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Naddika [18.5K]
3 years ago
7

Are all single bonds sigma bonds?

Chemistry
1 answer:
Nikolay [14]3 years ago
6 0
Single bonds are those that bond with one atom, and sigma bonds are the strongest type of covalent bonds that are single bonded. 

That means NO, not all single bonds are sigma bond, but all sigma bonds are single bonds. 
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Substances have a fixed arrangement of atoms. true or false (NO LINKS PLEASE)
KATRIN_1 [288]

Answer:

True

Explanation:

5 0
3 years ago
You find a ring with a mass of 107g . You fill a graduated cylinder up with 10ml of water and put the ring into the cylinder . T
Korvikt [17]

Answer:

5 mL

Explanation:

Given data:

mass of ring = 107 g

volume of water = 10 mL

increase in volume = 15 mL

How much water displace = ?

Solution:

V (ring) = V (water + ring) - V (water)

V (ring) = 15 mL -  10 mL

V (ring) = 5 mL

when the ring is put into cylinder, volume is increased by 15 mL. The volume of water was 10 mL so water is displaced by 5 mL and the volume 5mL is the voulme of ring.

3 0
3 years ago
What is the difference between a cation and an anion?
zmey [24]
An anion has a negative charge, while a cation has a positive charge.
3 0
3 years ago
Consider the reaction of gaseous hydrogen with gaseous oxygen to produce gaseous water. Given that the first picture represents
Bogdan [553]

The question is incomplete. There's missing the image, which is shown below.

Answer:

Volume of O₂ = 6 L, volume of mixture: 18 L, volume of H₂O = 12 L, molecule volume of H₂O = 0.667 molecule/L

Explanation:

The reaction between hydrogen gas and oxygen gas to form water is:

2H₂(g) + O₂(g) → 2H₂O(g)

So, for 1 mol of O₂ is necessary 2 moles of H₂ form 2 moles of H₂O. As the images below there's 8 molecules of H₂, 4 molecules of O₂, 12 molecules in the mixture, and 8 molecules of H₂O. Thus, there are stoichiometric values.

All the images are at the same temperature and pressure, so, by the ideal gas law:

PV= nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The number of moles and molecules are related, so let's substitute it in the equation. For the H₂:

P*12 = 8*RT

RT/P = 12/8 = 1.5

Thus, for O₂:

PV= nRT

V = n*(RT/P)

V = 4*1.5 = 6 L

For the mixture:

V = 12*1.5 = 18 L

For H₂O:

V = 8*1.5 = 12 L

The molecule volume is the number of molecules divided by the volume they occupy, thus for water: 8/12 = 0.667 molecules/L

6 0
3 years ago
a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill
baherus [9]

Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

Also,  

Moles = mass (m) / Molar mass (M)

Density (d)  = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

PM=dRT

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol  = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>

5 0
3 years ago
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