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Dominik [7]
3 years ago
13

What type of representative particle does KClO3 have?

Chemistry
1 answer:
bogdanovich [222]3 years ago
6 0

Answer:

formula units

Explanation:

$KClO_3$  is  potassium chlorate. It is an ionic compound containing potassium, chlorine and oxygen. The ion in potassium chlorate are held together by the ionic bonds in the lattice structure.

A representative particle may be defined as the smallest unit for which the substance exists in its natural form. For most of the elements, atom is the representative particle. Molecule is the representative particle for diatoms and molecular compounds. But for ionic compounds such as  $KClO_3$ , NaCl, etc. . the representative particle is its formula units.

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I'm stuck on this assignment, +20 PTS and BRAINLIEST for step by step:
dolphi86 [110]

Answer:

2. 2.74 L

3. 488 K

4. 7.47 L

5. 38.6 L

6. 2.85 mol

7. 319 K

8. 3.43 kPa

Explanation:

Generally speaking, you want to convert units to SI units, but in this case, we are working with ratios.  This makes up for using the units that wouldn't appropriate elsewhere.

2.  Use the equation P₁V₁ = P₂V₂.  Solve for V₂.  

(3.05 L)(870 kPa) = (969 kPa)(V₂)  

V₂ = 2.74 L.  

3.  Use the equation V₁/T₁ = V₂/T₂. Solve for T₂.  

(3.32 L)/(360 K) = (4.50 L)/(T₂)  

T₂ = 488 K.

4.  Do the same as above, but for V₂.  

(5.10 L)/(-56°C) = V₂/(-82°C)  

V₂ = 7.47 L

5.  Use the equation V₁/n₁ = V₂/n₂.  Solve for V₂.  

(37.2 L)/(0.750 mol) = (V₂)/(0.778 mol)  

V₂ = 38.6 L

6.  Do the same as above, but for n₂.

(86.0 L)/(2.65 mol) = (92.5 L)(n₂)  

n₂ = 2.85 mol

7.  Use the equation P₁/T₁ = P₂/T₂.  Solve for T₂.  

(3.00 atm)/(390 K) = (2.45 atm)/(T₂)  

T₂ = 319 K

8.  Do the same as above, but for P₂.  

In this specific case, however you will need to convert units.  Since both temperatures don't have the same sign, the ratio won't come out right.  Convert to Kelvin.  Add 273.15 to the temperature in Celsius to convert to Kelvin -12.3°C = 260.85 K  25°C = 298.15 K.

(3.00 kPa)/(260.85 K) = P₂/(298.15 K)

P₂ = 3.43 kPa

There is a lot in here... If you are confused about something, let me know!

6 0
3 years ago
"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the
soldier1979 [14.2K]

<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)       ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 46.7 g

m_2 = mass of cold water = 45.33 g

T_{final} = final temperature = 59.4°C

T_1 = initial temperature of hot water = 80.6°C

T_2 = initial temperature of cold water = 40.6°C

c_1 = specific heat of hot water = 4.184 J/g°C

c_2 = specific heat of cold water = 4.184 J/g°C

c_3 = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)

c_3=30.68J/g^oC

Hence, the specific heat of calorimeter is 30.68 J/g°C

6 0
4 years ago
Write a balenced chemical equation for the reaction: include abbreviation for the physical states-
Nuetrik [128]

Answer:

2Na(s) + 2H2O(l) ------> 2NaOH(aq) + H2(g)

Explanation:

5 0
3 years ago
Will Mark brainliest fro correct answer!!!!<br><br> Please answer in the picture chemistry
tiny-mole [99]

Answer:

look at the graph

Explanation:

We know that as temperature increases, solubility increases.So, when there is a rise in temperature, as more solute become dissolved, the saturation point will be lifted and more amount of solute will be needed to reach saturation.

Here, when the temperature was 20oC, 38 g of salt was needed for saturation. As the temperature is increased by 15oC, at 35oC more amount of salt was needed to reach saturation(45g). So a 15oC rise in temperature caused a 7 g rise in the amount of salt needed for saturation. So, if temperature is increased additionally through 10oC, an approximate 4.5 g of salt will be needed more to reach the saturation. That is at 45oC, the amount of salt at saturation will be approximately 49.5 g.

So, the temperature and solubility as well as temperature and amount of salt at saturation are linearly related(directly proportional)

6 0
3 years ago
A student weighs an empty flask and stopper and finds the mass to be 53.256 g. She then adds about 5 mL of an unknown liquid and
Delvig [45]

Answer:

a. The pressure in the flask open to the atmosphere during the vaporization of the unknown liquid is equal to the prevailing atmospheric pressure equivalent to 0.957734 atm

Explanation:

The mass of the empty flask and stopper, m = 53.256 g

The volume of the unknown liquid she adds = 5 mL

The temperature of the water in which she heats up the flask = 98.8 °C = 371.95 K

The mass of the flask and the condensed vapor = 53.780 g

The volume of the flask, V = 231.1 mL

The atmospheric pressure, P = 728 mm Hg

a. We note that the student stoppers the flask after all the liquid has evaporated. Therefore, given that the flask was open to the atmospheric pressure as the liquid evaporates, the pressure of the vapor in the flask is equal to the prevailing atmospheric pressure, or 728 mmHg

Using a calculator, 728 mm Hg is equivalent to 0.957734 atm.

3 0
3 years ago
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