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3 years ago

Compare the strength of intermolecular forces in no2 to the strength of intermolecular forces in n2

2 answers:

7 0

N2 is non-polar molecules,
NO2 is polar molecules that have slightly positive charge on N and slightly negative on O, so intermolecular forces are stronger in NO2

Crazy boy [7]3 years ago

4 0

Answer: The intermolecular forces "Dipole-dipole forces" in NO2 is more stronger than "London-dispersion forces" in N2.

Explanation:

Dipole-dipole forces is considered as an attractive force (5.0 kJ to 20.0 kJ) of the positive end of one polar molecule with the negative end of another polar molecule and it is found in NO2.
The London dispersion force is the weakest intermolecular forces. It results when the electrons in two neighboring atoms positions that make the atoms form temporary dipoles and it is found in N2. This force is sometimes called an induced dipole attraction.
So, The dipole-dipole forces are more stronger than that of London-dispersion forces.

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A beaker with 1.60×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and c

stich3 [128]

Answer:

The pH will change 0.16 ( from 5.00 to 4.84)

Explanation:

Step 1: Data given

volume of acetic acid buffer = 160 mL

The total molarity of acid and conjugate base in this buffer is 0.100 M

A student adds 7.10 mL of a 0.460 M HCl solution to the beaker.

The pKa of acetic acid is 4.740

pH = 5.00

Step 2: Calculate concentration of acid

Consider x = concentration acid

Consider y = concentration conjugate base

x + y = 0.100

5.00 = 4.740 + log y/x

5.00 - 4.740 = log y/x

0.26 = log y/x

10^0.26 =1.82 = y/x

1.82 x = y

Since x+y = 0.100

x + 1.82 x = 0.100

2.82 x = 0.100

x =0.0355 M = concentration acid

Step 3: Calculate concentration of conjugate base

y = 0.100 - x

0.100 - 0.0355 =0.0645 M= concentration conjugate base

Step 4: Calculate moles of acid

Moles = volume * molarity

moles acid = 0.160 L * 0.0355 M= 0.00568 moles

Step 5: Calculate moles of conjugate base

moles conjugate base = 0.0645 M * 0.160 L=0.01032 moles

Step 6: Calculate moles HCl

moles HCl = 7.10 * 10^-3 L * 0.460 M=0.003266 moles

Step 7: Calculate new moles

A- + H+ = HA

moles conjugate base = 0.01032 - 0.003266 =0.007054 moles

moles acid = 0.00568 + 0.003266=0.008946 moles

Step 8: Calculate the total volume

total volume = 160 + 7.10 = 167.1 mL = 0.1671 L

Step 9: Calculate the concentration of the acid

concentration acid = 0.008946/ 0.1671 =0.0535 M

Step 10: Calculate the concentration of conjugate base

concentration conjugate base = 0.007054/ 0.1671 =0.0422 M

Step 11: Calculate the pH

pH = 4.740 + log 0.0535/ 0.0422=4.84

change pH = 5.00 - 4.84=0.16

The pH will change 0.16

5 0

3 years ago

a nuclide of 64/29 cu absorbs a positron. witch is the resulting atom? (A) 65/29Cu (B) 63/29 CU (C) 64/28Ni (D) 64/30 Zn

Nesterboy [21]

Answer : Option D) $Zn^{64}_{30}$

Explanation : When a positron is getting absorbed it means it will be $e^{+1}_{0}$ so, the $Cu^{64}_{29}$ will get converted;

So, the whole reaction will be;

$Cu^{64}_{29}$ + $e^{+1}_{0}$ ----> $Zn^{64}_{30}$ .

This will convert the whole element of Cu will get changed into Zn. As, it absorbs by the positron, the atomic number gets increased from 29 to 30.

4 0

3 years ago

If you have 120. mL of a 0.100 M TES buffer at pH 7.55 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pKa of T

Simora [160]

Answer:

The new pH after adding HCl is 7.07

Explanation:

The formula for calculating pH of a buffer is

pH = pKa + log([Conjugate base]/[Acid])

Before adding HCl,

7.55 = 7.55 + log([Conjugate base]/[Acid])

⇔ log([Conjugate base]/[Acid]) = 0

⇔ [Conjugate base] = [Acid] = 1/2 x 0.100 = 0.05 M

⇒ Mole of Conjugate base = Mole of Acid = 0.05 M x 0.12 mL = 0.006 mol

After adding HCl (3.00 mL, 1.00 M)

⇒ Mole of HCl = 0.003 x 1 = 0.003 mol)

New volume solution is 120 m L+ 3 mL = 123 mL

HCl is a strong acid, it will convert the conjugate base to acid form, or we can express

Mole of new Conjugate base = 0.006 - 0.003 = 0.003 mol

⇒ Concentration = 0.003/0.123 M

Mole of new Acid form = 0.006 + 0.003 = 0.009 mol

⇒ Concentration = 0.009/0.123 M

Use the formula

pH = pKa + log([Conjugate base]/[Acid])

= 7.55 + log(0.003 / 0.009) = 7.07

8 0

3 years ago

Consider the chemical equation. CuCl2 + 2NaNO3 mc023-1.jpg Cu(NO3)2 + 2NaCl What is the percent yield of NaCl if 31.0 g of CuCl2

lana [24]

CuCl2 + 2NaNO3 ----> Cu(NO3)2 + 2NaCl

using molar masses:-
Theoretical yields:-
63.54 + 2(35.45) g of CuCl2 produces 2(22.98 + 35.45) g of NaCl
134.44 g .................................................... 116.86 g
31.0 g ....................................................31.0 * 116.86 /134.44=26.95g

So percentage yield is 21.2* 100 / 26.95 = 78.7% to nearest tenth

5 0

3 years ago

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Fynjy0 [20]

Answer:

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Explanation:

5 0

3 years ago