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gizmo_the_mogwai [7]
3 years ago
14

Question 9

Chemistry
1 answer:
Shkiper50 [21]3 years ago
7 0

Answer: There are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.

Explanation:

Given: M_{1} = 1.0 M,    V_{1} = ?

M_{2} = 0.1 M,    V_{2} = 200 mL

Formula used is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute values into the above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\1.0 M \times V_{1} = 0.1 M \times 200 mL\\V_{1} = \frac{0.1 M \times 200 mL}{1.0 M}\\= 20 mL

Thus, we can conclude that there are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.

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200.00 grams of an organic compound is known to contain 83.884 grams of carbon, 10.486
ololo11 [35]

The empirical formula of a given compound is C6H9ON5.

<u>Explanation</u>:

Step 1: Obtain the mass of each element present in grams

                  Element % = mass in g = m

Carbon = 83.884 grams, Hydrogen = 10.486 grams, Oxygen = 18.640 grams, Nitrogen = 86.99 grams.

Step 2: Determine the number of moles of each type of atom present

                m/atomic mass = Molar amount (M)

Molar amount of carbon = (83.884 1 mol ) / 12 g = 6.99

Molar amount of hydrogen = (10.486  1 mol) / 1 g = 10.49

Molar amount of oxygen = (18.64  1 mol) / 16 g = 1.17

Molar amount of nitrogen = (86.99  1 mol) / 14 g = 6.21

Step 3: Divide the number of moles of each element by the smallest number of moles

            M / least M value = Atomic Ratio (R)

Atomic radius of carbon = 6.99 / 1.17 = 5.9 = 6

Atomic radius of hydrogen = 10.49 / 1.17 = 8.9 = 9

Atomic radius of oxygen = 1.17 / 1.17 = 1

Atomic radius of nitrogen = 6.21 / 1.17 = 5

Step 4: Convert numbers to whole numbers. This set of whole numbers are the subscripts in the empirical formula.

            R * whole number = Empirical Formula

The empirical formula of a given compound is C6H9ON5.

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4 years ago
In the decomposition reaction, 1 mole of water (mw = 18.015 g/mol) was produced for every mole of cuo (mw = 79.545 g/mol) produc
natita [175]

Reactives -> Products

CuO and water are products.

I found this reaction which has CuO and water as products: decomposition of Cu(OH)2.

Cu(OH)2 -> CuO + H2O

Stoichiometry calculus involve the mole proportions you can see in the reaction: When 1 mole of Cu(OH)2 reacts, 1 mole of CuO and 1 mole of H2O are formed.

Considering the molar masses:

Cu(OH)2 = 83.56 g/mol

CuO = 79.545 g/mol

H2O = 18.015 g/mol

Then: When 83.56 g of Cu(OH)2 react, 79.545 g of CuO and 18.015 g H2O are formed.

You should use that numbers in the rule of three:

79.545 g CuO __________18.015 g water

3.327 g CuO__________ x =3.327*18.015 /79.545 g water 

x= 0.7535 g water




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