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Wittaler [7]
2 years ago
12

How to calculate atomic mass of sodium ?(in process)​

Chemistry
1 answer:
Vinvika [58]2 years ago
7 0

Answer:

Atomic mass of Sodium is 22.9897 u. Note that, each element may contain more isotopes, therefore this resulting atomic mass is calculated from naturally-occuring isotopes and their abundance. The unit of measure for mass is the atomic mass unit (amu). One atomic mass unit is equal to 1.66 x 10 -24 grams.

Explanation:

You might be interested in
A solution contains some or all of the ions Cu2+,Al3+, K+,Ca2+, Ba2+,Pb2+, and NH4+. The following tests were performed, in orde
xeze [42]

Answer:

See below explanation

Explanation:

When having a mixture of metals in solution, you may perform an analytical study (using selective chemical conditions), that may help you to determine whether a metal (cation) is present or not

Using selective analytes (or conditions), leads to consecutive precipitations, until most of the cations are separated in precipitates

With this technique, you may identify metals in different groups, each group will have its analyte (or condition), which will help to have a different precipitate:

- Group I: Ag⁺, Pb⁺², Hg⁺²;  Analyte: HCL ; Precipitate: AgCl (white) , PbCl₂, HgCl₂

- Group II: As⁺³ , Bi⁺³, Cd⁺², Cu⁺² , Sb⁺³, Sn⁺² ; Analyte: H₂S (g) with HCL ; Precipitate: As₂S₃ , Bi₂S₃ , CdS (yellow) , CuS (black), Sb₂S₃, SnS

- Group III: Co⁺², Fe⁺², Fe⁺³, Mn⁺², Ni⁺², Zn⁺², Al⁺³, Cr⁺³; Analyte: NaOH or NH₃ with (NH₄)₂S (ac) ; Precipitate: CoS (black) , FeS, MnS , NiS (black), ZnS (white) , Al(OH)₃ (white), Cr(OH)₃  

- Group IV: Mg⁺², Ca⁺², Sr⁺², Ba⁺²; Analyte: Na₂CO₃ (ac) or (NH₄)₂HPO₄ (ac); Precipitate: respective carbonate or phosphate MgCO₃/MgHPO₄, CaCO₃/CaHPO₄ , SrCO₃/SrHPO₄, BaCO₃/BaHPO₄

- Group V: Li⁺, K⁺, Na⁺, Rb⁺, Cs⁺, NH₄⁺ ; will remain all in final solution

According to the original statement:

A solution contains one or more of the following: Cu⁺², Al⁺³, K⁺, Ca⁺², Ba⁺², Pb⁺², NH₄⁺

1) Addition on HCl 6M produces no change: we can say the sample does not contain Pb⁺² (group I)

2) Addition of H₂S with 0.2 M HCL produced a black solid: we could say sample contains Cu⁺²(group II)

3) Addition of (NH₄)₂HPO₄ in NH₃ produces no reaction: we could say we don´t have Ca⁺² and /or Ba⁺²  (group IV)

4) The final supernatant, when heated produced a purple flame: in the final solution, we have K⁺ (group V), which produces a purple flame (based on its characteristic emission spectrum when subjected to flame)

This analysis will be inconclusive for NH₄⁺ (according to above describe technique)

6 0
3 years ago
A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution
Dmitrij [34]

Answer:

The change in internal energy is - 1.19 kJ

Explanation:

<u>Step 1:</u> Data given

Heat released = 3.5 kJ

Volume calorimeter = 0.200 L

Heat release results in a 7.32 °C

Temperature rise for the next experiment = 2.49 °C

<u>Step 2:</u> Calculate Ccalorimeter

Qcal = ccal * ΔT ⇒ 3.50 kJ = Ccal *7.32 °C

Ccal = 3.50 kJ /7.32 °C = 0.478 kJ/°C

<u>Step 3:</u> Calculate energy released

Qcal = 0.478 kJ/°C *2.49 °C = 1.19 kJ

<u>Step 4:</u> Calculate change in internal energy

ΔU =  Q + W       W = 0  (no expansion)

Qreac = -Qcal = - 1.19 kJ

ΔU = - 1.19 kJ

The change in internal energy is - 1.19 kJ

4 0
3 years ago
What is the volume in (a) liters and (b) cubic yards of a room that is 10. meters wide by 15 meters long and 8.0 ft high?
deff fn [24]

Answer:

V = 364500 L, 476.748 yard³

Explanation:

Given that,

The dimensions of a room are 10 meters wide by 15 meters long and 8.0 ft high.

l = 10 m, b = 15 m, h = 8 ft = 2.43 m

The volume of the room is :

V = lbh

So,

V = 10×15×2.43

V = 364.5 m³

As 1 m³ = 1000 L

364.5 m³ = 364500 L

Also, 1 m³ = 1.30795 yard³

364.5 m³ = 476.748 yard³

Hence, this is the required solution.

7 0
2 years ago
What is enthalpy?
White raven [17]
The awnser is A. Idek I looked it up so yeah that’s the awnser
7 0
3 years ago
Read 2 more answers
Acetic acid (HC2H3O2) is the active ingredient in vinegar. Calculate the mass percent composition of O in acetic acid.
skelet666 [1.2K]

Answer:

53.29% of acetic acid is Oxygen

Explanation:

Step 1: Given data

Acetic acid it's molecular formula is HC2H3O2. This means it consists of 3 elements Carbon, Hydrogen and oxygen.

Step 2: the molar masses

The molecular mass of acetic acid is:

4* H = 4* 1.01 g/mole

2* C = 2*12 g/mole

2*O = 2* 16 g/mole

Total molar mass = 4+ 24+32 = 60.052 g/mole

Step 3: Calculate the mass percent

32 g of the 60.052 g is Oxygen

(32/60.052) *100% = 53.29%

53.29% of acetic acid is Oxygen.

7 0
3 years ago
Read 2 more answers
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