CHEMISTRY HIGH SCHOOL

2 points) Now they feel it is best to have you identify an unknown gas based on its properties. Suppose 0.508 g of a gas occupies a volume of 0.175 L at a temperature of 25.0◦C is held at a pressure of 1.000 atm. What gas is this most likely?

Answers

Answer 1

Answer:

Answer: chlorine gas

Explanation:

mass = 0.508 g volume = 0.175 L

From ideal gas equation.

PV = nRT

(1.0 atm) × (0.175 L) = n × [0.0821 atm L / (mol K)] × [(273.2 + 25.0) K]

n = 1.0 × 0.175 / (0.0821 × 298.2) mol

n = 0.00715 mol

Molarmass= mass/mole

Molar mass of the gas = (0.508 g) / (0.00715 mol) = 71.0 g/mol

The molar mass of Cl₂ (chorine) = 35.5 × 2 g/mol = 71.0 g/mol

Hence, the gas is chlorine.

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. Detailed information about the possibilities and ways of adjusting cookies settings are available in the specify conditions of storing and accessing cookies in your brow

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Regular copyright is more restrictive than any other kind of license. true or flase

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The correct answer is true

An atom has 11 protons, 12 neutrons, and 11 electrons. what is the charge of the atoms nucleus?A) +11
B) -11
C) 0
D) 23

Answers

Answer:

Since sodium has 11 protons, the number of neutrons must be 23 – 11 = 12 neutrons.

Explanation:

When group 2A element from ion they A. lose two proton
B. gain two proton
C. lose two electron
D. gain two electron

Answers

Well, knowing that group 2A elements are metals, and metals lose electrons to gain a full electron shell, we can eliminate B and D. Knowing that these are 2 electrons, that the group 2A element provides to another atom, to form an ion.

The solution is C. Lose 2 electrons.

When two intermediate chemical equations are combined, the same substance that appears in the same phase can be canceled out, provided thatA) it is a reactant in one intermediate reaction and a catalyst in the other reaction.
B) it is a product in one intermediate reaction and a catalyst in the other reaction.
C) it is a reactant in one intermediate reaction and a product in the other reaction.
D) it is a reactant in both of the intermediate reactions.

Answers

Answer: Option (C) is the correct option.

Explanation:

When two intermediate chemical equations are combined, the same substance that appears in the same phase can be canceled out, provided that it is a reactant in one intermediate reaction and a product in the other reaction.

For example,

$2Al(s) + 2KOH(aq) + 6H_(2)O\rightarrow 2KAl(OH)_(4)(aq)+ 3H_(2)(g)$ ....(1)

$2KAl(OH)_(4)(aq) + 4H_(2)SO_(4)(aq) + 6H_(2)O(l)\rightarrow 2KAl(SO_(4))_(2)(aq)+ 12H_(2)O(s)$ .........(2)

Cancelling the common species in both the equations as follows.

$2Al(s) + 2KOH(aq) + 6H_(2)O\rightarrow \not{2KAl(OH)_(4)(aq)}+ 3H_(2)(g)$

$\not{2KAl(OH)_(4)(aq)}+ 4H_(2)SO_(4)(aq) + 6H_(2)O(l)\rightarrow 2KAl(SO_(4))_(2)(aq)+ 12H_(2)O(s)$

Therefore, on addition we get the equation as follows.

$2Al(s)+ 2KOH(aq) + 4H_(2)SO_(4)(aq) + 22H_(2)O(l)\rightarrow 3H_(2)(g)+ 2KAl(SO_(4))_(2) + 12H_(2)O(s)$

The right answer for the question that is being asked and shown above is that: "C) it is a reactant in one intermediate reaction and a product in the other reaction." When two intermediate chemical equations are combined, the same substance that appears in the same phase can be canceled out, provided that it is a reactant in one intermediate reaction and a product in the other reaction.

What volume of oxygen at STP is required for the complete combustion of 100.50 mL of C2H2?

Answers

The given substance combusts following the reaction:

C2H2 + (5/2)O2 -> 2CO2 + H2O

Assume C2H2 is an ideal gas. At STP, 1 mol of an ideal gas occupies 22.4 L. Given 100.50 mL of C2H2, this means that there is 4.4866 x 10^(-3) mol. Combusting 1 mol of C2H2 consumes (5/2) mol of O2, then combusting the given amount of C2H2 consumes 0.01121 mol of O2. At STP, this amount of O2 occupies 251.25 mL.

Answer: 251.25 ml

Explanation: $2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2+2H_2O$