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Radda [10]
4 years ago
12

Which procedure did early chemists use to determine which substances were elements?

Chemistry
1 answer:
Gnoma [55]4 years ago
6 0

C.
They broke down materials by chemical means
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To determine the concentration of chloride ion (Cl-) in a 100 mL sample of ground water, a chemist adds a large enough volume of
Orlov [11]

Answer:

[Cl^-]=232.3\frac{mgCl^-}{L}

Explanation:

Hello,

In this case, we can represent the chemical reaction as:

Cl^-(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NO_3^-(aq)

In such a way, since the mass of the obtained silver chloride is 93.9 mg, we can compute the chloride ions in the ground water by using the following stoichiometric procedure whereas the molar mass of chloride ions and silver chloride are 35.45 g/mol and 143.32 g/mol respectively:

m_{Cl^-}=93.3mgAgCl*\frac{1mmolAgCl}{143.32mgAgCl}*\frac{1mmolCl^-}{1mmolAgCl} *\frac{35.45mgCl^-}{1mmolCl^-} =23.23mgCl^-

Finally, for the given volume of water in liters (0.100L), we compute the required concentration:

[Cl^-]=\frac{23.2mgCl^-}{0.100L}\\

[Cl^-]=232.3\frac{mgCl^-}{L}

Best regards.

5 0
4 years ago
When alkanes react with chlorine in the presence of ultraviolet light, chlorine atoms substitute for one or more alkane hydrogen
Effectus [21]

Answer:

6

Explanation:

Alkanes undergo substitution reaction so the number of replacement reaction hydrogen is 6

5 0
3 years ago
Sulfuric acid (H2SO4) has a molar mass of 98.1 g/mol. How many oxygen atoms are found in 75.0 g of H2SO4? A. 1.15 × 10^23 B. 1.8
Margarita [4]

Answer:

B

Explanation:

first find the no. Of moles of H2SO4

=  75/98.1

= 0.7645

Next step

In on mole of H2SO4 the no.of oxygen atom is 4.

1 H2SO4 : 4 O

0.7645  : x                     x= 0.7645x4= 3.0581

next multiply avogadros constant

3.0581 x 6.022× 10^23 =1.84x10^24

3 0
4 years ago
Question (ii) answer
katrin [286]
The student stirs it because he wants the reaction to take place more quickly.
4 0
3 years ago
How many moles in 2.21 x10E24 atoms of aluminum
lisov135 [29]
<h3>Answer:</h3>

3.67 mol Al

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

2.21 × 10²⁴ atoms Al

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 2.21 \cdot 10^{24} \ atoms \ Al(\frac{1 \ mol \ Al}{6.022 \cdot 10^{23} \ atoms \ Al})
  2. Divide:                              \displaystyle 3.66988 \ mol \ Al

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

3.66988 mol Al ≈ 3.67 mol Al

7 0
3 years ago
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