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29 days ago

4. Which of the following statements correctly describe atoms? Select all that apply.

Chemistry

1 answer:

castortr0y [2.7K]29 days ago

6 0

Answer:

A, B, and C

Explanation:

Indeed, atoms possess mass and serve as the fundamental building blocks of chemical elements. While matter is composed of atoms, these particles themselves do not occupy physical space.

Atoms consist mostly of void, which excludes them from the other responses.

This confirms that A, B, and C are the right choices.

You might be interested in

Which of the following air pollutants is correctly paired with one of its major effects?

alisha [2704]

Response:

The accurate choice is;

Sulfur oxides linked to acid precipitation

Details:

Sulfur oxide in the atmosphere interacts with oxygen, water, and other chemicals leading to the creation of acidic precipitation known as acid rain.

Sulfur oxides react with moisture in clouds to generate sulfuric acid as follows;

The sulfur gas undergoes initial oxidation

SO₂ + OH → HOSO₂

The subsequent step involves producing sulfur trioxide

HOSO₂ + O₂ → HO₂ + SO₃

Finally, sulfur trioxide reacts with water to yield sulfuric acid

SO₃ + H₂O → H₂SO₄ (aq).

3 0

16 days ago

What percent, by mass, is Oxygen in the compound Fe(OH)3?

Anarel [2600]

Answer:

Oxygen's mass percent in Fe(OH)3 is 44.92%

Explanation: The mass percentage is a means of indicating the concentration of a specific element within a compound. It is determined through the ratio of the element's mass to the compound's total mass, multiplied by 100.

•First calculate the overall mass of the compound

•Fe's molar mass = 55.85 g/mol

•O's molar mass = 16 g/mol

•H's molar mass = 1 g/mol

Using these values, we can compute the molecular mass of Fe(OH)3 = 55.85 g/mol + (16 g/mol)3 + (1 g/mol)3

=55.85 g/mol + 48 g/mol + 3 g/mol

=106.85 g/mol

Mass percent of an element = mass of element/total mass of compound × 100

In the case of 3 oxygen atoms present within the compound, the mass of oxygen totals 48 g/mol

Mass percent of oxygen= 48 g/mol/106.85 g/mol × 100

= 0.4492×100= 44.92%

[[TAG_31]]Thus, the mass percent of oxygen in Fe(OH)3 amounts to 44.92%[[TAG_32]]

7 0

11 days ago

A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press

Tems11 [2400]

Answer: The vapor pressure of naphthalene within the flask remains at $2.906\times 10^{-4}$ atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

$C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)$

The formula employed to determine the enthalpy change for the reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

$\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})$

The provided information includes:

$\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol$

Substituting the values into the previous equation produces:

$\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol$

The formula utilized to compute Gibbs free energy change is of a reaction:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f(product)]-\sum [n\times \Delta G^o_f(reactant)]$

The equation for the enthalpy change for the reaction is:

$\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})$

The given factors include:

$\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol$

By inserting values from the above equation, we arrive at:

$\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol$

For the calculation of $K_1$ (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

$\Delta G^o=-RT\ln K_1$

where,

$\Delta G^o$ = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol (Conversion factor: 1kJ = 1000J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_1$ = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

$22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}$

To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

$\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$K_2$ = Equilibrium constant at 35°C =?

$K_1$ = Equilibrium constant at 25°C = $1.14\times 10^{-4}$

$\Delta H$ = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

R = Gas constant = $8.314J/K mol$

$T_1$ = Initial temperature = $25^oC=[273+25]K=298K$

$T_2$ = Final temperature = $35^oC=[273+35]K=308K$

By plugging values into the equation above, we obtain:

$\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}$

In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for $K_p$ , which is:

$K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)$

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for $K_2$ will equal $K_p$

$p_{C_{10}H_8}=2.906\times 10^{-4}$

Consequently, the partial pressure of naphthalene at 35°C is $2.906\times 10^{-4}$ atm.

3 0

8 days ago

How many grams of NaC2H3O2 are needed to prepare 350. mL of a 2.75 M solution? (molar mass of

eduard [2509]

Answer:

78.96 g of NaC2H3O2

Explanation:

The following information is provided:

The solution's volume is 350 mL
The solution's molarity is 2.75 M
The molar mass of NaC2H3O2 is 82.04 g/mol

We need to find the mass of the solute:

First, we calculate the number of moles:

Moles = Molarity × Volume

Thus;

Moles of solute = 2.75 M × 0.350 L

= 0.9625 moles

Next, we find the mass:

Mass = Moles × Molar mass

= 0.9625 moles × 82.04 g/mol

= 78.9635 g

= 78.96 g

Therefore, the amount of NaC2H3O2 required is 78.96 g

4 0

26 days ago

You have a balloon filled with hydrogen gas which keeps it at a

lorasvet [2515]

The resulting temperature is 46.5°C.

Details:

According to Charles's law, the volume of gas, while maintaining constant pressure, correlates directly with temperature in Kelvin.

The formula representing Charles's law is expressed as follows:

$$\frac{V}{T} = constant$

$$\frac{V1}{T1} = \frac{V2}{T2}$

We need to determine T2, thus:

$$T2 = \frac{V2T1}{V1}$

V1 = 736 ml = 0.736 L

T1 = 15 ° C

V2 = 2.28 L

Substituting the values gives us:

T2 = $$\frac{2.28 \times 15}{0.736}$

= 46.5°C

It is evident that as the volume increases, the temperature also rises.

5 0

24 days ago