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Mkey [24]
3 years ago
9

Which equation best describes the net changes based upon the observation that solid silver nitrate and solid potassium chloride

are soluble in water and these solutions react to form insoluble silver chloride and soluble potassium nitrate when mixed?AgNO3(s) + KCl(s) = AgCl(s) + KNO3(aq)Ag+(aq) + Cl-(aq) = AgCl(s)AgNO3(s) + KCl(s) = AgCl(s) + KNO3(s)AgNO3(aq) + KCl(aq) = AgCl(s) + KNO3(aq)
Chemistry
1 answer:
stealth61 [152]3 years ago
6 0

Answer:

The correct option is: AgNO₃(aq) + KCl(aq) = AgCl(s) + KNO₃(aq)  

Explanation:

Precipitation reaction is a chemical reaction that involves reaction between <em>two soluble salts to give an insoluble salt.</em> This <u>insoluble salt exists as a solid</u> and settles down.

Therefore, the solid formed in a precipitation reaction is known as the precipitate.  

As the solid silver nitrate (AgNO₃) and solid potassium chloride (KCl) are <u>soluble in water</u>, therefore, their aqueous solutions are represented as AgNO₃(aq) and KCl(aq), respectively.

The precipitation reaction of AgNO₃(aq) and KCl(aq) gives an <u>insoluble salt, silver chloride (AgCl) and a soluble salt, potassium nitrate (KNO₃).</u>

The insoluble salt, <u>AgCl is called the precipitate</u> and is represented as AgCl(s). Whereas, the <u>soluble salt</u>, KNO₃ is represented as KNO₃ (aq).

<u>Therefore, the chemical equation for this precipitation reaction is:</u>

AgNO₃(aq) + KCl(aq) → AgCl(s) + KNO₃(aq)

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Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container
Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

Explanation:

Here Pressure in a container is given as

P=\frac{1}{3} \rho

Here

  • P is the pressure which is to be calculated
  • ρ is the density of the gas which is to be calculated as below

                                         \rho =\frac{mass}{Volume of container}

        Here

                mass is to be calculated for 75 gas phase molecules as

                      m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times  10^{-21} g

              Volume of container is 0.5 lts

     So density is given as

                         \rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3

  • is the mean squared velocity which is given as

                                        =RMS^2

      Here RMS is the Root Mean Square speed given as 605 m/s so

                                      =RMS^2\\=(605)^2\\=366025

Substituting the values in the equation and solving

P=\frac{1}{3} \rho \\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa

So the average pressure in the container due to these 75 gas molecules is P=9.72 \times 10^{-16} Pa

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4 years ago
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3 years ago
Calculate the number of O atoms in 0.364 g of CaSO4 · 2H2O
Nikolay [14]

Answer:

<em>= 7.66 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

Explanation:

For problems like this posting, one needs an understanding of the following topics:

The definition of the mole

<u>1 mole of substance</u> = mass in grams of substance containing 1 Avogadro's Number ( = 6.023 x 10²³ ) of particles of the specified substance. This is generally one formula weight of the substance of interest. From this, the following equivalent relationships should be memorized:

<em>   1 mole = 1 formula weight = 1 mole weight (g)= 6.023 x 10²³ particles</em>

Converting grams to moles:

<em>Given grams => moles = grams/gram formula wt </em>

Converting moles to grams:

<em>Given moles => grams = moles x gram formula wt</em>

_________________________________________________________

<em>Calculate the number of O atoms in 0.364 g of CaSO₄ · 2H₂O.</em>

Given mass CaSO₄ · 2H₂O = 0.364 grams

Formula Wt CaSO₄ · 2H₂O = 172 g/mole

moles CaSO₄ · 2H₂O = mass <em>CaSO4 · 2H2O / formula Wt. CaSO₄ · 2H₂O</em>

<em>= 0.364 g CaSO₄·2H₂O </em><em>/ </em><em>172 g CaSO4·2H2O </em>

<em>= (0.364/172) mole CaSO₄·2H₂O </em>

<em>= 2.12 x 10⁻³ mole CaSO₄·2H₂O    </em>

<em>∴ number of Oxy (O) atoms in 0.364 grams CaSO₄·2H₂O </em>

<em>=  (2.12 x 10⁻³ mole CaSO₄ · 2H₂O)(6.023 x 10²³ molecules CaSO₄· 2H₂O/ mole)</em>

<em>= 1.276876 x 10²¹molecules CaSO₄· 2H₂O  CaSO₄2H₂O </em>

<em>= 1.276876 x 10²¹ molecules CaSO₄· 2H₂O   x   6 oxygen atoms / molecule</em>

<em>= 7.661256 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

<em>= 7.66 x 10²¹  oxygen atoms in 0.364 grams of  CaSO₄·2H₂O</em>

<em />

8 0
3 years ago
-60 points-
Firdavs [7]

Answer:

1.23e+27

Explanation:

6 0
3 years ago
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