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klemol [59]
3 years ago
14

AgCl (silver chloride)

Chemistry
1 answer:
m_a_m_a [10]3 years ago
8 0

Answer:

a. Silver; chlorine  

b. One atom of Ag; one atom of Cl

c. Two atoms total

Explanation:

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Determine the pressure change when a volume of gas at 760 mmHg is heated from 30.0 °C to 40.0 °C.
bekas [8.4K]

Answer:

<u>253.33 mmHg</u>

Explanation:

According to Charles' Law,

P₁ / T₁ = P₂ / T₂

P₁T₂ = P₂T₁

P₂ = P₁T₂ / T₁

= 760 x 40 / 30

= 760 x 4/3

= 1013.33 mmHg

Change in Pressure

= 1013.33 - 760

= <u>253.33 mmHg</u>

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Earth is approximately 149,600,000 kilometers away from the sun. How would this distance be expressed in scientific notation?
Ronch [10]

Answer:

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Explanation:

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7 0
3 years ago
A solution of sodium hydroxide was titrated against a solution of sulfuric acid. How many moles of sodium hydroxide would react
jeka57 [31]

Answer:

2 mole of Sodium hydroxide reacts with 1 mole of Sulfuric acid

Explanation:

Write down the equation in the beginning with reactants and products:

NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Now try to balance it. Try with Na first:

2NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Na atoms are balanced. There are 6 Oxygen atoms on the right and 5 on the left. Balance by increasing the H₂O moles:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

Check if H atoms are also balanced. They are. That means our final reaction is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

2 Moles of NaOH reacts with 1 mole of H₂SO₄

5 0
3 years ago
What name should be used for the ionic compound Cu(NO3)2
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<span>Copper(II) nitrate. Hope i cleared your doubt</span>
6 0
3 years ago
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The acid-dissociation constant of hydrocyanic acid (hcn) at 25.0 °c is 4.9 ⋅ 10−10. what is the ph of an aqueous solution of 0.0
vodomira [7]
According to the reaction equation:

and by using ICE table:

              CN-  + H2O ↔ HCN  + OH- 

initial  0.08                        0          0

change -X                        +X          +X

Equ    (0.08-X)                    X            X

so from the equilibrium equation, we can get Ka expression

when Ka = [HCN] [OH-]/[CN-]

when Ka = Kw/Kb

               = (1 x 10^-14) / (4.9 x 10^-10)

               = 2 x 10^-5

So, by substitution:

2 x 10^-5 = X^2 / (0.08 - X)

X= 0.0013

∴ [OH] = X = 0.0013 

∴ POH = -㏒[OH]

            = -㏒0.0013

            = 2.886 

∴ PH = 14 - POH

         = 14 - 2.886 = 11.11
5 0
3 years ago
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