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koban [17]
4 years ago
15

______________ forces are between different ATOMS of the SAME molecule. ______________ forces are between different MOLECULES of

a substance.
A. Intermolecular; Intramolecular
B. Intramolecular; Intrermolecular
C. Intermolecular; Polar
D. Polar; Intramolecular
Chemistry
2 answers:
777dan777 [17]4 years ago
5 0

Hey there!

Your answer is B. Intramolecular; intermolecular.

Intramolecular forces are the forces that keep the atoms together in the molecule.

Intermolecular forces are the forces that are different molecules interacting with each other.

You can remember this by thinking about the prefixes inter- and intra-. Inter- refers to open systems within groups. The groups are the different molecules which are interacting with each other.

Intra-, however, refers to the closed system inside of a single group. The single group is the one molecule, and the interactions are happening between the atoms within the molecule.

Hope this helps!

hichkok12 [17]4 years ago
4 0
It would be B that the answer!
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Chemistry
madam [21]

Answer: 1. 3.914 × ^10-4 | 2. 4.781 × ^10-1

Explanation:

7 0
3 years ago
A twenty-eight-liter volume of gas contains 11 g methane, 1.5-gram nitrogen and 16-gram carbon dioxide. Determine partial pressu
GREYUIT [131]

Based on Dalton's Law, for a mixture of gases, the total pressure is the sum of the partial pressure of each gas.

Partial pressure (p) of each gas is related to the total pressure (P) as follows:

p = X * P----------(1)

where X is the mole fraction of that gas

X = moles of a particular gas/total number of moles of all gases in the mixture--------------(2)

Step 1: Calculate the moles of each gas

Mass of methane, CH4 = 11 g

Mass of nitrogen, N2 = 1.4 g

Mass of carbon dioxide, CO2 = 16 g

# moles of CH4 = 11 g/16 gmol-1 = 0.6875

# moles of N2 = 1.4/28 = 0.05

# moles of CO2 = 16/44 = 0.3636

Total moles = 0.6875+0.05+0.3636 = 1.1011

Step2: Calculate mole fractions of each gas

Based on equation (2)

X(CH4) = 0.6875/1.1011 = 0.6244

X(N2) = 0.05/1.1011 = 0.0454

X(CO2) = 0.3636/1.1011 = 0.3302

Step 3: Calculate the total pressure

Based on ideal gas equation:

PV = nRT

given that;

V = 28 L

n = total moles = 1.1011

R = gas constant = 0.0821 Latm/mol-K

Since temp T is not given, let us consider room temperature of 25 C = 25 + 273 = 298 K

Now, P = nRT/V = 1.011*0.0821*298/28 = 0.962 atm

Step 3: Calculate partial pressures

Based on equation:

p(CH4) = 0.6244*0.962 atm = 0.601 atm

P(N2) = 0.0454*0.962 atm = 0.044 atm

P(CO2) = 0.3302*0.962 atm = 0.318 atm


4 0
4 years ago
Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6
liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 137ml+137ml=274ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

where,

q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 274 g

T_{final} = final temperature of water = 325.8 K

T_{initial} = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

q=274g\times 4.18J/g^oC\times (325.8-298)K

q=31839.896J=31.84KJ

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0
4 years ago
A sample of concrete contains cement, gravel, crushed rocks, sand, and water. Each of these contains different metals and minera
const2013 [10]

Answer:

The answer to your question is a heterogeneous mixture

Explanation:

The Matter is classified as Pure substances and Mixtures

Pure substances are elements or compounds which are not mixed with more substances.

Mixtures are several elements or compounds together, mixtures can be homogeneous or heterogeneous.

Homogeneous mixtures are when the different components can be identified.

Heterogeneous mixtures are when the different components can be identified just by looking at the mixture.

In concrete we can identify the different components like gravel, sand, crushed rocks, etc, so concrete is a heterogeneous mixture.

8 0
3 years ago
Addition of _____________ to pure water causes the least increase in conductivity. A) weak bases B) acetic acid C) ionic compoun
elena55 [62]

Answer:

c

Explanation:

3 0
3 years ago
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