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eimsori [14]
3 years ago
15

Use the picture to answer the question thanks

Chemistry
1 answer:
KiRa [710]3 years ago
8 0

Answer:the second one

Explanation:,Hydrogen bonding is a special type of dipole-dipole attraction between molecules ,not a covalent bond to a hydrogen atom. It results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom such as a N, O, or F atom and another very electronegative atom.

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The standard enthalpy of formation of BrCl(g) is 14.7 kJmol-1 . The standard enthalpies for the atomization of Br2(l) and Cl2(g)
natita [175]

Explanation:

Equation of the reaction:

Br2(l) + Cl2(g) --> 2BrCl(g)

The enthalpy change for this reaction will be equal to twice the standard enthalpy change of formation for bromine monochloride, BrCl.

The standard enthalpy change of formation for a compound,

ΔH°f, is the change in enthalpy when one mole of that compound is formed from its constituent elements in their standard state at a pressure of 1 atm.

This means that the standard enthalpy change of formation will correspond to the change in enthalpy associated with this reaction

1/2Br2(g) + 1/2Cl2(g) → BrCl(g)

Here, ΔH°rxn = ΔH°f

This means that the enthalpy change for this reaction will be twice the value of ΔH°f = 2 moles BrCl

Using Hess' law,

ΔH°f = total energy of reactant - total energy of product

= (1/2 * (+112) + 1/2 * (+121)) - 14.7

= 101.8 kJ/mol

ΔH°rxn = 101.8 kJ/mol.

8 0
3 years ago
Only some particles split up into smaller particles
Makovka662 [10]

Answer:

Everything around you can be broken down into smaller particles called atoms. The particles of one substance are all the same and different substances are made up of different particles.

Explanation:

4 0
3 years ago
Type of reaction<br> PbSO. →<br> PbSO3 + __02<br> Type of reaction:
Marysya12 [62]

Answer:

1.

work out the mean mode median and range

Explanation:

6 0
2 years ago
A cylindrical piece of metal is 4.5 dm in height with radius of 5.50 x 10^-5 km.
adell [148]

Answer:

a) V=4.3x10^3mL

b) V=4.3x10^6mm^3

c) \rho=1.5x10^5g/L

Explanation:

Hello,

a) In this case, the given height in cm is:

h=4.5dm*\frac{1m}{10dm}* \frac{100cm}{1m}=45cm

And the radius in cm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{100cm}{1m}=5.5cm

Thus, the volume in cubic centimeters which is also equal in mL (1cm³=mL) is:

V=\pi (5.5cm)^2*45cm\\\\V=4.3x10^3cm^3=4.3x10^3mL

b) In this case, the given height in mm is:

h=4.5dm*\frac{1m}{10dm}* \frac{1000mm}{1m}=450mm

And the radius in mm is:

r=5.50x10^{-5}km*\frac{1000m}{1km}*\frac{1000mm}{1m}=55mm

Thus, the volume in cubic millimeters is:

V=\pi (55mm)^2*450mm\\\\V=4.3x10^6mm^3

c) Finally, since 1000 mL equal 1 L, the required density in g/L turns out:

\rho=\frac{m}{V}=\frac{6.54x10^5g}{4.3x10^3mL}*\frac{1000mL}{1L}\\   \\\rho=1.5x10^5g/L

Best regards.

8 0
3 years ago
What is the importance of cell membrane​
Vilka [71]
It protects the nucleus and
5 0
3 years ago
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