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alukav5142 [94]
4 years ago
13

Pablo's experiments confirm the relationship between pressure and the number of gas molecules. as the pressure in a system incre

ases, the number of gas molecules increases. pablo conducted many trials. what can you conclude about the relationship between pressure and the number of gas molecules?
a.pressure and the number of gas molecules are not related.
b.pressure and the number of gas molecules remain constant.
c.pressure and the number of gas molecules are directly related.
d.pressure and the number of gas molecules are indirectly related.
Chemistry
2 answers:
horrorfan [7]4 years ago
6 0
The correct answer is:  [C]:  
___________________________________________________________
          "<span>pressure and the number of gas molecules are directly related."
___________________________________________________________
<u>Note</u>:  The conclusion was: "</span> as the pressure in a system increases, the number of gas molecules increases" — over the course of many trials.
          This means that the "pressure" and the "number of gas molecules" are directly related.

Furthermore, this conclusion is consistent with the "ideal gas law" equation:

     " PV = nRT " ; 
____________________________________________________________
in which:

    "P = Pressure" ; 
    "n = number of gas molecules" ; 
___________________________________________________________
        All other factors held equal, when "n" (the "number of gas molecules") 
 increases in value (on the "right-hand side" of the equation), the value for "P" (the "pressure" — on the "left-hand side" of the equation), increases.
___________________________________________________________
Tomtit [17]4 years ago
3 0

Answer is C .............

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If an atom suffers from a collision, that causes an electron to jump from a lower to higher state, it is called collisional excitation 
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What is colligative property? why does theproperty changes the way that it does when the amount of solute is increased. State an
elena-14-01-66 [18.8K]

Colligative properties are those substances that depend on the number of substances in the solution, not in the identity of that substance. The property changes the way that it does when the amount of solute is increased because it enables the solute to be scattered more. For example, the freezing point of salt water is lower than that of the pure water due to the salt ions present in water.

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3 years ago
What is the gram formula mass of Ca3(PO4)2
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Mass Molar of Ca_{3} (PO_4)_2
Ca = 3*40 = 120 amu
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O = (16*4)*2 = 64*2 = 128 amu
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Mass Molar of Ca_{3} (PO_4)_2 = 120 + 62 + 128 = 310 g/mol

Therefore: <span>What is the gram formula mass of Ca3(PO4)2 ?
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6 0
3 years ago
1) For the following reaction, 8.44 grams of carbon (graphite) are allowed to react with 9.63 grams of oxygen gas.C(graphite)(s)
Yuki888 [10]

Answer:

The answers to the question are

(a) 13.24 g

(b) (O₂)

(c) 4.8252 g

(2) 0.662 M/L

Explanation:

(a) To solve the question we write the equation as follows

C + O₂ → CO₂

That is one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide

number of moles of graphite = 8.44/12 = 0.703 moles

number of moles of oxygen = 9.63/32 = 0.3009

However since one mole of graphite reacts with one mole of oxygen to form one mole of carbon dioxide, therefore, 0.3009 moles of oxygen will react with 0.3009 moles of  carbon to fore 0.3009 moles of  CO₂

The maximum mass of carbon dioxide that  can be formed = mass = moles × molar mass

= 0.3009×44 = 13.24 g

(b) The formula for  the limiting reagent (O₂)

Finding the limting reagent is by checking the mole balance of the reactants available to the moles specified in th stoichiometry of the reaction and selecting the reagent with the list number of moles

(c) The mass of excess reagent = 0.703 moles - 0.3009 moles = 0.4021 moles

mass of excess reagent = 0.4021 × 12 = 4.8252 g

(2) The molarity is given by number of moles per liter of solution, therefore

molar mass of mgnesium iodide = 278.1139 g/mol, number of moles of magnesium iodide in 23 g = 23g/ 278.1139 g/mol= 8.3 × 10⁻² M

Therefore the moles in 125 mL = (8.3 × 10⁻² M)/(125 mL) = (8.3 × 10⁻² M)/(0.125 L) = 0.662 M/L

3 0
3 years ago
A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling
aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}        .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL

Putting values in above equation, we get:

0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g

Hence, the mass of glucose in final solution is 0.420 grams

3 0
3 years ago
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