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Assoli18 [71]
2 years ago
6

Which of the following is a precipitation reaction? a) Zn (s) + 2 AgNO3 (aq) → 2 Ag (s) + Zn(NO 3) 2 (aq) b) 2 LiI (aq) + Hg 2(N

O3)2 (aq) → Hg 2I2 (s) + 2 LiNO3 (aq) c) HCl (aq) + KOH (aq) → KCl (aq) + H 2O (l) d) NaCl (aq) + LiI (aq) → NaI (aq) + LiCl (aq) e) None of the above.
Chemistry
1 answer:
lawyer [7]2 years ago
4 0

Answer: 2LiI(aq)+Hg_2(NO_3)_2(aq)\rightarrow Hg_2I_2(s)+2LiNO_3(aq)

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

Precipitation is a type of displacement reaction in which one the products is formed in the solid state.

2LiI(aq)+Hg_2(NO_3)_2(aq)\rightarrow Hg_2I_2(s)+2LiNO_3(aq)

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Calculate the mass of hydrogen formed when 27 g of aluminum reacts with excess hydrochloric acid according to the balanced equat
Alex

6.069 grams is the  mass of hydrogen formed when 27 g of aluminum reacts with excess hydrochloric acid.

Explanation:

Balanced equation for the reaction:

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

data given:

mass of aluminum = 27 grams

atomic mass of one mole of aluminum = 26.89 grams/mole

formula to calculate number of moles:

number of moles = \frac{mass}{atomic mass of one mole}

number of moles = \frac{27}{26.89}

                              = 1.004 moles of aluminum will react

from the balanced equation:

2 moles of Al reacted to form 3 moles of H2

1.004 moles of Al will produce x moles of H2

\frac{3}{2} = \frac{x}{1.004}

x = 3.012 moles of H2 will be formed.

mass will be calculated as number of moles multiplied by atomic weight

mass of 3.012 moles of hydrogen ?(atomic weight of one mole H2 = 2.015 grams)

= 3.012 x 2.015

= 6.069 grams of H2 will be formed.

6 0
2 years ago
Use Q = mcAT
mr Goodwill [35]

Answer:

1).....for the specific heat capacity(c) of water is 4200kg/J°C..

....guven mass(m)=320g(0.32kg)

...change in temperature(ΔT) =35°C

from the formula

Q=mcΔT

Q=0.32Kg x 4200kg/J°C x 35°C

Q=47,040Joules

5 0
3 years ago
Your task is to create a buffered solution. You are provided with 0.10 M solutions of formic acid and sodium formate. Formic aci
Anton [14]

Answer:

15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.

Explanation:

To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid</em>

<em />

4.25 = 3.75 + log [A⁻] / [HA]

0.5 = log [A⁻] / [HA]

3.162 = [A⁻] / [HA] <em>(1)</em>

<em></em>

As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:

0.10M  * 20x10⁻³L =

2x10⁻³moles = [A⁻] + [HA] <em>(2)</em>

Replacing (2) in (1):

3.162 = 2x10⁻³moles - [HA] / [HA]

3.162 [HA] = 2x10⁻³moles - [HA]

4.162[HA] = 2x10⁻³moles

[HA] = 4.805x10⁻⁴ moles

[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles

That means, to create the buffer you must add:

[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =

<h3>15.2mL of the 0.10M sodium formate solution</h3>

[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =

<h3>4.8mL of the 0.10M formic acid solution</h3>
4 0
3 years ago
Name all 7 elements classified as metalloids. <br> PLEASE HELP ASAP
Strike441 [17]

Answer:

Explanation

Boron (B)

Silicon (Si)

Germanium (Ge)

Arsenic (As)

Antimony (Sb)

Tellurium (Te)

Polonium (Po

6 0
3 years ago
A balloon contains 5 L of oxygen at a pressure of 90 kPa. How much oxygen will the balloon contain if the pressure is lowered to
laila [671]
Answer is: volume will be 6,7 L.
Boyle's Law: the pressure volume law - <span> volume of a given amount of gas held  varies inversely with the applied pressure when the temperature and mass are constant.
p</span>₁V₁ = p₂V₂.
90 kPa · 5 L = 67 kPa · V₂.
V₂ = 90 kPa · 5 L / 67 kPa.
V₂ = 6,7 L, but same amount of oxygen.


7 0
3 years ago
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