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3 years ago

Equation balancing

Chemistry

1 answer:

meriva3 years ago

4 0

Answer:

For a: The balanced equation is $2S(s)+3O_2(g)\rightarrow 2SO_3(g)$

For c: The balanced equation is $2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

Explanation:

A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.

For (a):

The given unbalanced equation follows:

$S(s)+O_2(g)\rightarrow SO_3(g)$

To balance the equation, we must balance the atoms by adding 2 infront of both $S(s)$ and $SO_3$ and 3 in front of $O_2$

For the balanced chemical equation:

$2S(s)+3O_2(g)\rightarrow 2SO_3(g)$

For (b):

The given balanced equation follows:

$2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)$

The given equation is already balanced.

For (c):

The given unbalanced equation follows:

$2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+H_2O(l)$

To balance the equation, we must balance the atoms by adding 2 infront of $H_2O(l)$

For the balanced chemical equation:

$2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

For (d):

The given balanced equation follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

The given equation is already balanced.

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6 0

2 years ago

A 1.00 g sample of a hydrogen peroxide (H2O2) solution is placed in an Erlenmeyer flask and diluted with 20 mL of 1 M aqueous su

Minchanka [31]

Following reaction is involved in present system:

2KMnO4 + 5H2O2 + 3H2SO4 → 2MnSO4 + K2SO4 + 5O2 + 8H2O

From the above balance reaction, it can be seen that 2 moles of KMnO4 is consumed for every 5 moles of H2O2.

Now, percent by mass of hydrogen peroxide in the original solution can be estimated as follows:
percent by mass = $\frac{\text{mass of H2O2(g)}}{\text(volume of H2SO4(ml))}X 100$
∴percent by mass = $\frac{\text{1}}{\text(25)}X 100$
= 4 %

5 0

3 years ago

A solution of phosphoric acid was made by dissolving 10.8 g of H3PO4 in 133.00 mL of water. The resulting volume was 137 mL. Cal

Nesterboy [21]

Answer:

Density is: 1.05 g/ml

Mole fraction solute: 0.015

Mole fraction solvent: 0.095

Molarity: 0.80 M

Molality: 0.82 m

Explanation:

A typical excersise of solution.

It is more confortable to make a table for this.

| masss | volume | mol

solute | | |

solvent | | |

solution | | |

Let's complete, what we have.

| masss | volume | mol

solute | 10.8g | |

solvent | | 133 mL |

solution | | 137 mL |

We can first, know how many moles are 10.8 g

Molar Mass H3PO4 = 97.99 g/mol

Mass / Molar mass = mol

10.8 g / 97.99 g/m = 0.110 mol

Density of water is 1 g/ml (it is a very knowly value)

From this data, we can know water mass, solvent.

Density = mass / volume

1 g/ml = mass / 133 mL

Mass = 133 g

We can also have the moles, by the molar mass of water 18 g/m

133 g / 18 g/m = 7.39 mol

| masss | volume | mol

solute | 10.8g | | 0.110 mol

solvent | 133g | 133 mL | 7.39 mol

solution | 143.8g | 137 mL | 7.50 mol

Mass of solution will be solute mass + solvent mass

Moles of solution will be solute moles + solvent moles

Now we can calculate everything.

Molarity means mol of solute in 1 L of solution. (mol/L)

We have to convert 137 mL in L (/1000)

0.137L so → 0.110 m / 0.137L = 0.80 M

Molality means mol of solute in 1kg of solvent.

We have to convert 133g in kg (/1000)

0.133 kg so → 0.110 m/0.133 kg = 0.82 m

Density is mass / volume

Solution density will be solution mass / solution volume

143.8 g/137 mL = 1.05 g/m

Molar fraction is : solute moles / total moles or solvent moles/total moles.

You can also (x 100%) to have a percent of them.

Remember sum of molar fraction = 1

Molar fraction of solute = 0.110 mol / 7.50mol = 0.015

Molar fraction of solvent = 7.39 mol / 7.50 mol = 0.985

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3 years ago

What group of elements are the most reactive nonmetals?

alexdok [17]

Halogens I believe because it included fluorine which is the most reactive nonmetal

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4 years ago

The element silver (Ag) has two naturally occurring isotopes: 109Ag and 107Ag with a mass of 106.905 amu. Silver consists of 51.

algol [13]

Answer:

108.76 amu is the mass of 109Ag

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 109Ag:

% = 51.82 %

Mass = x amu

For second isotope, 107Ag:

% = 100 - x = 100 - 51.82 % = 48.18 %

Mass = 106.905 amu

Given, Average Mass = 107.868 amu

Thus,

$107.868=\frac{51.82}{100}\times {x}+\frac{48.18}{100}\times {106.905}$

Solving for x, we get that:

x = 108.76 amu

108.76 amu is the mass of 109Ag

6 0

3 years ago