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3 years ago

The electron configuration belonging to the atom with the highest second ionization energy is ________.

Chemistry

1 answer:

Gemiola [76]3 years ago

5 0

Answer:

Element Lithium

Explanation:

The element with the highest second ionization energy is lithium. It belongs to the alkaline metal group I.e group one metals

It has the highest second ionization energy because it is very difficult to remove the electron from the 1s orbital.

Its atomic number is 3. The electronic configuration is 1s2 2S1

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Vanadium crystallizes in a body-centered cubic lattice, and the length of the edge of a unit cell is 305 pm. What is the density

Anestetic [448]

Answer:

5.96 g/cm^3

Explanation:

Corner atom = 1/8

Atoms in center = 1

Atoms in face of the cube= 1/2

Molar mass of V = 50.94 g/mol (from period table)

1 mole = 6.02x10^23

In BCC unit cell:

(8 x 1/8)+ 1=2 per 1 unit cell

Mass: 2(50.94g)/6.02x10^23 = 1.69x10^-22 g/unit cell

305pm=(305x10^-12m÷10^-2m) x (1mL÷1cm^3)

= 2.837 x 10^-23 mL

1pm=10^-12m

1cm=10^-2m

1mL=1cm^3

density=mass/volume

density of V = 1.69x10^-22g÷2.837x10^-23mL

=5.957g/mL

=5.96g/cm^3

5 0

3 years ago

1 The process of a solid turning into a liquid is called

katrin [286]

Answer:

Melting

Explanation:

Once heat is added it will turn to a liquid

7 0

3 years ago

Read 2 more answers

How many moles of gold, Au, are in 3.60 x 10^-5 g of gold?

zlopas [31]

<h3>Answer:</h3>

1.83 × 10⁻⁷ mol Au

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

3.60 × 10⁻⁵ g Au (Gold)

Step 2: Identify Conversions

Molar Mass of Au - 196.97 g/mol

Step 3: Convert

Set up: $\displaystyle 3.60 \cdot 10^{-5} \ g \ Au(\frac{1 \ mol \ Au}{196.97 \ g \ Au})$
Multiply: $\displaystyle 1.82769 \cdot 10^{-7} \ mol \ Au$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

1.82769 × 10⁻⁷ mol Au ≈ 1.83 × 10⁻⁷ mol Au

4 0

3 years ago

(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction

ale4655 [162]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

$Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2$

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

$m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2$

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

$m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu} \\\\m_{Cu}=0.380gCu$

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

$m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3} \\\\m_{AgNO_3}=2.03gAgNO_3$

Best regards!

5 0

3 years ago

Plete the following problems significant figures: 6 × 0.30

jeka57 [31]

6= Only the digits 1 and 6 are the actual measured values. Therefore we have only 2 significant figures.

0.3= Zeros used as placeholders are not significant. Zeros that come before non-zero integers are never significant. Example 5: The zeros in 098, 0.3, and 0.000000000389 are not significant because they are all in front of non-zero integers. c. If the zeros come after non-zero integers and are followed by a decimal point, the zeros are significant.

3 0

3 years ago