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tensa zangetsu [6.8K]
4 years ago
14

Which of the following is needed to calculate the density of a substance?

Chemistry
2 answers:
mixas84 [53]4 years ago
8 0
D. Mass and volume. The equation to find density is Density=Mass/Volume
Ad libitum [116K]4 years ago
4 0
Mass divided by volume 
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Gasoline is a mixture of hydrocarbons, a major component of which is octane, CH3CH2CH2CH2CH2CH2CH2CH3. Octane has a vapor pressu
Nitella [24]

Answer:

\Delta \:H_{vap}=40383.88\ J/mol

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c

Where,  

P is the vapor pressure

ΔHvap  is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)

Given:

P_1 = 13.95 torr

P_2 = 144.78 torr

T_1 = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K

T_1 = 298.15 K

T_2 = 75°C  = 348.15 K

So,

\ln \:\left(\:\frac{13.95}{144.78}\right)\:=\:\frac{\Delta \:H_{vap}}{8.314}\:\left(\:\frac{1}{348.15}-\:\frac{1}{298.15}\:\right)

\Delta \:H_{vap}=\ln \left(\frac{13.95}{144.78}\right)\frac{8.314}{\left(\frac{1}{348.15}-\frac{1}{298.15}\right)}

\Delta \:H_{vap}=\frac{8.314}{\frac{1}{348.15}-\frac{1}{298.15}}\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)

\Delta \:H_{vap}=\left(-\frac{863000.86966\dots }{50}\right)\left(\ln \left(13.95\right)-\ln \left(144.78\right)\right)

\Delta \:H_{vap}=40383.88\ J/mol

4 0
3 years ago
Convert the pressure 0.8 atm to kPa.
Ludmilka [50]

Answer:

81.0

Explanation:

mutiply the value by 101.325 to get the answer a. 81.1

3 0
3 years ago
Read 2 more answers
Rank the following acids in order of increasing acid strength. Key: Weakening of hydrogen bond and stability of resulting anion.
raketka [301]

Explanation:

The given compounds are oxyacids and in these compounds more is the electronegativity of the central atom more will be its acidic strength.

This is because more is the electronegativity of the central atom more will be the polarity of OH bond. As a result, the compound can readily lose H^{+} ion.

Also, more is the electronegativity of central atom more will be the stability of conjugate base formed.

Thus, we can conclude that given compounds are arranged in increasing acid strength as follows.

       HOI < HOBr_{2} < HOCl_{3} < HOF

8 0
3 years ago
Box 1
Alex73 [517]

10g

Explanation:

Box 1, Mass of A = 10g

Box 2, Mass of B = 5g

Box 3, = 1A + 1B

Unknown:

Mass of B that would combine with mass of 20g of A

Solution:

   Mass ratio of A to B:

   \frac{mass of A}{mass of B} = mass ratio

           \frac{10}{5} = mass ratio

        The mass ratio of A to B = 2: 1

Now, number of B that will combine with 20g of A;

       

           \frac{mass of A}{mass of B} = mass ratio

               \frac{20}{mass of B} = \frac{2}{1}

                     Mass of B = 10g

10g of B would combine with 20g of A

learn more:

Rate brainly.com/question/8677367

#learnwithBrainly

7 0
3 years ago
Read 2 more answers
The heat of combustion of ethane is -337.0kcal at 25 degrees celsius. what is the heat of the reaction when 3g of ethane is burn
Ghella [55]

Answer:

Q = -33.6kcal .

Explanation:

Hello there!

In this case, according to the equation for the calculation of the total heat of reaction when a fixed mass of a fuel like ethane is burnt, we can write:

Q=n*\Delta _cH

Whereas n stands for the moles and the other term for the enthalpy of combustion. Thus, for the required total heat of reaction, we first compute the moles of ethane in 3 g as shown below:

n=3g*\frac{1mol}{30.08g}=0.1mol

Next, we understand that -337.0kcal is the heat released by the combustion of 1 mole of ethane, therefore, to compute Q, we proceed as follows:

Q=0.1mol*-337.0\frac{kcal}{mol}\\\\Q=-33.6kcal

Best regards!

8 0
3 years ago
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