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3 years ago

What is the strongest type of intermolecular force present in nh2ch3?

1 answer:

3 0

Correct Answer: Hydrogen bonding

Reason:
In methyl amine (i.e. CH3NH2) several inter-molecular forces of interaction may be operable. This includes:
1) Dipole-Dipole interaction
2) Dipole-induced dipole intraction
3) van der Waal's interaction
4) Hydrogen bonding

Among all the listed interactions, hydrogen bonding is the strongest. With reference to hydrogen bonding, it exist in system when 'H' atom is bonded to electronegative element like N, O and F.

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The final marks in a statistics course are normally distributed with a mean of 70 and a standard deviation of 10. The professor

Alinara [238K]

Use inversenormal on a calculator and type in .1 , 70 , 10 . That percent is for A and will determine at what mark an A will be. Do the same for the rest of the grades but change the first argument in the calculation

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3 years ago

2. If 5.4 g Al are reacted with excess HCI, how many moles of H2 will be produced?

Tanya [424]

Answer:

0.3 moles of H2 will be produced.

Explanation:

Balanced reaction: 2 Al + 6 HCl → 2 AlCl3 + 3 H2

Given Al mass = 5.4 g, we know Mr of Al is 27

Using the formula:

moles = mass / Mr

moles of Al = 5.4 / 27

moles of Al = 0.2 moles

Here we see that "2" molar mass before Al; so in this reaction, "1" molar mass will have [(0.2)/2] = 0.1 moles. He see "3" molar mass before H2, therefore moles of H2 is (3 * 0.1) = 0.3 moles.

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2 years ago

The Law of Superposition is most relevant when studying which type of rock?

bagirrra123 [75]

Answer:

igneous or metamorphic

Explanation:

Those two are sorta relvant

8 0

3 years ago

Read 2 more answers

Assume that 50.0mL 50.0mL of 1.0MNaCl(aq) 1.0MNaCl(aq) and 50.0mL 50.0mL of 1.0M AgNO 3 (aq) 1.0MAgNO3(aq) were combined. Accord

S_A_V [24]

Answer:

The amount of precipitate formed would 7.175 grams of silver chloride.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of NaCl = n

Volume of NaCl solution = 50.0 mL = 0.050 L

Molarity of the hydrogen peroxide = 2.0 M

$n=2.0 M\times 0.050 L=0.100 mol$

Moles of silver nitarte = n'

Volume of silver nitrate solution = 50.0 mL = 0.050 L

Molarity of the silver nitrate = 1.0 M

$n'=1.0 M\times 0.050 L=0.050 mol$

$NaCl(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of NaCl

This means that silver nitrate is in limiting amount and NaCl is in excessive amount.

So, the amount of AgCl depends upon amount of silver nitrate.

According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.

Then 0.050 moles of silver nitrate will give;

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of AgCl

Mass of 0.050 moles of AgCl ;

$0.050 mol\times 143.5 g/mol=7.175 g$

The amount of precipitate formed would 7.175 grams of silver chloride.

8 0

3 years ago

2Al + 6HCl --> 2AlCl3 + 3H2 Aluminium reacts with hydrochloric acid. How many grams of aluminum are necessary to produce 11 L

dem82 [27]

Answer:

8.8g of Al are necessaries

Explanation:

Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.

To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:

Moles H2:

PV = nRT; PV/RT = n

Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP

Replacing:

1atm*11L/0.082atmL/molK*273.15K = n

n = 0.491 moles of H2 must be produced

Moles Al:

0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required

Mass Al -Molar mass: 26.98g/mol-:

0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries

7 0

4 years ago