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3 years ago

How structural formulas differ from ball-and stick models

1 answer:

8 0

The difference between them is that instead of sticks representing bonds, as in the ball and stick model, the spheres that represent atoms are fused together and the size of the sphere represents the size of the actual atom.

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Write your question here (Keep it simple and clear to get the best answer) calculate the atomic weight of a metal when 125ml of

lana [24]

Answer:

The answer to your question is 24.32 g

Explanation:

Data

Atomic weight = ?

HCl volume = 125 ml

Molarity = 0.2

mass of metal = 0.304 g

Balanced chemical equation

M + 2HCl ⇒ MCl₂ + H₂

Process

1.- Calculate the moles of HCl

Molarity = moles / volume (L)

- Solve for moles

moles = Molarity x volume

moles = 0.2 x 0.125

= 0.025

2.- Calculate the moles of the Metal

1 mol of M ----------------- 2 moles of HCl

x ----------------- 0.025 moles of HCl

x = (0.025 x 1) / 2

x = 0.0125 moles of HCl

3.- Calculate the atomic weight of the metal

atomic weight ---------------- 1 mol

0.304 g ---------------0.0125 moles

Atomic weight = (1 x 0.304) / 0.0125

Atomic weight = 24.32 g

7 0

3 years ago

Read 2 more answers

At 298 K, what is the Gibbs free energy change (ΔG) for the following reaction?

9966 [12]

Answer:

(a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

Explanation:

Given that,

Temperature = 298 K

Suppose, density of graphite is 2.25 g/cm³ and density of diamond is 3.51 g/cm³.

$\Delta H\ for\ diamond = 1.897 kJ/mol$

$\Delta H\ for\ graphite = 0 kJ/mol$

$\Delta S\ for\ diamond = 2.38 J/(K mol)$

$\Delta S\ for\ graphite = 5.73 J/(K mol)$

(a) We need to calculate the value of $\Delta G$ for diamond

Using formula of Gibbs free energy change

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G= (1897-0)-298\times(2.38-5.73)$

$\Delta G=2895.3$

$\Delta G=2.895\ kJ$

The Gibbs free energy change is positive.

(b). When it is compressed isothermally from 1 atm to 1000 atm

We need to calculate the change of Gibbs free energy of diamond

Using formula of gibbs free energy

$\Delta S=V\times\Delta P$

$\Delta S=\dfrac{m}{\rho}\times\Delta P$

Put the value into the formula

$\Delta S=\dfrac{12\times10^{-6}}{3.51}\times999\times10130$

$\Delta S=34.59\ J/mole$

(c). Assuming that graphite and diamond are incompressible

We need to calculate the pressure

Using formula of Gibbs free energy

$\beta= \Delta G_{g}+\Delta G+\Delta G_{d}$

$\beta=V(-\Delta P_{g})+\Delta G+V\Delta P_{d}$

$\beta=\Delta P(V_{d}-V_{g})+\Delta G$

Put the value into the formula

$0=\Delta P(\dfrac{12\times10^{-6}}{3.51}-\dfrac{12\times10^{-6}}{2.25})\times10130+2895.3$

$0=-0.0194\Delta P+2895.3$

$\Delta P=\dfrac{2895.3}{0.0194}$

$\Delta P=14924\ atm$

(d). Here, $C_{p}=0$

So, The value of $\Delta H$ and $\Delta S$ at 900 k will be equal at 298 K

We need to calculate the Gibbs free energy of diamond relative to graphite

Using formula of Gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Put the value into the formula

$\Delta G=(1897-0)-900\times(2.38-5.73)$

$\Delta G=4912\ J$

Hence, (a). The Gibbs free energy change is 2.895 kJ and its positive.

(b). The Gibbs free energy change is 34.59 J/mole

(c). The pressure is 14924 atm.

(d). The Gibbs free energy of diamond relative to graphite is 4912 J.

7 0

3 years ago

How many milliliters (mL) of 0.610 M NaOH solution are needed to neutralize 20 mL of a 0.245 M H2SO4 solution?

jenyasd209 [6]

There is to know the b

6 0

3 years ago

How many sigma bonds does NBr3 have?

Kisachek [45]

Answer:

It has 3 sigma bonds

Explanation:

A sigma bond is a covalent interaction between two atoms, these bonds are the strongest type of covalent bonds and they are formed by the overlap of atomic orbitals.

The nitrogen is bond to the bromine atoms through sigma bonds, which means that exist 3 sigma bonds in the molecule.

this molecule has a trigonal pyramid geometry due to the free pair of nitrogen electrons.

5 0

3 years ago

An aqueous solution that is 20.0 percent sulfuric acid (h2so4) by mass has a density of 1.105 g/ml. determine the molarity of th

GuDViN [60]

The definition of molarity is:

Molarity = moles of solute / total volume of solution in Liters

In this case, our solute is Sulfuric Acid. To solve this problem, let us assume a basis for calculation.

Let us say that,

Total mass of solution = 100 g

Therefore mass of H2SO4 = 20 g

The molar mass of H2SO4 = 98.079 g / mol

Calculating for number of moles of H2SO4:

number of moles = 20 g / (98.079 g / mol)

number of moles = 0.204 mol

Calculating for the volume of solution using the density:

total volume of solution = 100 g / (1.105 g / mL)

total volume of solution = 90.50 mL = 0.0905 L

Therefore the molarity is:

Molarity = 0.204 mol / 0.0905 L

Molarity = 2.2542 mol / L

Molarity = 2.25 M

3 0

4 years ago