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DiKsa [7]
3 years ago
11

the reaction of zinc with copper surface is exothermic. how can you tell from the students results that the reaction is exotherm

ic?
Chemistry
1 answer:
Airida [17]3 years ago
6 0

Answer:

the tempeture increases

Explanation:

Zn(s) + CuSO4(aq) � ZnSO4(aq) + Cu(s) An important aspect of these experiments is that they are exothermic.

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At what part of most rivers dose the water flow quickly
DedPeter [7]
Where the width is narrow and the bottom steep
5 0
3 years ago
A) If Kb for NX3 is 4.5×10^−6, what is the pOH of a 0.175 M aqueous solution of NX3 ?
sukhopar [10]

3.08 is the pOH of a 0.175 M aqueous solution of NX_3.

0.215% is the per cent ionization of a 0.325 M aqueous solution of NX_3

<h3>What is pH?</h3>

pH is a measure of how acidic/basic water is.

A)

NX_3 + H_2O →NHX_3^+ + OH^-

Kb = 4.5 x10^-6

Kb = {concentration of (NH₄⁺) x concentration of (OH⁻)} ÷ concentration of (NH₃).

concentration of (NH₄⁺) = concentration of (OH⁻) = x.

x² = Kb x concentration of (NH₃)

x² = 4.5 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = concentration of (OH⁻) = √(7.0 × 10⁻⁷)

= 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

=- log ( 8.367 × 10⁻⁴)

= 3.08

B)

Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.

Concentration of (NX₃) = 0.325 M.

Kb = 4.5 x 10⁻⁶.

[NX₃H⁺] = [OH⁻] = x.

[NX₃] = 0.325 M - x.

Kb = [NX₃H⁺] x [OH⁻] ÷  [NX₃].

4.5 x 10⁻⁶ = x² ÷ (0.325 M - x).

x = 0.0007 M.

Per cent of ionization:

α = 0. 0007 M ÷ 0. 325 M x 100%

= 0.215%.

Hence,

3.08 is the pOH of a 0.175 M aqueous solution of NX_3.

0.215% is the per cent ionization of a 0.325 M aqueous solution of NX_3

Learn more about pH here:

brainly.com/question/12353627

#SPJ1

5 0
2 years ago
Combustion of hydrocarbons such as butane (C4H10) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's a
sweet-ann [11.9K]

Answer: 2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

Explanation

Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The balanced combustion reaction for butane is,:

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

7 0
3 years ago
4) Calculate the heat needed to melt 40.0 g of ice at 0°C?
Sav [38]
A lot of heat like a lot no cap it’s gonna he a lot
3 0
3 years ago
A reaction proceeds with 2.72 moles of magnesium chlorate and 3.14 moles of sodium hydroxide. this is the equation of the reacti
den301095 [7]

The theoretical amount of each product obtained are:

  • 1.57 moles of Mg(OH)₂
  • 3.14 moles of NaClO₃

<h3>Balanced equation</h3>

Mg(ClO₃)₂ + 2NaOH —> Mg(OH)₂ + 2NaClO₃

<h3>How to determine the limiting reactant</h3>

From the balanced equation above,

1 mole of Mg(ClO₃)₂ reacted with 2 moles of NaOH

Therefore,

2.72 moles of Mg(ClO₃)₂ will react with = 2.72 × 2 = 5.44 moles of NaOH

From the calculation above, we can see that a higher amount (5.44 moles) of NaOH than what was given (3.14 moles) is needed to react completely with 2.72 moles of Mg(ClO₃)₂

Therefore, NaOH is the limiting reactant

<h3>How to determine the theoretical yield of Mg(OH)₂</h3>

From the balanced equation above,

2 moles of NaOH reacted to produce 1 mole of Mg(OH)₂

Therefore,

3.14 moles of NaOH will react to produce = 3.14 / 2 = 1.57 moles of Mg(OH)₂

Thus, the theoretical yield of Mg(OH)₂ is 1.57 moles

<h3>How to determine the theoretical yield of NaClO₃</h3>

From the balanced equation above,

2 moles of NaOH reacted to produce 2 mole of NaClO₃

Therefore,

3.14 moles of NaOH will also react to produce 3.14 moles of NaClO₃

Thus, the theoretical yield of NaClO₃ is 3.14 moles

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8 0
2 years ago
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