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shepuryov [24]
3 years ago
13

The following are formed as white precipitates in solution except

Chemistry
2 answers:
Firdavs [7]3 years ago
7 0

Answer:

i think it is iron

Explanation:

(III) hydroxide

aniked [119]3 years ago
6 0

Answer:

A. Iron (III) hydroxide

Explanation:

Orange-brown or rust

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Pls explain ionization energy​
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Answer:the energy required by an atom to lose or gain electron (e-) to be able to form ions.either cation or anion.

Explanation:

to lose or gain e- an atom has to donate or receive e- to to this it requires the more easily it can lose an electron the more electro positive it is and the more easily it gains electron the more electro negative it is.

4 0
3 years ago
Calculate the percent error in calculating the pH of a 0.040 M HCl solution that is also 0.010 M in HNO3 using ionic strength an
ankoles [38]

Answer:

O2+ e-→O2-εo’= -0.040 V+ 0.046 V= -0.925 Vb. Q = 1/0.02 = 50,the number of electrons transferred νe= 1, ε’=εo’-(0.0591V/νe)*logQ = -0.971V –0.0591V*log50 = -1.071 V

Explanation:

7 0
3 years ago
Balance the following equations:
neonofarm [45]
Hope this helps! If you dont understand balancing equations in general, say so in the comments, I’m happy to help

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B5%7D%20%2B%20%20%5Cfrac%7B2%7D%7B5%7D%20" id="TexFormula1" title=" \frac{1
Travka [436]
3/4 cause its addition
8 0
3 years ago
A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an
erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0
3 years ago
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