively charge ion

An anion is a negatively charged ion.

In an atom, we have sub-atomic particles:

 Protons are the positively charged particles

 Electrons are negatively charged

  Neutrons carry no charges

Only electrons can be lost or gained in chemical processes that forms cations and anions.

When a neutral atom gains electron, it has more electrons than protons. This makes it negatively charged and we call it an anion.

When a neutral atom loses an electron, the number of protons is more. We call it a cation.

Learn more:

Cations brainly.com/question/4670413

#learnwithBrainly

7 0
3 years ago
Can somebody help me please? it’s urgent
Anettt [7]

Answers:

See attached table

Explanation:

The isotope symbol will have the mass number and chemical symbol.  The atomic number will be equal to the number of protons because the atomic number is the number of protons in an atom.  The electrons will be equal to the protons because the atom is neutral.  The neutrons can be found by subtracting the number of protons from the mass number.  The mass number can be found by adding the protons and neutrons.  It is also in the name of the isotope.

4 0
3 years ago
Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for
abruzzese [7]

Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

Ca^{2+ + y^{4- ⇄  CaY^{2-

Formation constant Kf

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) = 5.0 × 10¹⁰

Now,

[y^{4-] = \alpha _4CH_4Y; ∝₄ = 0.35

so the equilibrium is;

Ca^{2+ + H_4Y ⇄  CaY^{2- + 4H⁺

Given that; CH_4Y = Ca^{2+     { 1 mol Ca^{2+  reacts with 1 mol H_4Y  }

so at equilibrium, CH_4Y = Ca^{2+ = x

∴

Ca^{2+ + y^{4- ⇄  CaY^{2-

x        + x         0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) =  CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] )  = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x²  = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

8 0
2 years ago
What generic products are produced by an Acid/Base reaction?
gavmur [86]

Answer:

Salt NaCl

Explanation:

5 0
3 years ago
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