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Xelga [282]
3 years ago
5

A generic gas, X , is placed in a sealed glass jar and decomposes to form gaseous Y and solid Z . 2 X ( g ) − ⇀ ↽ − Y ( g ) + Z

( s ) How are these equilibrium quantities affected by the initial amount of X ( g ) placed in the container? Assume constant temperature.
Chemistry
1 answer:
tamaranim1 [39]3 years ago
4 0

Answer:

The equilibrium will go right, increasing the quantities of Y(g) and Z(s), when the amount of X(g) increases.

Explanation:

Equilibrium is a state in which the velocity of the formation of the products is equal to the velocity of the formation of the reactants. Thus, the concentrations remain constant.

By Le Chatêlier's principle, when the equilibrium is disturbed, the equilibrium must shift to reestablish the equilibrium. Thus, when X is put in the container, it intends to decompose and form Y and Z, thus, as higher is the initial X, as higher is the Y and Z formed.

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The number of neutrons in an atom can be found by subtracting the atomic number from the what?
Svetradugi [14.3K]

Answer:

This approximation of mass can be used to easily calculate how many neutrons an element has by simply subtracting the number of protons from the mass number. Protons and neutrons both weigh about one atomic mass unit or amu. Isotopes of the same element will have the same atomic number but different mass numbers.

Explanation:

3 0
3 years ago
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Write one common thing between condensation and hydrolysis ?​
serious [3.7K]

Answer:

The common thing is the compound water

Explanation:

in condensation h2O is expelled while in hydrolysis water is used or added

8 0
3 years ago
In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe
const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = 10^{-9} m.

Hence, 0.283 nm = 0.283 \times 10^{-9} m

  • Formula for coulombic energy is as follows.

             U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}

where,   e = 1.6 \times 10^{-19} C

            \epsilon_{o} = 8.85 \times 10^{-12}

          U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}

                         = 1.423 \times 10^{-18} J

  • As 1 eV = 1.6 \times 10^{-19} J

So,       1 J = \frac{1 eV}{1.6 \times 10^{-19}}

Hence,    U = \frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}

                   = 8.9 eV

  • Also,   1 J = \frac{10^{-3} kJ}{6.022 \times 10^{23}mol}

                = 1.67 \times 10^{-27} kJ/mol

Therefore, U = 1.423 \times 10^{-18} J \times 1.67 \times 10^{-27} kJ/mol

                     = 2.37 \times 10^{-45} kJ/mol

7 0
3 years ago
What mass of butane in grams is necessary to produce 1.5×103 kJ1.5×103 kJ of heat? What mass of CO2CO2 is produced? Assume the r
saul85 [17]

32.8 g of Butane is required and 99.3 g of CO₂ is produced

<u>Explanation:</u>

The above mentioned reaction can be written as,

C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g)     where ΔH (rxn)= -2658 kJ

It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.

That is

$\frac{1500}{2658}=0.564 \text { moles }    of butane reacted

Now this moles is converted into mass by multiplying it with its molar mass  = 0.564 mol × 58.122 g / mol

                     = 32.8 g of butane.

Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol

                                        = 99.3 g of CO₂

Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced

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katen-ka-za [31]
Aluminium is a good conductor of heat
physical properties is what you can see about the element or object all other options are chemical properties
6 0
3 years ago
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