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blagie [28]
4 years ago
7

If equal amounts of silver ion are added to silver bromide and silver chloride, why does the bromide mixture precipitate first?

(Ksp of silver bromide = 5.3 mc022-1.jpg 10-13 and Ksp of silver chloride = 1.8 mc022-2.jpg 10-10)
Chemistry
1 answer:
VARVARA [1.3K]4 years ago
5 0
Answer is: precipitation requires fewer Ag⁺<span> ions in AgBr than in AgCl.
Chemical reactions:
Ksp(KBr) = 5,3</span>·10⁻¹³.
Ksp(KCl) = 1,8·10⁻¹⁰.
Ksp is <span>solubility product constant. The higher the Ksp value, substance is more soluble. KBr has lower Ksp, so it is easier to form precipitant of KBr than KCl.</span>
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IF 500 mL of air contains 2.50x10^22 particles (atoms and molecules). how much particles do you inhale in one day if you breath
vaieri [72.5K]

Answer:

I think it is 0.00192

Explanation:

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Number of molecules in 4HCl​
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Answer:

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6 0
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Assume that a daily diet of 2000 calories (i.e. 8.37 x 106 J) is converted completely to body heat.
slava [35]

Answer:

(a) the mass of the water is 3704 g

(b) the mass of the water is 199, 285.7 g

Explanation:

Given;

Quantity of heat, H= 8.37 x 10⁶ J

Part (a) mass of water (as sweat) need to evaporate to cool that person off

Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg

H = m x Lvap.

m = \frac{H}{L._{vap}} =\frac{8.37 * 10^6.J}{2.26*10^6\ \frac{J}{kg}} = 3.704 \ kg

mass in gram ⇒ 3.704 kg x 1000g = 3704 g

Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J

specific heat capacity of water, C, 4200 J/kg.°C

H = mcΔθ

where;

Δθ is the change in temperature = 35 - 25 = 10°C

m =\frac{H}{c* \delta \theta} = \frac{8.37 *10^6}{4200*10} = 199.2857 kg

mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g

5 0
3 years ago
1. Silver nitrate will react with aluminum metal, yielding aluminum nitrate and silver metal. If you start with 0.223 moles of a
Archy [21]

Answer:

Explanation:

1)

Given data:

Number of moles of aluminium = 0.223 mol

Mass of silver produced = ?

Solution:

Chemical equation:

3AgNO₃  +   Al  →  3Ag + Al(NO₃)₃

Now we will compare the moles of Al with silver.

                               Al           :            Ag

                                1            :             3

                                0.223   :         3×0.223= 0.669 mol

Grams of silver:

Mass = number of moles × molar mass

Mass = 0.669 mol × 107.87 g/mol

Mass = 72.2 g

2)

Given data:

Number of moles of mercury(II) oxide produced = 3.12 mol

Mass of mercury = ?

Solution:

Chemical equation:

2Hg + O₂  →   2HgO

Now we will compare the moles of mercury with mercury(II) oxide.

                         HgO         :         Hg

                            2            :          2

                          3.12          :       3.12

Mass of Hg:

Mass = number of moles × molar mass

Mass = 3.12 mol × 200.59 g/mol

Mass = 625.84 g

3)

Given data:

Number of moles of dinitrogen pentoxide = 12.99 mol

Mass of oxygen = ?

Solution:

Chemical equation:

2N₂  + 5O₂   →  2N₂O₅

Now we will compare the moles of N₂O₅ with oxygen.

                 N₂O₅          :           O₂

                     2            :             5

                    12.99      :         5/2×12.99 = 32.48 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 32.48 mol × 32 g/mol

Mass = 1039.36 g

4)

Given data:

Number of moles of benzene = 0.103 mol

Mass of carbon dioxide = ?

Solution:

Chemical equation:

2C₆H₆  + 15O₂   →  12CO₂ + 6H₂O

Now we will compare the moles of N₂O₅ with oxygen.

                  C₆H₆         :           CO₂

                     2            :             12

                    0.103      :         12/2×0.103 = 0.618 mol

Mass of carbon dioxide:

Mass = number of moles × molar mass

Mass = 0.618 mol × 44 g/mol

Mass = 27.192 g

3 0
3 years ago
El agua oxigenada se descompone rápidamente en agua y oxígeno cuando se le agrega una peueña cantidad de dióxido de manganeso, q
katovenus [111]

Answer:

IDK

Explanation:

I dont speak spanish

6 0
4 years ago
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