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3 years ago

Give an example of each of the following chemical changes.

Chemistry

1 answer:

Andrews [41]3 years ago

7 0

Answer:

Give an example of each of the following chemical changes.

(a) A photochemical reaction involving

(i) silver salt

(ii) water

(b) A reaction involving

(i) blue solution

(ii) formation of dirty green precipitate

(d) A reaction where colour change is noticed.

Explanation:

a) A photochemical reaction involving silver salt is used in black and white photography.

AgCl breaks down and converts into Ag during this photochemical reaction.

(ii) Photochemicalreaction involving water takes place in plants during the photosynthesis process.

Plants prepare food(carbohydrate) by using sunlight water and CO2 gas.

(b) A reaction involving

(i) blue solution:

For example reaction of copper sulfate solution with an iron nail.

When an iron nail is placed in CuSO4 blue color solution, then it changes to green color and reddish-brown solid deposits at the bottom of the container.

(ii) Reaction of ferrous sulfate with NaOH forms a dirty green precipitate of ferrous hydroxide.

For example when HCl gas reacts with ammonia gas, then a white solid of ammonium chloride will be formed.

(d) A reaction where the color change is noticed.:

When an iron nail is placed in CuSO4 blue color solution, then it changes to green color and reddish-brown solid deposits at the bottom of the container.

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FrozenT [24]

Thank you :))) for the points have a great day

8 0

3 years ago

Read 2 more answers

Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. The Swedish chemist Alfr

Ganezh [65]

Answer:

a. $4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

b. 146.0 g

Explanation:

Question 1 (a). Just as the problem states, liquid nitroglycerin decomposes into nitrogen gas $N_2$ , oxygen gas $O_2$ , water vapor $H_2O$ and carbon dioxide $CO_2$ . Let's write the decomposition of nitroglycerin into these 4 components:

$C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + CO_2 (g)$

Now we need to balance the equation. Firstly, notice we have 3 carbon atoms on the left and 1 on the right, so let's multiply carbon dioxide by 3:

$C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)$

Now, we have 3 nitrogen atoms on the left and 2 on the right, so let's multiply nitrogen on the right by $\frac{3}{2}$ :

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)$

We have 5 hydrogen atoms on the left, 2 on the right, so let's multiply the right-hand side by $\frac{5}{2}$ :

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)$

Finally, count the oxygen atoms. We have a total of 9 on the left. On the right we have (excluding oxygen molecule):

$\frac{5}{2} + 6 = 8.5$

This leaves $9 - 8.5 = 0.5 = \frac{1}{2}$ of oxygen. Since oxygen is diatomic, we need to take one fourth of it to get one half in total:

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + \frac{1}{4} O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)$

To make it look neater without fractional coefficients, multiply both sides by 4:

$4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

Question 2 (b). Now we can make use of the balanced chemical equation and apply it for the context of this separate problem. We're given the following variables:

$V_{CO_2} = 41.0 L$

$T = -14.0^oC + 273.15 K = 259.15 K$

$p = 1 atm$

Firstly, we may find moles of carbon dioxide produced using the ideal gas law $pV = nRT$ .

Rearranging for moles, that is, dividing both sides by RT (here R is the ideal gas law constant):

$n_{CO_2} = \frac{pV_{CO_2}}{RT} = \frac{1 atm\cdot 41.0 L}{0.08206 \frac{L atm}{mol K}\cdot 259.15 K} = 1.928 mol$

According to the stoichiometry of the balanced chemical equation:

$4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

4 moles of nitroglycerin (ng) produce 12 moles of carbon dioxide. From here we can find moles o nitroglycerin knowing that:

$\frac{n_{ng}}{4} = \frac{n_{CO_2}}{12} \therefore n_{ng} = \frac{4}{12}n_{CO_2} = \frac{1}{3}\cdot 1.928 mol = 0.6427 mol$

Multiplying the number of moles of nitroglycerin by its molar mass will yield the mass of nitroglycerin decomposed:

$m_{ng} = n_{ng}\cdot M_{ng} = 0.6427 mol\cdot 227.09 g/mol = 146.0 g$

3 0

3 years ago

Encountering problems during separation/purification is a very common issue in organic synthesis, but these steps are crucial to

Rus_ich [418]

True is the correct answer! your welcome! :)

8 0

3 years ago

The natural abundances of elements in the human body, expressed as percent by mass, are: oxygen (O),65 percent; carbon (C), 18 p

almond37 [142]

Answer:

The mass in grams of each element in the body of 62 Kg are

Oxygen 40300 g

Carbon 11160 g

Hydrogen 6200 g

Nitrogen 1860 g

Calcium 992 g

Phosphorus 744 g

Other elements 744 g

It is possible to check it adding all the components, the total is 62000 g.

Explanation:

Firstly, we have a total mass of 62 Kg and the composition as a percentage of each component.

Percentage ( %)

Oxygen 65.0

Carbon 18.0

Hydrogen 10.0

Nitrogen 3.0

Calcium 1.6

Phosphorus 1.2

Other elements 1.2

Taking the definition of a percentage as part of hundred total, each percentage can be expressed as a decimal number. We should divide the percentage with 100 of total. For example: Oxygen 65%, it means 65/100 = 0.65. So, Oxygen is 0.65 in decimal numbers. We do the same for each component.

Percentage ( %) Decimal number

Oxygen 65.0 65/100 = 0.65

Carbon 18.0 18/100 = 0.18

Hydrogen 10.0 10/100 = 0.10

Nitrogen 3.0 3/100 = 0.03

Calcium 1.6 1.6/100 = 0.016

Phosphorus 1.2 1.2/100 = 0.012

Other elements 1.2 1.2/100 = 0.012

Finally. We can multiply decimal number of each component taking the total. 1 Kg is equivalent to 1000 g.

So, 62 Kg = 62 * 1000 g = 62000 g

After that we get the mass in grams of each element, by multiplying 62000 g and the decimal number of each component as follows:

Percentage ( %) Decimal number Mass in grams (g)

Oxygen 65.0 65/100 = 0.65 0.65*62000=40300

Carbon 18.0 18/100 = 0.18 0.18*62000= 11160

Hydrogen 10.0 10/100 = 0.10 0.10*62000= 6200

Nitrogen 3.0 3/100 = 0.03 0.03*62000= 1860

Calcium 1.6 1.6/100 = 0.016 0.016*62000= 992

Phosphorus 1.2 1.2/100 = 0.012 0.012*62000=744

Other elements 1.2 1.2/100 = 0.012 0.012*62000=744

5 0

3 years ago

The specific heat of gold is 0.031 calories/gram°C. If 10.0 grams of gold were heated and the temperature of the sample

IgorLugansk [536]

Answer:

6.2 calories

Explanation:

Data Given:

change in temperature = 20 °C

specific heat of gold = 0.031 calories/gram °C

mass of gold = 10.0 grams

Amount of Heat = ?

Solution:

Formula used

Q = Cs.m.ΔT

Where:

Q = amount of heat

Cs = specific heat of gold = 0.031 calories/gram °C

m = mass

ΔT = Change in temperature

Put values in above equation

Q = 0.031 calories/gram °C x 10.0 g x 20 °C

Q = 6.2 calories

So option A is correct = 6.2 calories

6 0

3 years ago