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Paul [167]
3 years ago
15

What were the first 16 elements known in 1760

Chemistry
2 answers:
BartSMP [9]3 years ago
4 0
The discovery of the first basic elements in nature was sufficient to begin the era of industrial revolution through combining two or more of these elements in order to speed up the production processes.

The first discovered elements in nature were:
sulphur
chlorine
argon
neon
carbon
oxygen
helium
silicon
calcium
hydrogen
nitrogen 
sodium
fluorine
lithium
aluminum
Volgvan3 years ago
4 0
<span>This is a simple matter of looking up the discovery date of the elements.  And the results are Known since ancient times Copper, Sulfur, Silver, Tin, Antimony, Gold, Mercury, Lead Then somewhere around 1400, Bismuth. 1669 Phosphorus 1694 Carbon 1735 Iron 1735 Platinum 1751 Nickel, Cobalt 1755 Magnesium The list in alphabetical order is: Antimony, Bismuth, Carbon, Cobalt, Copper, Gold, Iron, Lead, Magnesium, Mercury, Nickel, Phosphorus, Platinum, Silver, Sulfur, and Tin</span>
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Br+ is isoelectronic with noble gas?
ArbitrLikvidat [17]

Answer:

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I believe the answer is A. However, I would double check the formula.

4 0
3 years ago
The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p
Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

    Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g  

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}

(ii) Solve the relationship

 Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

    56x + 18(7.5 - x) = 267.3

      56x + 135 - 18x = 267.3

                        38x = 132.3

                            x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

4 0
3 years ago
A balloon contains 7.36 g of oxygen gas (02). How many oxygen molecules
sergey [27]
Answer D. 15 molecules
3 0
2 years ago
An acetate buffer solution is prepared by combining 50. mL of a 0.20 M acetic acid, HC2H3O2 (aq), and 50. mL of 0.20 M sodium ac
Gala2k [10]

Answer:

B. CH3COOH pH > 4.7 (4.8)

Explanation:

  • CH3COOH + NaOH ↔ CH3COONa + H2O
  • CH3COONa + NaOH ↔ CH3COONa

∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol

⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol

⇒ <em>C</em> CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)

⇒ <em>C</em> CH3COOH = 0.0905 M

∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol

⇒ <em>C</em> CH3COONa =  (0.01 mol + 5 E-4 mol) / (0.105 L )

⇒ <em>C</em> CH3COONa = 0.1 M

∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5

⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])

⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]

⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0

⇒ [H3O+] = 1.5835 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = - Log (1.5835 E-5)

⇒ pH = 4.8004 > 4.7

7 0
3 years ago
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