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Lelu [443]
3 years ago
5

Magnesium bromide is prepared in a laboratory using the reaction below. Typically, the magnesium bromide is recoveted as a solid

by heating way the remaining water
Mg(s) + 2 HB-(aq) - MgBrac) + H200)
If 0.940 mole of HBr in solution is added to 16.5 grams of Mgo in a beaker and then is heated until dry, what are the expected contents of the beaker?
Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
6 0
Can you give more examples?
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In recent decades rainforests throughout the world have been cut down for various reasons. In some cases the wood has been used;
zloy xaker [14]
C) one is the correct option
7 0
3 years ago
Read 2 more answers
6 Calculate the number of molecules in 8 g of oxygen.
Mariana [72]

Answer:

If you mean the number of atoms in 8g of oxygen, it's 3.011 x 10^23 atoms.

Explanation:

Convert the grams to moles. 8 grams of oxygen is 0.5 moles. Then multiply the number of moles by Avogadro's number: 6.022 x 10^23.

3 0
3 years ago
What is the overall enthalpy of reaction for the equation shown below?
Rudiy27

Answer:

ΔH₁₂ = -867.2 Kj

Explanation:

Find enthalpy for 3H₂ + O₃ => 3H₂O given ...

2H₂ + O₂ => 2H₂O      ΔH₁ = -483.6 Kj

        3O₂ => 2O₃        ΔH₂ = + 284.6 Kj

_____________________________

3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O       (multiply by 3 to cancel O₂)

6H₂ + 3O₂ => 6H₂O        ΔH₁ = 3(-483.6 Kj) = -1450.6Kj

          2O₃ => 3O₂           ΔH₂ = -284.6Kj              (reverse rxn to cancel O₂)

_______________________________

6H₂ + 2O₃ => 6H₂O         ΔH₁₂ = -1735.2 Kj       (Net Reaction - not reduced)

________________________________

divide by 2 => target equation (Net Reaction - reduced)

3H₂ + O₃ => 3H₂O            ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj    

4 0
3 years ago
Tap water at room temperature has a pH of 7.2 and carbonate alkalinity of 200 mg/L (= 40 mmol/L). What is the concentration of b
UNO [17]

Answer:

the concentration of bicarbonate is <em>[HCO₃⁻] = 0,03996 M </em>and carbonate is <em>[CO₃²⁻] = 3,56x10⁻⁵ M.</em>

Explanation:

Carbonate-bicarbonate is:

HCO₃⁻ ⇄ CO₃²⁻ + H⁺ With pka = 10,25

Using Henderson-Hasselbalach formula:

pH = pka + log₁₀\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}

7,2 = 10,25 + log₁₀\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}

8,91x10⁻⁴ = \frac{[CO_{3}^{2-}]}{[HCO_{3}^-]} <em>(1)</em>

Also:

0,040 M = [CO₃²⁻] + [HCO₃⁻] <em>(2)</em>

Replacing (2) in 1:

<em>[HCO₃⁻] = 0,03996 M</em>

Thus:

<em>[CO₃²⁻] = 3,56x10⁻⁵ M</em>

I hope it helps.

4 0
3 years ago
What is the temperature change that 725 g of aluminum will undergo when 2.35 x 104
Rashid [163]

Answer:

36°C

Explanation:

Given parameters:

Mass of aluminum = 725g

Quantity of heat  = 2.35 x 10⁴J

Unknown:

Temperature change  = ?

Solution:

To solve this problem, we simply use the expression below:

  The quantity of energy is given as:

          Q  = m C Δt

Q is the quantity of energy

m is the mass

C is the specific heat capacity of aluminum  = 0.9J/g°C

Δt is the change in temperature

  The unknown is Δt;

             Δt  = \frac{Q}{mc}    = \frac{2.35 x 10^{4} }{725 x 0.9}    = 36°C

5 0
2 years ago
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