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Katyanochek1 [597]
3 years ago
8

1.How many milliliters of an aqueous solution of 0.157 M aluminum bromide is needed to obtain 13.4 grams of the salt?

Chemistry
1 answer:
trasher [3.6K]3 years ago
4 0

Answer:

The answer to your question is:

1.- 320 ml

2.- 0.63 M

Explanation:

Data

V = ?   AlBr₃ = 0.157 M

mass = 13.4 g

Process

MW ALBr₃ = 27 + (80x3) = 267 g

                                267 g -------------------- 1 mol

                                13.4 g   ------------------  x

                                x = (13.4 x 1) / 267

                               x = 0.05 moles

Volume = 0.05/ 0.157

             = 0.320 l or 320 ml

2.-

NaNO₃ mass = 20.2 g

V = 375 ml

Molarity = ?

MW = 23 + 14 + 48 = 85g of NaNO₃

                          85 g of NaNO₃ -----------------   1 mol

                          20.2 g               -----------------   x

                         x = (20.2 x 1) / 85

                         x = 0.24 mol

Molarity = 0.24 / 0.375

             = 0.63                

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inna [77]

Answer:

Explanation:

The two green substances are not the same.  Most of their properties are different, except some are alike.  Since not all of their properties are matching, they are not the same substance.

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7 0
3 years ago
Reaction intermediates differ from activated complexes in that A. they are stable molecules with normal bonds and are frequently
vova2212 [387]

Answer:

they are molecules with normal bonds rather than partial bonds and can occasionally be isolated.

Explanation:

In chemistry, reaction intermediates are species that are formed from reactants and are subsequently being transformed into products as the reaction progresses. In other words,  reaction intermediates are species that do not appear in a balanced reaction equation but occur somewhere along the reaction mechanism of a non-elementary reaction. They are usually short lived species that possess a high amount of energy. They may or may not be isolated.

They are often molecular species with normal bonds unlike activated complexes that are sometimes hypervalent species.

3 0
3 years ago
C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w
Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

<h3>Hess's Law</h3>

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>ΔH in this case</h3>

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃   ΔH = –183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O     ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

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7 0
2 years ago
How is mass calculated given density and volume? (1 point) Question 8 options: 1) Volume divided by volume 2) Sum of volume and
laila [671]

Answer:

3) Density multiplied by volume

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

So by rearranging this formula mass can be determine.

 d = m/v

m = d×/v

Unit of mass when volume in mL and ddensity is g/mL

m = g/mL×mL

m = g

Example:

density = 5g/mL

volume = 3 mL

mass = ?

m = d×v

m = 5g/mL ×3 mL

m = 15 g

8 0
3 years ago
Read 2 more answers
An object has a mass of 20 kg and a volume of 5 mL. What is the object's density?
e-lub [12.9K]
Good grief, this stuff got caught in a black hole somewhere. It is terribly dense.
1 mL = 1 cc under normal conditions.

d = mass / volume
m = 20 kg
v = 5 mL

d = 20kg / 5 mL
d = 4 kg / mL
d = 4 kg / cc

A <<<<answer
7 0
3 years ago
Read 2 more answers
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