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Katena32 [7]
3 years ago
15

Scoring: Your score will be based on the number of correct matches minus the number of incorrect matches. There is no penalty fo

r missing matches. Use the References to access important values if needed for this question. Indicate whether each of the following compounds will gave an acidic, basic or neutral solution when dissolved in water.
1. ammonium bromide
2. potassium cyanide
3. sodium chloride
4. potassium iodide
Chemistry
1 answer:
IRINA_888 [86]3 years ago
4 0

Answer:

Explanation:

1 ) Ammonium bromide is a salt of ammonium hydroxide and hydrogen bromide . The former is a weak base and the later is a strong acid so the salt will make acidic solution in water. It happens due to salt hydrolysis.

2 ) Potassium cyanide is  salt of potassium hydroxide and hydrogen cyanide . The former is strong base and the later is weak acid so its salt will be basic in nature .

3 ) sodium chloride is a salt of sodium hydroxide and hydrogen chloride . The former is strong base and later is strong acid. So the salt is neutral .

4 ) Potassium iodide is a salt of potassium hydroxide and hydrogen iodide . The former is a strong base and the later is a strong acid . So the salt is neutral or a bit basic.

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The smell of dirty gym socks is caused by the compound caproic acid(contains H, C, O). Combustion of 0.844 g of caproic acid pro
adoni [48]

<u>Answer:</u> The molecular formula for the given organic compound is C_2H_{32}O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=0.784g

Mass of H_2O=1.92g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 0.784 g of carbon dioxide, \frac{12}{44}\times 0.784=0.214g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.92 g of water, \frac{2}{18}\times 1.92=0.213g of hydrogen will be contained.

Mass of oxygen in the compound = (0.844) - (0.214 + 0.213) = 0.417 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{0.214g}{12g/mole}=0.0134moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.213g}{1g/mole}=0.213moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.417g}{16g/mole}=0.0261moles

  • <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 0.0134 moles.

For Carbon = \frac{0.0134}{0.0134}=1

For Hydrogen = \frac{0.213}{0.0134}=15.89\approx 16

For Oxygen = \frac{0.0261}{0.0134}=1.95\approx 2

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

The ratio of C : H : O = 1 : 16 : 2

The empirical formula for the given compound is CH_{16}O_2

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 = 116.2 g/mol

Mass of empirical formula = 60 g/mol

Putting values in above equation, we get:

n=\frac{116.2g/mol}{60g/mol}=1.94\approx 2

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

C_{(1\times 2)}H_{(16\times 2)}O_{(2\times 2)}=C_2H_{32}O_4

Hence, the molecular formula for the given organic compound is C_2H_{32}O_4

5 0
3 years ago
When a forest burns and all that's left is some ash,
dalvyx [7]

Answer:

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + H₂O

Explanation:

When a forest burns and all that's left is some ash,  most of the mass of the trees go to the atmosphere, in the form of carbon dioxide.

In a total combustion process of organic matter,<em> the two molecules produced are carbon dioxide and water. </em>(CO₂ and H₂O)

The equation for the combustion of glucose is:

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + H₂O

8 0
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kykrilka [37]
Density = mass/volume
D = 500 g /200 cm^3
D = 2.5 g/cm^3

hope this helps!
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How does it’s atomic radius compare to chlorine’s?
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I do not have a clue.
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Question: What is shown on the photo above?

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<em>Hope this helps!.</em>

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3 years ago
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