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Vlada [557]
2 years ago
15

What properties of a substance determine how that substance will react when combined with other substances

Chemistry
1 answer:
Semmy [17]2 years ago
6 0

Answer:

Chemical Properties

Explanation:

Chemical properties are properties that are observed during chemical reactions. Some examples of chemical properties are reactivity, flammability and chemical stability.

You might be interested in
The raw water supply for a community contains 18 mg/L total particulate matter. It is to be treated by addition of 60 mg alum (A
s344n2d4d5 [400]

Solution :

Given :

The steady state flow = 8000 $ m^3 /d $

                                    $= 80 \times 10^5 \ I/d $

The concentration of the particulate matter = 18 mg/L

Therefore, the total quantity of a particulate matter in fluid $= 80 \times 10^5 \ I/d \times 18 \ mg/L $

$= 144 \times 10^6 \ mg/g$

$= 144 \ kg/d $

If 60 mg of alum $ [Al_2(SO_4)_3.14 H_2O] $ required for one litre of the water treatment.

So Alum required for  $ 80 \times 10^5 \ I/d $

$= 80 \times 15^5 \ I/d  \times 60 \ mg \ alum /L$

$= 480 \times 10^6 \ mg/d $

or 480 kg/d

Therefore the alum required is 480 kg/d

1 mg of the alum gives 0.234 mg alum precipitation, so 60 mg of alum will give $ = 60 \times 0.234 \text{ of alum ppt. per litre} $

      $= 14.04 $ mg of alum ppt. per litre

480 kg of alum will give = 480 x 0.234 kg/d

                                        = 112.32 kg/d ppt of alum

Daily total solid load is  $= 144 \ kg/d + 112.32 \ kg/d$

                                       = 256.32 kg/d

So, the total concentration of the suspended solid after alum addition $= 18 \ mg/L + 60 \times 0.234 $

= 32.04 mg/L

Therefore total alum requirement = 480 kg/d

b). Initial pH = 7.4

 The dissociation reaction of aluminium hydroxide as follows :

$Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-} $

After addition, the aluminium hydroxide pH of water will increase due to increase in $ OH^- $ ions.

Therefore, the pH of water will be acceptable range after the addition of aluminium hydroxide.

c). The reaction of $CO_2$ and water as follows :

$CO_2 (g) + H_2O (l) \rightarrow H_2CO_3$

For the atmospheric pressure :

$p_{CO_2} = 3.5 \times 10^{-4} \ atm $

And the pH is reduced into the range of 5.9 to 6.4

6 0
2 years ago
The modern-day quantum model of the atom is better than John Dalton’s model because it
Licemer1 [7]
The modern day model of an atom has a lot of questions answered about it answered and is very accurate while John's version was a very early model before technology was advanced enough to get more information on it so it was very basic and not as accurate as today's model. Hope this helps!
7 0
3 years ago
Read 2 more answers
0. A liquid solution of LiCl in water at 25°C contains 1 mol of LiCl and 7 mol of water. If 1 mol of LiCl⋅3H2O(s) is dissolved i
kupik [55]

Answer:

19.488 kJ

Explanation:

The overall reaction mechanism shows the reaction between LiCl and H₂O

LiCl.3H_2O ------> Li +\frac{1}{2} Cl_2+3H_2+\frac{3}{2} O_2     -------- (1)

2Li+Cl_2+10H_2O ----->LiCl.10H_2O              -------- (2)

3H_2+\frac{3}{2}O_2 -----> 3H_2O                                     --------- (3)

LiCl.7H_2O ------>  Li + \frac{1}{2} Cl_2+7H_20             ---------- (4)

The overall reaction =

LiCl.7H_2O +LiCl.3H_2O ------>  2LiCl.10H_2O

The heat effects of the above reactions from 1-4 respectively are in the order ; 11311.34 kJ, -857.49 kJ, -873.61 kJ and 439.288kJ respectively

The overall enthalpy change is:

\delta H = \delta H _{LiCl.3H_2O}+\delta H_{3H_2O}  + \delta  H_{2(LiCl.5H_2O)} + \delta  H _{liCl.7H_2O}

\delta H =Q at constant pressure;

Thus; Q = 1311.3 (kJ)  857.49 (kJ) -873.61 (kJ) + 439.288 (kJ)

Q = 19.488 kJ

Thus, the heat effect = 19.488 kJ after the addition of  1 mol of LiCl⋅3H2O(s)

4 0
3 years ago
Which is true according to the kinetic theory?
iragen [17]

Answer: C. all particles of matter move very quickly

Explanation:

Heat is how fast the molecules are moving in something. This means even  a solid's molecules are moving.

8 0
3 years ago
Read 2 more answers
Question 2
guajiro [1.7K]

These questions all involve special cases of the ideal gas law, namely Boyle's, Charles', and Gay-Lussac's Laws. The ideal gas law relates together the absolute pressure (P), volume (V), the absolute temperature (T), and number of moles (n) of a gas by the following:

PV = nRT

where R is the universal gas constant.

The special cases of the ideal gas law are obtained by holding constant all but two of the variables of a gas.

Boyle's Law relates the pressure and volume of a given mass of gas at a constant temperature: PV = k or P₁V₁ = P₂V₂.

Charles' Law relates the volume and temperature of a given mass of gas at a constant pressure: V/T = k or V₁/T₁ = V₂/T₂.

Gay-Lussac's Law relates the pressure and temperature of a given mass of gas at a constant volume: P/T = k or P₁/T₁ = P₂/T₂.

Depending on what we're given and instructed to find in each question, we can figure out which law to use.

---

Question 2:

We are given the volume of a gas at some pressure, and we're to find the new volume of the gas at a different pressure. Here, we use Boyle's Law: P₁V₁ = P₂V₂ where P₁ = 60 atm, V₁ = 20.0 L, and P₂ = 30 atm. We want to find V₂, which we can determine by rearranging the equation into the form V₂ = P₁V₁/P₂. Note that pressure and volume are inversely related according to Boyle's Law; since we're decreasing the pressure, the new volume of the gas should be greater than the initial volume of 20.0 L.

V₂ = (60 atm)(20.0 L)/(30.0 atm) = 40.0 L.

So, at 30 atm, the balloon will have a volume of 40.0 L.

---

Question 3:

This is another Boyle's Law question. The standard pressure (our initial pressure) is 1 atm. Here, we are decreasing the volume of the gas, and we want to find the new pressure; the pressure of the gas should thus increase proportionally (the pressure will be greater than 1 atm). Rearranging Boyle's Law to solve for P₂, we get P₂ = P₁V₁/V₂.

P₂ = (1 atm)(8.00 L)/(3 L) = 2.67 atm.

So, the new pressure of the gas is 2.67 atm (or 3 atm if we're considering V₂ to comprise one significant figure).

---

Question 4:

Here, we are increasing the temperature of a gas at a known pressure, and we want to determine what the new pressure will be. This is a Gay-Lussac's Law question; from the law, we see that pressure and temperature are directly proportional. Since we're increasing the temperature of the gas, we should expect the pressure of the gas to be greater than the initial 200 atm. Gay-Lussac's Law rearranged to solve for P₂ gives us P₂ = P₁T₂/T₁. When working with gas laws, temperatures must be in Kelvin (°C + 273.15 = K). So, T₁ = 300.15 K, T₂ = 350.15 K, and P₁ = 200 atm.

P₂ = (200 atm)(350.15 K)/(300.15 K) = 233 atm.

So, if the temperature is increased from 27 to 77 °C, the pressure of the gas in the tennis ball will be 233 atm. Here, it's ambiguous how many sig figs to use; if we use one sig fig per P₁, then our P₂ would equal P₁, which I think would be an absurd for a question to ask for. I would stick with either 233 atm or 230 atm (following the two sig figs of the temperatures), or you may go with however you've been instructed.

---

Question 5:

This is a Charles' Law question; we're looking for the new volume of a gas when the temperature of the gas is increased. As was the case in Gay-Lussac's Law, the two parameters in Charles' Law—volume and temperature—are directly proportional. Since the temperature of the gas is increased, we should expect the new volume of the gas to also increase (V₂ will be greater than 5.00 L). Temperatures should be in Kelvin.

V₂ = V₁T₂/T₁ = (5.00 L)(300.15 K)/(250.15 K) = 5.99 L.

---

Question 6:

Another Charles' Law question. As with question 5, we want to find the new volume of the gas after a change in temperature. This time, the final temperature is lower than the initial temperature, so we should expect that V₂ will be less than the initial 0.5 L. Again, temperatures in Kelvin.

V₂ = V₁T₂/T₁ = (0.5 L)(313.15 K)/(493.15 K) = 0.317 L.

So, the volume of the balloon when it is fully cooled by your refrigerator will be 0.317 L.

---

Question 7:

This is yet another Charles' Law question, and, again, we are solving for V₂ after a change in temperature. Since the final temperature is greater than the initial temperature, V₂ should be greater than 2.2 L. Again, the temperatures should be in Kelvin.

V₂ = V₁T₂/T₁ = (2.2 L)(653.15 K)/(453.15 K) = 3.17 L.

The new volume of the gas is 3.17 L ≈ 3.2 L (two sig figs).

---

Question 8:

We return to Gay-Lussac's Law here; pressure and temperature are directly proportional, and the temperature of the gas is increased. Thus, P₂ should be greater than 3 atm. Again, remember that temperatures must be in Kelvin.

P₂ = P₁T₂/T₁ = (3 atm)(298.15 K)/(288.15 K) = 3.1 atm.

So, the pressure inside the can after the temperature rise is 3.1 atm. Not a big increase, but an increase nonetheless.

6 0
3 years ago
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