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uysha [10]
3 years ago
15

D. potassium + iodine →

Chemistry
1 answer:
aliina [53]3 years ago
5 0

Answer:

The missing reagents are.

Potassium + Iodine =<u> Potassium iodide</u>

<u>Calcium</u> + oxygen = Calcium oxide

Beryllium +<u> Bromine</u> = Beryllium bromide

<u>Copper + Oxygen</u> = Copper oxide

Explanation:

The balanced equation can be written as:

1.Potassium + Iodine =<u> Potassium iodide</u>

2K(s)+I_{2}(s)\rightarrow 2KI(s)

Here K = potassium

I2 = Iodine

KI = potasssium iodide.

2.<u>Calcium</u> + oxygen = Calcium oxide

2Ca(s)+O_{2}(g)\rightarrow 2CaO(s)

Ca = calcium

O2 = oxygen

CaO = Calcium Oxide

3.Beryllium +<u> Bromine</u> = Beryllium bromide

Be(s)+Br_{2}(g)\rightarrow BeBr_{2}(s)

Here,

Be = beryllium

Br2 = bromine

BeBr2 = Beryllium Bromide

4. Copper + Oxygen = Copper oxide

2Cu(s)+O_{2}(g)\rightarrow 2CuO(s)

Cu = Copper

O2 = oxygen

CuO = Copper Oxide

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nonmetal

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For hydrocyanic acid, HCN, Ka = 4.9 × 10-10. Calculate the pH of 0.20 M NaCN. What is the concentration of HCN in the solution?
gogolik [260]
By using the ICE table :

initial    0.2 M            0           0
change -X                 + X           +X
Equ     (0.2 -X)            X               X

when Ka = (X) (X) / (0.2-X)
so by substitution:

4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X 
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and PH = -㏒[H+]
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3 0
3 years ago
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What is the empirical formula for a substance that is composed of 40.66% carbon, 8.53% Hydrogen,23.72% Nitrogen, and 27.09% Oxyg
prohojiy [21]

Answer:

THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO

Explanation:

The steps involved in calculating the empirical formula of this substance in shown in the table below:

Element                            Carbon           Hydrogen          Nitrogen         Oxygen

1. % Composition              40.66              8.53                 23.72                27.09

2. Mole ratio =

%mass/ atomic mass       40.66/12         8.53/1          23.72/14            27.09/16

                                       =  3.3883            8.53              1,6943             1.6931

3. Divide by smallest

value (0.6931)          3.3883/1.6931    8.53/1.6931    1.6943/1.6931   1.6931/1.6931

                               =      2.001                  5.038           1.0007                      1

4. Whole number ratio        2                       5                   1                               1

The empirical formula = C2H5NO

7 0
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Explanation:estro man

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