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tamaranim1 [39]
3 years ago
10

Van der Waals forces hold molecules together by: A. moving electrons from one molecule to another. B. attracting a lone pair of

electrons to the positive charge of a hydrogen. C. inducing temporary dipoles that attract each other. D. sharing electrons between atoms.
Chemistry
1 answer:
Rasek [7]3 years ago
6 0

The forces that are existing between molecules are known as intermolecular forces. These forces which are weaker than ionic and covalent bonds are classified into three types'

--> dipole-dipole attraction

--> VAN DER WAALS FORCES and

--> hydrogen bonding.

Van Der Waals forces was postulated by a Dutch physicist known as Van Der Waals. He postulated the existence of weak, short-range forces of attraction, which are independent of normal bonding forces, between non-polar molecules. He came to this conclusion after studying the behaviour of real gases at low temperatures and high pressures that:

--> electrons in a non-polar molecule such as hydrogen are close to one nucleus as to the other, although momentary concentration at one end of the molecule may occur,

--> this momentary concentration of electron cloud on one side create a temporary dipole in the hydrogen molecule, that is, one side of the molecule acquires a partial negative charge while the other side acquires a partial positive charge of equal magnitude,

--> the temporary dipole induces a similar dipole in an adjacent molecule,

--> this results in a temporary dipole-induced dipole attraction between the positive and negative ends of the adjacent molecules.

This is how weak Van Der Waals forces are set up. Therefore, option C is CORRECT which states that VAN DER WAALS forces hold molecules together by inducing temporary dipoles that attract each other.

Learn more about Van Der Waals forces here:

brainly.com/question/11457190

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Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly
trapecia [35]

Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of Ag_2O = 25.0 g

Mass of C_{10}H_{10}N_4SO_2 = 50.0 g

Molar mass of Ag_2O = 231.7 g/mole

Molar mass of C_{10}H_{10}N_4SO_2 = 250.3 g/mole

Molar mass of AgC_{10}H_{9}N_4SO_2 = 357.1 g/mole

First we have to calculate the moles of Ag_2O and C_{10}H_{10}N_4SO_2.

\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles

\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)

From the balanced reaction we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 1 mole of Ag_2O

So, 0.1998 moles of C_{10}H_{10}N_4SO_2 react with \frac{0.1998}{2}=0.0999 moles of Ag_2O

From this we conclude that, Ag_2O is an excess reagent because the given moles are greater than the required moles and C_{10}H_{10}N_4SO_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgC_{10}H_9N_4SO_2

From the reaction, we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 2 mole of AgC_{10}H_9N_4SO_2

So, 0.1998 mole of C_{10}H_{10}N_4SO_2 react with 0.1998 mole of AgC_{10}H_9N_4SO_2

Now we have to calculate the mass of AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=\text{ Moles of }AgC_{10}H_9N_4SO_2\times \text{ Molar mass of }AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=(0.1998moles)\times (357.1g/mole)=71.35g

Therefore, the mass of silver sulfadiazine produced can be, 71.35 grams.

8 0
3 years ago
Which equation represents single replacement reaction? A) CH4 + 2 02 → CO2 + 2 H2O B) CaCO3 + CaO + CO2 C) 2 Na + 2 H2O → 2 NaOH
loris [4]

Answer:

C) 2 Na + 2 H2O → 2 NaOH + H2.

Explanation:

Hello there!

In this case, according to the given chemical reactions, it is possible to firstly understand that a single displacement reaction is characterized by the presence of a single element as the first reactant and a compound as the second one, thus, yielding a compound as the first product and a single element as the second one.

In such a way, according to the given choices, it possible to note that C) 2 Na + 2 H2O → 2 NaOH + H2 is the only one with the aforementioned condition as the element at the reactants side is Na and at the products side is H2.

Best regards!

6 0
3 years ago
Which of the following situations most likely indicates the lowest relative humidity? situation A situation B situation C situat
statuscvo [17]

Answer:

Situation B indicates the significant depression of temperature in the wet bulb thermometer.

Answer Statement: B

Explanation:

Temperature Depression & Relative Humidity

As we can see that there are four cases to study the affect of relative humidity  on the temperature measured by thermometers, set up in different situations i.e., one is dry and the other one is dry.

As situation B describes the most admissible temperature drop in the wet bulb thermometer with a relative humidity of approximately 17% i.e., this can be the appropriate to reflect the lowest relative humidity with a fact that the relative humidity reduces with a larger count of temperature depression. Whereas, the relative humidity for other situations are resulting much higher.

5 0
3 years ago
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"Rate" is related to: <br> volume <br> time <br> distance <br> percentage
wariber [46]

Answer:

Time

Explanation:

8 0
3 years ago
The normal freezing point of water is 0.00 ⁰C. What is the freezing point of a solution containing450.0 mg of ethylene glycol (M
anyanavicka [17]

Answer:

Freezing T° of solution = - 8.98°C

Explanation:

We apply Freezing point depression to solve this problem, the colligative property that has this formula:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m

Kf = 1.86°C/m, this is a constant which is unique for each solvent. In this case, we are using water

m = molality (moles of solute / 1kg of solvent)

We convert the mass of solvent from g to kg

1.5 g . 1kg/1000g = 0.0015 kg

We convert the mass of solute, to moles. Firstly we make this conversion, from mg to g → 450mg . 1g/1000mg = 0.450 g

0.450 g. 1mol / 62.07g = 7.25×10⁻³ moles

Molality → 7.25×10⁻³ mol / 0.0015 kg = 4.83 m

- Freezing T° of solution = 1.86°C /m . 4.83 m - Freezing T° of pure solvent

-Freezing T° of solution = 1.86°C /m . 4.83 m - 0°C

Freezing T° of solution = - 8.98°C

8 0
3 years ago
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