The children’s playful acts made their parents roar with laughter

Please help me like now please

777dan777 [17]

Answer:

1-1) NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

1-2) 0.5 mole of CO2

2-1) 2C4H10 + 13O2 --> 8CO2 + 10H2O

2-2) 4 mol CO2

Explanation:

Question 1

NaHCO3 + CH3COOH --> NaCH3COO + H2O + CO2

To balance the equation, count the number of atoms on both sides of the equation

(1 Na, 1+3+1H, 1+1+1C, 3+2Oxygen) --> (1 Na, 1+1+1C, 3+2H, 2+1+2Oxygen)

Combining the pluses will give you the following

(1 Na, 5H, 3C, 5Oxygen) --> (1 Na, 3C, 5H, 5Oxygen)

Both sides are the same, therefore the chemical equation is balanced (originally).

From the equation, we can see that 1 mole of NaHCO3 produces 1 mole of CO2.

So that means 0.5 mole of NaHCO3 would produce 0.5 mole of CO2.

Question 2

C4H10 + O2 --> CO2 + H2O

Again, count the number of atoms on both sides of the equation

(4C, 10H, 2O) --> (1C, 2H, 3O) This time left does not equal right side

You now need to find factors that can make both sides equal.

C4H10 + O2 --> 4CO2 + H2O Now the C is balanced, let's recount 

(4C, 10H, 2Oxygen) --> (4C, 8+1Oxygen, 2H) H&O is still not balanced

C4H10 + O2 --> 4CO2 + 5H2O Now the H is balanced, let's recount

(4C, 10H, 2Oxygen) --> (4C, 8+5Oxygen, 10H) O is still not balanced

C4H10 + (13/2)O2 --> 4CO2 + 5H2O Now the O is balanced

(4C, 10H, 13Oxygen) --> (4C, 13Oxygen, 10H)

But because 13/2 is a fraction, we want to eliminate that by multiplying every reactant and product by 2 (the denominator).

2C4H10 + 13O2 --> 8CO2 + 10H2O Now it's completely balanced!

(8C, 20H, 28Oxygen) --> (8C, 28Oxygen, 20H) Yayy! It's balanced.

Now, 2 mol C4H10 produces 8 mol CO2.

So 1 mol C4H10 produces 4 mol CO2.

6 0

2 years ago

Determine the pH of a 2.8 ×10−4 M solution of Ca(OH)2.

shepuryov [24]

Answer:

pH = 10.75

Explanation:

To solve this problem, we must find the molarity of [OH⁻]. With the molarity we can find the pOH = -log[OH⁻]

Using the equation:

pH = 14 - pOH

We can find the pH of the solution.

The molarity of Ca(OH)₂ is 2.8x10⁻⁴M, as there are 2 moles of OH⁻ in 1 mole of Ca(OH)₂, the molarity of [OH⁻] is 2*2.8x10⁻⁴M = 5.6x10⁻⁴M

pOH is

pOH = -log 5.6x10⁻⁴M

pOH = 3.25

pH = 14-pOH

3 0

2 years ago

Determine the mass of CuSO4 • 5H20 that must be used to prepare 250mL of 2.01 M CuSO4(aq).

mario62 [17]

Given parameters:

Volume of CuSO₄ = 250mL

Concentration of CuSO₄ = 2.01M

Unknown:

Mass of CuSO₄.5H₂O = ?

To solve this problem, we must write the chemical relationship between both species.;

CuSO₄.5H₂O → CuSO₄ + 5H₂O

Now that we know the expression, it is possible to solve for the unknown mass.

First find the number of moles of CuSO₄;

Number of moles = Concentration x Volume

Take 250mL to L so as to ensure uniformity of units;

Volume = 250 x 10⁻³L

Input the parameters and solve for number of moles;

Number of moles = 250 x 10⁻³ x 2.01 = 0.5mol

From the equation;

1 mole of CuSO₄ is produced from 1 mole of CuSO₄.5H₂O

So 0.5 moles of CuSO₄ will be produced from 0.5 moles of CuSO₄.5H₂O

Now let us find the molar mass of CuSO₄.5H₂O = 63.6 + 32 + 4(16) + 5(2x1 + 16) = 249.6g/mole

Mass of CuSO₄.5H₂O = number of moles x molar mass

= 0.5 x 249.6

= 124.8g

The mass of CuSO₄.5H₂O is 124.8g

5 0

3 years ago

6. Салыстырып

Annette [7]

Answer:

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Explanation:

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7 0

2 years ago

What is the volume of 0.250 m hydrochloric acid required to react completely with 20.0 ml of 0.250 m ca(oh)2?

Soloha48 [4]

The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol / 0.250 mol/L = 40.0 mL
40.0 mL of 0.250 M HCl is required

5 0

3 years ago