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1 year ago

A college student drives 40 miles roundtrip each weekday to attend class. His car gets 18 miles per

1 answer:

3 0

In four weeks of commuting to school, it is expected the student saves a total of $13.32 if the prices went down from $2.49/gallon to $2.19.

<h3>How many gallons of gasoline does the student need each day and each week?</h3>

It is known, the total distance each day is 40 miles; moreover, the car requires 1 gallon of gasoline per 18 miles. Based on this, let's calculate the number of gallons of gasoline required:

40 miles / 18 miles = 2.22 gallons of gasoline

Now, if the student requires 2.22 gallons of gasoline everyday, the total number of gallons of gasoline per week is:

2.22 x 5 days = 11.1 gallons of gasoline
11.1 x 4= 44.4 gallons per month

<h3>How much will the student save?</h3>

Money spent with the regular gasoline price

44.4 gallons of gasoline x $2.49 = $110.55

Money spend with the new price:

44.4 gallons of gasoline x $2.19 = $97.23
$110.55-$97.23 =$13.32

Learn more about gasoline price in: brainly.com/question/13898342

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When 2.3 × 10^3 g of CaCO3 are heated, the actual yield of CaO is 1.09 × 10^3g. What is

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The percent yield : 4. 84.58%

<h3>Further explanation</h3>

Reaction

CaCO₃ ⇄ CaO+CO₂

mass CaCO₃ = 2.3 × 10³ g

mol CaCO₃ (MW=100.0869 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{2.3\times 10^3}{100,0869}\\\\mol=22.98$

From the equation, mol CaCO₃ : CaO = 1 : 1, so mol CaO=22.98

mass CaO(MW=56.0774 g/mol)⇒ (theoretical) :

$\tt mass=mol\times MW\\\\mass=22.98\times 56,0774\\\\mass=1288.659~g$

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{1090}{1288.659}\times 100\%\\\\\5yield=84.58\%$

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a it rarely reach with other elements

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Why would the federal government and city government find it necessary regulate water standard?

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They do this so that when you drink it or use it the contaminates that used to be in it dont make you sick

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3 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

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<u>Answer:</u>

<u>For A:</u> The $K_p$ for the given reaction is $4.0\times 10^1$

<u>For B:</u> The $K_c$ for the given reaction is 1642.

<u>Explanation:</u>

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

<u>For A:</u>

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

<u>For B:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

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