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lyudmila [28]
3 years ago
3

A sample of crystalline compound when heated in an open test tube, produced several droplets of water on the upper region of the

tube. The residue dissolved in water forming a yellow brown solution. Why should you conclude that the compound is not a true hydrate
Chemistry
2 answers:
Yanka [14]3 years ago
6 0

The difference in color shows that there has been a chemical change on heating so the water formed could be produced in that chemical change. Alternatively the material could still be a true hydrate and the anhydrous material may have undergone a change on heating so you can't really conclude anything.

Taya2010 [7]3 years ago
4 0

Answer is: If the compound were a true hydrate it would form aqueous solutions of identical color both before and after heating.

For example, copper(II) sulfate pentahydrate (CuSO₄·5H₂O) form aqueous solutions of bright blue both before and after heating.

Hydrate is a substance that contains water or its constituent elements.

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One gram is equal to ____ nanograms?
Katen [24]

Answer:

A

Explanation:

1g=1000mg

1g= 1000 000 micro gram

1g= 1000 000 000 nm = 10^9

5 0
3 years ago
Write the equilibrium constant expression for this reaction: h2so4 (aq) → 2h+ (aq) + so−24 (aq)
ddd [48]

when Kc = [the concentration of products]^(no.of its mol in the balanced equation)     /  [the concentration of the reactants]^(no.of mol in the balanced equation)
∴ The equilibrium constant expression for this reaction is:
Kc= [H+]^2 * [So4-2] / [H2SO4]
8 0
3 years ago
Read 2 more answers
The following reaction can be written as the sum of two reactions, one of which relates to ionization energy and one of which re
Karo-lina-s [1.5K]
A)
First ionization energy of Li:
Li(g) + 520 kJ/mol → Li⁺(g) + e⁻ 

B)
Electron affinity of Fluorine:
F(g) + e⁻ →  F⁻(g) + 328 kJ/mol
8 0
3 years ago
The reaction 2HgO (s)→2Hg (I)+O2 (g) has a percent yield of 50%. You want to produce 100 g of Hg.
nevsk [136]

The mass of HgO needed for the reaction is 216 g

The correct answer to the question is Option C. 216 g

We'll begin by calculating the theoretical yield of Hg.

  • Actual yield of Hg = 100 g
  • Percentage yield = 50%
  • Theoretical yield of Hg =?

Theoretical yield = Actual yield / percentage yield

Theoretical yield = 100 / 50%

Theoretical yield of Hg = 200 g

Finally, we shall determine the mass of HgO needed for the reaction.

2HgO → 2Hg + O₂

Molar mass of HgO = 201 + 16 = 217 g/mol

Mass of HgO from the balanced equation = 2 × 217 = 434 g

Molar mass of Hg = 201 g/mol

Mass of Hg from the balanced equation = 2 × 201 = 402 g

From the balanced equation above,

402 g of Hg were produced from 434 g of HgO.

Therefore

200 g of Hg will be produce by = (200 × 434) / 402 = 216 g of HgO.

Thus, 216 g of HgO is needed for the reaction.

Learn more about stoichiometry:

brainly.com/question/24426334

4 0
2 years ago
Read 2 more answers
A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.
Alexxandr [17]

<u>Answer:</u> The empirical and molecular formula of the compound is C_4H_4O and C_8H_8O_2 respectively

<u>Explanation:</u>

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.

For Carbon = \frac{0.313}{0.078}=4.01\approx 4

For Hydrogen  = \frac{0.316}{0.078}=4.05\approx 4

For Oxygen  = \frac{0.078}{0.078}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is C_4H_4O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

n=\frac{130g/mol}{68g/mol}=1.9\approx 2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2

Hence, the empirical and molecular formula of the compound is C_4H_4O and C_8H_8O_2 respectively

4 0
3 years ago
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