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vagabundo [1.1K]

2 years ago

5

a- If an Lithium (atomic number 3) atom has a -2 charge, how many electrons does it have? b- Why do you do it this way? c- Is th

is an anion or a cation? d- Why? (4 points)

1 answer:

Darina [25.2K]2 years ago

6 0

Answer:

5

Explanation:

A lithium atom with an atomic number of 3 and a charge of -2 will have a total number of electrons.

The atomic number is the number of protons in an atom. For a neutral atom that has not gained or lost electrons, the atomic number is the same as the number of electrons.

neutral atom: atomic number of electrons

For neutral atom of lithium, we will have 3 electrons

Now, lithium with a charge of -2 has gained two more electrons to the original 3 this now makes it 5.

ii.

It is an anion because it is negatively charged.

Negatively charged particles are called anion.

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Balance this equation-<br> _P4 + _02 --> _P203

ikadub [295]

Answer:

Explanation:

If you insist on filling in the first blank you can put a one.

3 0

3 years ago

How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 4.46 atm and 45°C in the reaction shown below?

natka813 [3]

Answer:

8.1433 g of XeF₆ are required.

Explanation:

Balanced chemical equation;

XeF₆ (s) + 3H₂ (g) → Xe (g) + 6HF (g)

Given data:

Volume of hydrogen = 0.579 L

Pressure = 4.46 atm

Temperature = 45 °C (45+273= 318 k)

Solution:

First of all we will calculate the moles of hydrogen

PV = nRT

n = PV/ RT

n = 4.46 atm × 0.579 L / 0.0821 atm. dm³. mol⁻¹. K⁻¹ × 318 K

n = 2.6 atm . L / 26.12 atm. dm³. mol⁻¹

n = 0.0995 mol

Mass of hydrogen:

Mass = moles × molar mass

Mass = 0.0995 mol × 2.016 g/mol

Mass = 0.2006 g

Now we will compare the moles of hydrogen with XeF₆ from balance chemical equation.

H₂ : XeF₆

3 : 1

0.0995 : 1/3× 0.0995 = 0.0332 mol

Now we will calculate the mass of XeF₆.

Mass = moles × molar mass

Mass = 0.0332 mol × 245.28 g/mol

Mass = 8.1433 g

4 0

3 years ago

mixas84 [53]

1) Area of one side of each cube

The area of one side of each cube can be calculated by multiplying the length times the width:

$A=L\cdot W$

where

L = length

W = width

For cube A: $L=1 cm, W=1 cm$

So the area is: $A_A = (1)(1)=1 cm^2$

For cube B: $L=2 cm, W=2 cm$

So the area is: $A_B=(2)(2)=4 cm^2$

For cube C: $L=3 cm, W=3 cm$

So the area is: $A_C=(3)(3)=9 cm^2$

2) Surface area of each cube

The surface area of a cube is given by the sum of the areas of all its faces. Since a cube has 6 identical faces, this means that the total surface area is equal to 6 times the area of one face:

$A=6 A_1$

where

$A_1$ is the area of one face of the cube

For cube A, the area of one face is $A_A=1 cm^2$

So the surface area of cube A is

$A'_A = 6A_A=6(1)=6 cm^2$

For cube B, the area of one face is $A_B=4 cm^2$

So the surface area of cube B is

$A'_B = 6A_B=6(4)=24 cm^2$

For cube C, the area of one face is $A_C=9 cm^2$

So the surface area of cube C is

$A'_C = 6A_C=6(9)=54 cm^2$

3) Volume of each cube

The volume of a cube is obtained by multiplying its length, its width and its height:

$V=L\cdot W \cdot H$

Where:

L = length

W = width

H = height

Moreover for a cube, all the sides have equal length, so $L=W=H$

So the volume can be rewritten as

$V=L^3$

For cube A: L = 1 cm

So the volume is $V_A=(1 cm)^3 = 1 cm^3$

For cube B: L = 2 cm

So the volume is $V_B=(2 cm)^3 = 8 cm^3$

For cube C: L = 3 cm

So the volume is $V_C = (3 cm)^3 = 27 cm^3$

4) Ratio surface area/volume for each cube

In this part, we have to calculate the ratio between surface area and volume of each cube:

$r=\frac{A}{V}$

where

A is the surface area

V is the volume

For cube A, we have:

$A_A = 6 cm ^2$ (surface area)

$V_A=1 cm^3$ (volume)

So the ratio for cube A is:

$r=\frac{6}{1}=6$

For cube B, we have:

$A_B = 24 cm ^2$ (surface area)

$V_B=8 cm^3$ (volume)

So the ratio for cube B is:

$r=\frac{24}{8}=3$

For cube C, we have:

$A_C = 54 cm ^2$ (surface area)

$V_C=27 cm^3$ (volume)

So the ratio for cube C is:

$r=\frac{54}{27}=2$

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3 years ago

If an atom of thorium-234 were to undergo beta emission twice, which atom would result?

AnnZ [28]

Answer:

The answer to your question is U-234

Explanation:

Data

Thorium-234

beta emission twice

Definition

Beta emission is when a beta particle (electron) is emitted from an atomic nucleus.

First beta emission

²³⁴₉₀Th ⇒ ²³⁴₉₁Pa + e⁻

Second beta emission

²³⁴₉₁ Pa ⇒ ²³⁴₉₂U + e⁻

The atom will be Uranium-324

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Inessa [10]

Answer:

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Explanation:

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2 years ago

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