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dsp73
2 years ago
14

The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all

of the boxes that apply
Chemistry
1 answer:
Brilliant_brown [7]2 years ago
6 0

<u>Full Question:</u>

The list below includes some of the properties of butane, a common fuel. Identify the chemical properties in the list. Check all of the boxes that apply.

denser than water

burns readily in air  

boiling point of –1.1°C  

odorless

does not react with water

burns readily in air

does not react with water

<h3><u>Explanation:</u></h3>

The type of alkane that is used in many products includes Butane. It is found as a natural gas in the environment. It is found on the deeper part of ground. It can be obtained by drilling process and gets used up in many of the products that is used for commercial purposes.

Molecular mass that is associated with butane is 58.12 g/mol. The boiling point of butane is -1 degree Celsius and -140 degree Celsius is its melting point. It is a liquefied gas and does not react with water. It will burn in air more readily.

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If 5.61 g of ammonium nitrate (NH4NO3) is dissolved in 155 mL of water to make an instant cold pack, what is the molarity of the
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Answer:

0.452 M

Explanation:

Remember that to get molarity, you have to go thru an equation:

M=mol/V (in liters)

So...

5.61 g NH4NO3 x 1 mol NH4NO3/80.043 g

=0.0701 mol NH4NO3

Convert mL to L

=155--->0.155 L

Plug in and divide.

0.0701/0.155

=0.452 M

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Based on the results of this lab, write a short paragraph that summarizes how to distinguish physical changes from chemical chan
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The highest principal enegy level of period 2 elements is 2. Period 3 elements all have six 3p electrons. Period 4 elements have
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Explanation:

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The highest principle energy level of elements of any period is equal to period number. So the given statement is true.

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For example, sodium and sulfur are period 3 elements but do not contain six 3p electrons. Only Argon of period possesses six p-electrons.

Period 4 elements have an inner electron configuration of [Ar]- (True)

Atomic number of elements belonging to period 4 have atomic number more than 18 (atomic number of Ar). So they must have an inner electron configuration of [Ar]

The valence electrons of group 5A elements are in the 6s subshell – [FALSE]

Elements of group 5A are nitrogen, arsenic, antimony and bismuth. They are p-block elements. Therefore, valence electrons are present in p-subshell.

Group 8A elements have full outer principal s and p subshells –(True)  

Group 8A elements are also known as noble gas element. Octets of noble gases are complete and hence, have full outer principal s and p subshells.

The valence electrons of group 2A elements are in an s subshell – (True)

Elements of group 2A are barium, magnesium, calcium, etc. They belong to s-block elements. Therefore, their valence electrons will be in s-subshell.

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6 0
2 years ago
The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T
kifflom [539]

Answer : The equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

Explanation :

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

First we have to calculate the standard free energy of reaction (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]

where,

\Delta G^o = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]

\Delta G^o=-90.72kJ/mol

Now we have to calculate the value of K_p

\Delta G^o=-RT\ln K_p

where,

\Delta G_^o =  standard Gibbs free energy  = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

K_p = equilibrium constant  = ?

Now put all the given values in this expression, we get:

-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p

K_p=7.98\times 10^{15}

Now we have to calculate the value of K_p.

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

The expression for equilibrium constant will be :

K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium SO_2 pressure be, x

Pressure of SO_2 = Pressure of H_2S = x

Now put all the given values in this expression, we get

7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}

x=3.93\times 10^{-5}torr

Thus, the equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

4 0
2 years ago
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