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3 years ago

BRAINLIESTTT ASAP!! PLEASE HELP ME :)

Chemistry

2 answers:

deff fn [24]3 years ago

7 0

Answer:

True

Explanation:

There are two types of Van der waals forces which are intermolecular bonds. These bonds are London dispersion forces and Dipole-dipole attractions.

The london dispersion forces are weak attractions found between non-polar molecues and noble gas.

The attractions here is as a result of the fact that non-polar molecules or atoms sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant. This leads to the formation of a temporary dipole or instantaneous dipole. The dipole can induce the neighbouring molecules to be distorted and they form dipoles as well.

BartSMP [9]3 years ago

4 0

Answer:

True

Explanation:

When the electrons migrate to one end of a molecule or atom, the other end has a tiny positive pole, whereas when the electrons travel, the other end has a tiny negative pole. For that moment, a dipole is formed.

I am joyous to assist you anytime.

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Teeth and long bones are good indicators of age in children; ribs and joints are better for adults.

attashe74 [19]

Answer:

True

Explanation:

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2 years ago

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According to kinetic-molecular theory, which of the following would not be considered an ideal gas? Check all that apply.

Vikki [24]

Answer:

are a gas at very low volumes, when gas particles are very close together

a gas at very low temperatures, when gas particles have very little kinetic energy

a gas with highly polar molecules that have very strong intermolecular forces

Explanation:

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3 years ago

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Which of the following ions would represent the ion of an element from Group 2 on the Periodic Table? A2+ Y3+ X2- Z+

Luba_88 [7]

A2+. Group 2 elements form cations with 2+ charge.

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3 years ago

Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm

8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

$\frac{dV}{dt}=10\frac{cm^3}{s}$

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

$\frac{dh}{dt} = ?*\frac{dV}{dt}$

Of course, $?=\frac{dh}{dV}$ .

Now, since the volume of a cone is $V=\pi r^2h/3$ and the ratio $r/h=15/20=3/4$ or $r=3/4h$ , the volume becomes:

$V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3$

We proceed to its differentiation:

$\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}$

Then, we compute $\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}$

Finally, at h=2:

$\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}$

Best regards.

4 0

3 years ago

What is the freezing point of a solution of 498mL of water (solute) dissolved in 2.50 L of ethanol (solvent), C2H5OH? The densit

jok3333 [9.3K]

Answer:

Freezing T° of solution is -142.4°C

Explanation:

This excersise is about colligative properties, in this case freezing point depression,

ΔT = Kf . m . i

Where ΔT = Freezing T° of solvent - Freezing T° of solution

Kf = Cryoscopic constant

m = mol/kg (molality)

i = Number of ions dissolved.

Water is not ionic, so i = 1

Let's find out m.

We determine mass of water, by density

498ml . 1 g/mL = 498 g

We convert the mass of water to moles → 498 g . 1mol/18g = 27.6 moles

We determine mass of solvent by density

2500 mL . 0.789 g/mL = 1972.5 g

Notice, we had to convert L to mL to cancel units.

1 cm³ = 1 mL

We convert the mass from g to kg → 1972.5 g . 1kg /1000g = 1.97kg

We determine m = mol/kg → 27.6mol / 1.97kg = 13.9 m

Kf for ethanol is: 1.99 °C/m

Freezing T° for ethanol is: -114.6°C

We replace at formula: - 114.6°C - Freezing T° solution = 1.99 °C/m . 13.9 m . 1

- 114.6°C - Freezing T° solution = 27.8 °C

- Freezing T° solution = 27.8°C + 114.6°C

Freezing T° Solution = - 142.4 °C

7 0

2 years ago