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3 years ago

draw a representation that shows the interactions between the solute and solvent in an aqueous solution of potassium bromide

1 answer:

3 0

The visualization of the solute and solvent in an aqueous solution is such that the solute molecules are surrounded by solvent molecules, which in this case is water.

The potassium bromide will be found in the form of potassium and bromide ions. Both of these ions will be surrounded by water molecules. In the case of potassium, because it is a positively charged ion, the oxygen atoms of the water molecules will be close to the ion and the hydrogen atoms will be on the other side. The opposite will occur with the bromide ions due to their negative charge.

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Sulfuric acid is essential to dozens of important industries from steelmaking to plastics and pharmaceuticals. More sulfuric aci

Pavlova-9 [17]

Given:

K = 0.71 = Kp

The reaction of sulphur with oxygen is

S(s) + O2(g) ---> SO2(g)

initial Pressure 6.90 0

Change -x +x

Equilibrium 6.90-x x

Kp = pSO2 / pO2 = 0.71 = x / (6.90-x)

4.899 - 0.71x = x

4.899 = 1.71x

x = 2.86 atm = pressure of SO2 formed

temperature = 950 C = 950 + 273.15 K = 1223.15 K

Volume = 50 L

Let us calculate moles of SO2 formed using ideal gas equation as

PV = nRT

R = gas constant = 0.0821 L atm / mol K

putting other values

n = PV / RT = 2.86 X 50 / 1223.15 X 0.0821 = 1.42 moles

Moles of Sulphur required = 1.42 moles

Mass of sulphur required or consumed = moles X atomic mass of sulphur

mass of S = 1.42 X 32 = 45.57 grams or 0.04557 Kg of sulphur

6 0

3 years ago

A certain redox reaction takes place in a basic solution . which species must you include to balance the equation for this react

Arte-miy333 [17]

OH

*inserts random text so that this answer can be posted reaching the needed amount of words.*

3 0

3 years ago

Read 2 more answers

For each of the following reactions, identify the missing reactant(s) or products(s) and then balance the resulting equation. No

Pachacha [2.7K]

Answer:

A. The reactants are= Li and O2

The balanced equation is:

4Li + O2 —> 2Li2O

B. The products are: MgCl2 and O2

The balanced equation is:

Mg(ClO3)2 —> MgCl2 + 3O2

C. The products are: Ca(NO3)2 and H2O

The balanced equation is:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 +

2H2O

D. The products are: CO2 and H2O

The balanced equation is:

C5H12 + 8O2 —> 5CO2 + 6H2O

Explanation:

A. ____ —> Li2O

The reactants are Li and O2. Thus the equation is given below:

Li + O2 —> Li2O

Thus the equation is balanced as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Li2O as shown below:

Li + O2 —> 2Li2O

Now, we have 4 atoms of Li on the right side and 1 atom on the left. It can be balance by putting 4 in front of Li as shown below:

4Li + O2 —> 2Li2O

Now the equation is balanced

B. Mg(ClO3)2 —> __

The products are MgCl2 and O2

The equation is given below:

Mg(ClO3)2 —> MgCl2 + O2

The equation can be balance as follow:

There are 6 atoms of O on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

Mg(ClO3)2 —> MgCl2 + 3O2

Now the equation is balanced

C. HNO3 + Ca(OH)2 —>___

The products are: Ca(NO3)2 and H2O

The equation is given below:

HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

The equation is balanced as follow:

There are 2 atoms of NO3 on the right side and 1 atom on the left. It can be balance by putting 2 in front of HNO3 as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

There are a total of 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + 2H2O

Now the equation is balanced

D. C5H12 + O2 —>__

The products are: CO2 and H2O

The equation is given below:

C5H12 + O2 —> CO2 + H2O

The equation can be balance as follow:

There are 5 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 5 in front of CO2 as shown below:

C5H12 + O2 —> 5CO2 + H2O

There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:

C5H12 + O2 —> 5CO2 + 6H2O

Now, there are a total of 16 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 8 in front of O2 as shown below:

C5H12 + 8O2 —> 5CO2 + 6H2O

Now we can see that the equation is balanced.

4 0

3 years ago

Read 2 more answers

Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following

bearhunter [10]

Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the reaction:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the number of moles that we have in the 110.0 g of carbon dioxide, to this, we have to know the atomic mass of each atom:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the molar mass of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$

In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the molar ratio between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the grams of magnesium carbonate:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

I hope it helps!

8 0

3 years ago

the elements carbon hydrogen and oxygen are all part of the same blank on the periodic table........ i i think GROUP diangol row

Mrrafil [7]

The elements Carbon, Hydrogen, and Oxygen are all part of Non-Metals.

Hope This Helps! :)

6 0

3 years ago