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WARRIOR [948]
2 years ago
8

Given the reaction: A + B <--> C + D

Chemistry
1 answer:
Lady_Fox [76]2 years ago
6 0

Answer:

A.) 4.0

Explanation:

The general equilibrium expression looks like this:

K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }

In this expression,

-----> K = equilibrium constant

-----> uppercase letters = molarity

-----> lowercase letters = balanced equation coefficients

In this case, the molarity's do not need to be raised to any numbers because the coefficients in the balanced equation are all 1. You can find the constant by plugging the given molarities into the equation and simplifying.

K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }                                       <----- Equilibrium expression

K = \frac{[2 M] [2 M]}{[1 M] [1 M] }                                     <----- Insert molarities

K = \frac{4}{1  }                                                <----- Multiply

K = 4                                                <----- Divide

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Which quantity must be determined experimentally in order to determine the bonding atomic radius of an atom?.
spin [16.1K]

Only by measuring the separation between the nuclei of two contacting atoms and halving that distance can one determine the radius of an atom. The illustrations show how the environment around an atom can cause it to have a varied radius, even if it is the same atom.

<h3>How can the atomic size of an element be determined?</h3>

By measuring the distance between the two atoms after two are united, we may determine the atomic size of the new molecule. The third technique for determining an element's atomic size is to create a single covalent link between two atoms and measure the space between them.

Atomic radii in the periodic table change from left to right across a row and from top to bottom along a column. Because of these two patterns, the periodic table's bottom left corner contains the biggest atoms, while the upper right corner has the smallest.

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7 0
2 years ago
Will NaCl be soluble or insoluble
attashe74 [19]

NaCl is salt to it is obviously SOLUBLE :)

6 0
4 years ago
A bicycle tire is filled with air to a pressure of 100. psi at a temperature of 19C. Riding the bike on asphalt on a hot day inc
kirill [66]

Answer:

108.9897 psi

Explanation:

Using Ideal gas equation for same mole of gas as

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

Given ,  

Let V₁ = x units

The new volume increases by 4 % which means

V₂ = x + 0.04x = 1.04 x

P₁ = 100 psi

P₂ = ?

T₁ = 19 ºC

T₂ = 58 ºC

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

T(K) = T( °C) + 273.15  

So,  

T₁ = (19 + 273.15) K = 292.15 K  

T₂ = (58 + 273.15) K = 331.15 K  

Using above equation as:

\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}

\frac {{100}\times {x}}{292.15}=\frac {{P_2}\times {1.04x}}{331.15}

Solving for P₂ , we get:

<u>P₂ = 108.9897 psi</u>

6 0
3 years ago
What is the percent of Cu in CuSO4?
Alinara [238K]

The mass percentage of the copper (Cu) in the compound CuSO₄ is 39.81%.

<h3>What is the mass percentage?</h3>

The percentage of an element in a compound can be determined as the number of parts by mass of that element present in 100 parts by mass of the given compound.

First, calculate the molecular mass of the given compound by the addition of the atomic masses of all the present elements in the molecular formula. Then, the percentage of the elements can be determined by dividing the total mass of the element by the molar mass of the compound multiplied by 100.

Given, the atomic mass of copper, sulfur, and oxygen is 63.55 g, 32.07 g, and 16.0g respectively.

The molecular mass of CuSO₄ =  63.55 + 32.07 + 4(16.0)  = 159.62 g.

The mass percentage of the Sulphur = (63.55/159.62) × 100 = 39.81%.

Therefore, the mass percentage of Copper is equal to 39To learn.

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7 0
1 year ago
You have 505 mL of a 0.160 M HCl solution and you want to dilute it to exactly 0.100 M. How much water should you add
liraira [26]

This problem is providing information about the volume and concentration of HCl as 505 mL and 0.160 M respectively, and it is asking for the water that have to be added to prepare a 0.100-M solution.

In such a way, we work over the assumption of constant moles in dilution processes, so that we are able to write:

M_1V_1=M_2V_2

Which relates de volume and concentration at the beginning and end of the experiment. This means we can solve for the resulting volume of the solution as a first calculation:

V_2=\frac{M_1V_1}{M_2}\\\\V_2=\frac{0.160M*505mL}{0.100M}=808mL

Hence, the following amount of water must be added:

V_w=808mL-505mL=303mL

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