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kap26 [50]
3 years ago
11

What is the mass of 2.50 moles of NaCl

Chemistry
1 answer:
garik1379 [7]3 years ago
8 0

Answer:

The mass of 2,50 moles of NaCl is 146, 25 g.

Explanation:

First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate the mass of 2.50 moles of compound, making a simple rule of three:

Weight NaCl= Weight Na + Weight Cl=  23 g+ 35,5 g= 58, 5 g/ mol

1 mol ------ 58, 5 g

2,5 mol---x= (2,5 mol x 58, 5 g)/ 1 mol = <u>146, 25 g</u>

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Answer:

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Consider the combustion of octane (C8H18), a primary component of gasoline.
kicyunya [14]

Answer:

a. 2562 L

b. 4082 L

c. 413,6 kJ

d. 0,115 kWh

e. 1,3 cents

f. 0,34 %

Explanation:

a. To calculate the volume we should obtain the moles of Octane to know the reactant moles and produced moles. Then, with ideal gas law obtain the change in volume:

3,784 L ≡ 3784 mL octane × (0,703 g / 1 mL) = <em>2660 g Octane</em>

<em>                                                    density</em>

2660 g octane × ( 1 mol / 114,23 g octane) = <em>23,29 mol octane</em>

                              <em>molar mass of C₈H₁₈</em>

With the combustion reaction of octane we can know how many moles are produced from 23,29 mol of octane, thus, in (1) :

2 C₈H₁₈ (l) + 25 O₂(g) ---> 18 H₂O(g) + 16 CO₂ (g) <em>(1)</em>

23,29 mol octane × ( 25 mol O₂ / 2 mol octane) = <em>291,1 mol O₂ -reactant moles-</em>

23,29 mol octane × ( 18 + 16 produced mol  / 2 mol octane) = 395,9 moles produced

Ideal gas formula says:

V = nRT/ P

Where:

n = Δmoles number (produced-reactant) → 104,8 moles

R = Ideal gas constant → 0,082 atm·L/mol·K

T = Temperature → 25°C, 298,15 K

P = Pressure → 1 atm

Thus, replacing in the equation:

ΔV = 2562 L

b. To calculate the gas volume we should use the same values of ideal gas formula just changing the temperature value for 475 K -Because the produced moles of gas and presure are the same and R is constant.

Thus, the volume of produced gases is:

ΔV = 4082 L

c. The work, w, is equal to -pressure times Δ Volume:

w = - P×ΔV

The pressure is 1 atm and ΔV in the system is 4082 L

So, w = 4082 atm·L (101,325 J / 1 atm·L) = 413,6 kJ

d. As kJ is equal to kWs, 413,6 kJ ≡ 413,6 kWs × ( 1 hour / 3600 s) =

0,115 kWh

e. In Seattle 1kWh cost 11,35 cents. So, 0,115 kWh cost:

0,115 kWh × (11,35 cents/ 1kWh) = 1,3 cents

f. The energy calculated in part C, <em>413,6 kJ</em> is due to the work done by the system in gas expansion but total of heat produced in (1) are 1,2 ×10⁵ kJ. Thus, the proportion of work in gas expansion in total energy in combustion of octane is:

413,6 kJ / 1,2×10 ⁵ kJ × 100 = 0,34 %

I hope it helps!

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Solution A has a concentration of 3 M. If solution B is 5 times more concentrated than A what is it’s concentration?
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Assume 1.0g of CO2 was gotten as a product of the experiment

2) For the theoretical yield

The mass of NaHCO3 was not stated, but for the purpose of this solution, I would assume 2g of NaHCO3 was used for the experiment. You can substitute any parameter by following the steps I follow

Number of mole of NaHCO3 = Mass of NaHCO3/ Molar Mass of NaHCO3

Number of mole of NaHCO3 = 2/84.01 = 0.0238 moles

1 mole of NaHCO3 yielded 1 mole of CO2

0.0238 moles of NaHCO3 will yield 0.0238 moles of CO2

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The value of log₂(x/4) is 22. Using the properties of the logarithm, the required value is calculated.

<h3>What are the required properties of the logarithm?</h3>

The required logarithm properties are

logₐx = n ⇒ aⁿ = x; and logₐ(xⁿ) = n logₐ(x);

Where a is the base of the logarithm.

<h3>Calculation:</h3>

It is given that,

log₄(x) = 12;

On applying the property logₐx = n ⇒ aⁿ = x; here a = 4;

So,

log₄(x) = 12 ⇒ 4¹² = x

⇒ x = (2²)¹² = 2²⁴

Then, calculating log₂(x/4):

log₂(x/4) = log₂(2²⁴/4)

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On applying the property logₐ(xⁿ) = n logₐ(x);

log₂(x/4) = 22 log₂2

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log₂(x/4) = 22(1)

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Learn more about the properties of logarithm here:

brainly.com/question/12049968

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