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2 years ago

How can you tell if a molecule is polar or nonpolar?.

Chemistry

1 answer:

Semmy [17]2 years ago

3 0

$\huge \underline\frak \red{Answer}$

1) If the arrangement is symmetrical and the arrows are of equal length the molecule is nonpolar

2) If the arrows are of different lengths and if they do not balance each other the molecule is polar

3) If the arrangement is asymmetrical the molecule is polar

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Object a has a mass of 12 G and a volume of 8 cm3 object be has a mass of 20 G and a volume of 8 cm3 which object has a greater

solniwko [45]

Answer:

Option C = object B by 1 gram per cubic cm.

Explanation:

Given data:

Mass of object A = 12 g

Volume of object A = 8 cm³

Mass of object B = 20 g

Volume of object B = 8 cm³

Densities = ?

Solution:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Density of object A:

d = m/v

d = 12 g/ 8 cm³

d = 1.5 g/cm³

Density of object B:

d = m/v

d = 20 g/ 8 cm³

d = 2.5 g/cm³

object b has high density.

5 0

3 years ago

The partial pressure of hellium gas in a gaseous mixture of hellium and hydrogen is<br>

RUDIKE [14]

Answer:

The partial pressure of helium gas in a gaseous mixture of helium and hydrogen is the pressure that the helium would exert in the absence of the hydrogen. equal to the total pressure divided by helium's molar mass. O equal to the total pressure divided by the number of helium atoms present.

Hope this Helps (✿◡‿◡)

8 0

3 years ago

After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108. Expre

Marina CMI [18]

Answer:

$\large \boxed{1.77 \times 10^{-5}\text{ mol/L}}$

Explanation:

Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.

1. Moles of Ni²⁺

$n = \text{135 mL} \times \dfrac{\text{0.0147 mmol}}{\text{1 mL}} = \text{1.984 mmol}$

2. Moles of NH₃

$n = \text{190 mL} \times \dfrac{\text{0.250 mmol}}{\text{1 mL}} = \text{47.50 mmol}$

3. Initial concentrations after mixing

(a) Total volume

V = 135 mL + 190 mL = 325 mL

(b) [Ni²⁺]

$c = \dfrac{\text{1.984 mmol}}{\text{325 mL}} = 6.106 \times 10^{-3}\text{ mol/L}$

(c) [NH₃]

$c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}$

3. Equilibrium concentration of Ni²⁺

The reaction will reach the same equilibrium whether it approaches from the right or left.

Assume the reaction goes to completion.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0

C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0

E/mol·L⁻¹: 0 0.1095 6.106×10⁻³

Then we approach equilibrium from the right.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 0 0.1095 6.106×10⁻³

C/mol·L⁻¹: +x +6x -x

E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}$

Kf is large, so x ≪ 6.106×10⁻³. Then

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}$

3 0

3 years ago

The town of Eatonville has decided to replace all of the incandescent light bulbs in their public buildings with more energy-eff

astraxan [27]

Answer:A

Explanation:becuse they waste less heat energy

6 0

2 years ago

How many quarts of pure antifreeze must be added to 5 quarts of a 20 ntifreeze solution to obtain a 50 ntifreeze solution?

Ede4ka [16]

Quarts of pure antifreeze must be added to 5 quarts of a 20 antifreeze solution to obtain a 50 antifreeze solution is 3.

Antifreeze solution is a solute that lowers the freezing point of the liquid in a solution.

For calculating the amount of quarts of pure antifreeze solution added can be given by:

5×20/100 + x × 100% = (x+5) × 50/100

Where, x is the amount of antifreeze required for addition and pure antifreeze is 100% of the antifreeze taken.

1 + x = (x+5) 0.5

1 + x = 0.5x + 2.5

0.5x = 1.5

x = 3

So, 3 quarts of pure antifreeze must be added to 5 quarts of a 20 antifreeze solution to obtain a 50 antifreeze solution.

Learn more about Antifreeze solution here, brainly.com/question/14619701

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5 0

2 years ago