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White raven [17]
3 years ago
15

Help please !

Chemistry
2 answers:
Veronika [31]3 years ago
6 0
The best answer is E=63 times 10^12
nikklg [1K]3 years ago
4 0

mass  \: defect \: is \:  \:  given \: to \: be \:  6.9986235 \times  {10}^{ - 29}  \: which \: is \: approx \: 7 \times  {10}^{ - 29}

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One of the reasons that stars are so large is because _____.
ValentinkaMS [17]

Answer:

They are composed of hydrogen and helium

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3 years ago
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What is the chemical name for K2Se
Oduvanchick [21]

Answer:

Potassium selenide

Explanation:

Potassium selenide (K2Se)

6 0
2 years ago
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What amount of sucrose (C12H22O11) should be added to 5.83 mol water to lower the vapor pressure of water at 50 °C to 72.0 torr?
Triss [41]

Answer:

The amount of sucrose that must be added is 1.66 moles

Explanation:

Colligative property of lowering vapor pressure has this formula:

Vapor pressure of pure solvent (P°) - Vapor pressure of solution = P° . Xm

We have both vapor pressure (pure solvent and solution9, so let's determine the ΔP

ΔP = 92.6 Torr - 72 Torr = 20.6 Torr

Let's add the data in the formula

20.6 Torr = 92.6 Torr . Xm

Xm = Mole fraction of solute → (mol of solute/ mol of solute + mol of solvent)

Mol of solvent = 5.83 mol (data from the problem)

Therefore Xm = 20.6 Torr / 92.6 Torr → 0.222

Let's find out the moles of solute (our unknown value)

0.22 = moles of solute / moles of solute + 5.83 moles of solvent

0.222 (moles of solute + 5.83 moles of solvent) = moles of solute

0.222 moles of solute + 1.29 moles of solvent = moles of solute

1.29 moles of solvent = moles of solute - 0.222 moles of solute

1.29 moles = 0.778 moles of solute

1.29 / 0.778 = moles of solute → 1.66 moles

3 0
3 years ago
Extend the aufbau sequence through an element that has not yet been identified, but whose atoms would completely fill 7p orbital
Thepotemich [5.8K]

Answer:

<u>Number of electrons</u> = 118

<u>Electronic configuration</u>: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

Explanation:

Given: A chemical element that has completely filled 7p orbital.

According to this, the principal occupied electron shell or the <u>valence shell of such an element is 7p.</u>

⇒ the principal quantum number <u>(n) for the valence shell is 7.</u>

∴ this element belongs to the period 7 of the periodic table.

Also, an element that has completely filled p-orbital belongs to the group 18 of the p-block.

Therefore, an element that belongs to the group 7 and period 18 of the periodic table, should have a <u>completely filled 5f, 6d, 7s and 7p orbitals</u>.

Therefore, the electronic configuration should be: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶

And, the <u>number of electrons</u> = atomic number of radon (Rn) + 14 + 10 + 2 + 6 = 86 + 32 = <u>118</u>.

<u>Therefore, the given element has atomic number 118 and has the electronic configuration: [Rn] 5f¹⁴ 6d¹⁰ 7s² 7p⁶. Thus the given element can be Oganesson.</u>

7 0
3 years ago
Fill in the blanks:
Misha Larkins [42]

Answer:

1. smaller. 2. smaller. 3. greater

Explanation:

1. H−O−H angle is 104.45  and H−C−H angle is 109.5

2. O−S−O angle is 119 and F−B−F angle is 120

3. The F−S−F bond angle in SF₆ is 90 and F−Br−F bond angle in BrF₅ is 84.8

8 0
3 years ago
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