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2 years ago

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Q#5 Balance and list the coefficients from reactants to products.

2 answers:

skad [1K]2 years ago

8 0

A. $2Fe_2O_3(s) +3C(s)$ → $4Fe(s) + 3CO_2(g)$

B. $2Al(s) + 3FeO(s)$ → $Fe(s) + 3Al_2O_3(s)$

C. $2Al(s) +3H_2SO_4(aq)$ → $Al_2(SO_4)_3(aq) + 3H_2(g)$

What is a balanced chemical equation?

An equation that has an equal number of atoms of each element on both sides of the equation is called a balanced chemical equation.

A. $2Fe_2O_3(s) +3C(s)$ → $4Fe(s) + 3CO_2(g)$

B. $2Al(s) + 3FeO(s)$ → $Fe(s) + 3Al_2O_3(s)$

C. $2Al(s) +3H_2SO_4(aq)$ → $Al_2(SO_4)_3(aq) + 3H_2(g)$

Learn more about the balanced chemical equation here:

brainly.com/question/15052184

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nika2105 [10]2 years ago

4 0

A. $2\text{Fe}_{2}\text{O}_{3}+3\text{C} \longrightarrow 4\text{Fe} +3\text{CO}_{2}$

B. $2\text{Al}+3\text{FeO} \longrightarrow \text{Al}_{2}\text{O}_{3}+3\text{Fe}$

C. $2\text{Al}+3\text{H}_{2}\text{SO}_{4} \longrightarrow \text{Al}_{2}(\text{SO}_{4})_{3}+3\text{H}_{2}$

You might be interested in

What Type of chemical reaction is:<br> BaCl2 + 2 AgNO3 --> 2 AgCl + Ba(NO3)2

romanna [79]

Answer:

This is a precipitation reaction: AgCl is the formed precipitate.

8 0

3 years ago

How many molecules are in 68.0 g of H2S

Kryger [21]

The correct answer is 12.044 × 10²³ molecules.

The molecular mass of H₂S is 34 gram per mole.

Number of moles is determined by using the formula,

Number of moles = mass/molecular mass

Given mass is 68 grams, so no of moles will be,

68/34 = 2 moles

1 mole comprises 6.022 × 10²³ molecules, therefore, 2 moles will comprise = 6.022 × 10²³ × 2

= 12.044 × 10²³ molecules.

3 0

3 years ago

How many molecules of carbon dioxide are dissolved in 0.550 L of water at 25 °C if the pressure of CO2 above the water is 0.250

Grace [21]

<u>Answer:</u> The number of molecules of carbon dioxide gas are $2.815\times 10^{21}$

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{CO_2}=K_H\times p_{CO_2}$

where,

$K_H$ = Henry's constant = $0.034mol/L.atm$

$C_{CO_2}$ = molar solubility of carbon dioxide gas

$p_{CO_2}$ = pressure of carbon dioxide gas = 0.250 atm

Putting values in above equation, we get:

$C_{CO_2}=0.034mol/L.atm\times 0.250atm\\\\C_{CO_2}=8.5\times 10^{-3}M$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of carbon dioxide = $8.5\times 10^{-5}M$

Volume of solution = 0.550 L

Putting values in above equation, we get:

$8.5\times 10^{-3}M=\frac{\text{Moles of }CO_2}{0.550L}\\\\\text{Moles of }CO_2=(8.5\times 10^{-3}mol/L\times 0.550L)=4.675\times 10^{-3}mol$

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules

So, $4.675\times 10^{-3}$ moles of carbon dioxide will contain = $(6.022\times 10^{23}\times 4.675\times 10^{-3})=2.815\times 10^{21}$ number of molecules

Hence, the number of molecules of carbon dioxide gas are $2.815\times 10^{21}$

3 0

3 years ago

Read 2 more answers

Consider a solution that is 0.05 M HCl. Your goal is to neutralize 1 L of this solution (i.e. bring the pH to 7). You also have

Ilia_Sergeevich [38]

Answer:

The volume of NaOH required is - 0.01 L

Explanation:

At equivalence point ,

Moles of $HCl$ = Moles of NaOH

Considering :-

$Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}$

Given that:

$Molarity_{NaOH}=5\ M$

$Volume_{NaOH}=?\ L$

$Volume_{HCl}=1\ L$

$Molarity_{HCl}=0.05\ M$

So,

$Molarity_{HCl}\times Volume_{HCl}=Molarity_{NaOH}\times Volume_{NaOH}$

$0.05\times 1=5\times Volume_{NaOH}$

$Volume_{NaOH}=\frac{0.05\times 1}{5}=0.01\ L$

<u>The volume of NaOH required is - 0.01 L</u>

3 0

3 years ago

ch sulfide (ZnS) occurs in the zinc blende crystal structure, (a) If 254 g of ZnS contains 170 g of Zn, what is the mass ratio o

Aleonysh [2.5K]

Answer:

The mass ratio of zinc to sulfide is 85:42.

2.5559 kg of Zn are in 3.82 kg of ZnS.

Explanation:

a) Mass of zinc sulfide = 254 g

Mass of zinc in a zinc sulfide sample = 170 g

Mass of sulfide in zinc sulfide sample = x

254 g = 170 g+ x

x = 84 g

The mass ratio of zinc to sulfide:

$\frac{170 g}{84 g}=\frac{85}{42}$

b) Mass of zincsulfide sample = 3.83 kg

The mass ratio of zinc to sulfide is 85:42.

Let the mass of zinc and sulfide be 85x and 42x respectively:

85 x+ 42 x=3.82 kg

x =0.03007 kg

Mass of an zinc= 85x=85 × 0.03007 kg= 2.5559 kg

8 0

3 years ago