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DerKrebs [107]
3 years ago
11

I kinda need this question answered fast please and thank you :)

Chemistry
1 answer:
Aloiza [94]3 years ago
6 0
D. A chemical bond

A mutual attraction between the nuclei and electrons in two different atoms is called a chemical bond, therefore the answer is D
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0.785 moles of N2,fill a balloon at 1.5 atm and 301 K.<br> What is the volume of the balloon?
Tcecarenko [31]

Answer:

V = 12.93 L

Explanation:

Given data:

Number of moles = 0.785 mol

Pressure of balloon = 1.5 atm

Temperature = 301 K

Volume of balloon = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will put the values.

V = nRT/P

V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm

V = 19.4 L /1.5

V = 12.93 L

7 0
3 years ago
How many milliliters of water are needed to prepare a 3.5M solution of NaOH if you have .5mol of the solute
Mamont248 [21]

Answer:

1335.12 mL of H2O

Explanation:

To calculate the mililiters of water that the solution needs, it is necessary to know that the volume of the solution is equal to the volume of the solute (NaOH) plus the volume of the solvent (H2O).

From the molarity formula we can first calculate the volume of the solution:

M=\frac{solute moles}{solution volume}

Solutionvolume=\frac{solute moles}{M} =\frac{5mol}{3.5\frac{mol}{L} } =1.429L

The volume of the solution as we said previously is:

Solution volume = solute volume + solvent volume

To determine the volume of the solute we first obtain the grams of NaOH through the molecular weight formula:

MW=\frac{mass}{mol}

Mass=MW*mol=39.997\frac{g}{mol} *5mol=199.985g

Now with the density of NaOH the milliliters of solute can be determined:

d=\frac{mass}{volume}

Volume=\frac{mass}{d} =\frac{199.985g}{2.13\frac{g}{mL} } =93.88mL of NaOH

Having the volume of the solution and the volume of the solute, the volume of the solvent H2O can be calculated:

Solvent volume = solution volume - solute volume

Solvent volume = 1429 mL -  93.88 mL = 1335.12 mL of H2O

7 0
3 years ago
Quiz
castortr0y [4]

Answer:

I'm assuming atmospheric pressure, since it says she is measuring pressure exerted my atmospheric gases

7 0
3 years ago
Summarize the five points of Dalton’s atomic theory, and explain at least one change that occurred to the atomic theory because
Umnica [9.8K]
The postulates of Dalton's theory were:
1) Elements are made of extremely small particles called atoms
2) <span>Atoms of a given element are identical in size, mass, and other properties
</span>3) <span>Atoms cannot be subdivided, created, or destroyed</span>
4) Atoms combine in whole number ratio to form compounds
5) Chemical reactions are the rearrangement of atoms
The third postulate has been disproved by modern science, in which the atom has been split and been subdivided into smaller parts such as the neutron, proton and electron, which are further subdivided into quarks, gluons, and kaons.
The second postulate was also disproved upon the discovery of isotopes.
4 0
3 years ago
Read 2 more answers
Consider the reaction FeO (S) + CO(g) &lt;-----&gt; Fe(s) + CO2(g) for which KP is found to have the following values:
Svetach [21]

Explanation:

\Delta G^o=-RT\ln K_1

where,

R = Gas constant = 8.314J/K mol

T = temperature = 600^oC=[273.15+600]K=873.15 K

K_p = equilibrium constant at 600°C =  0.900

Putting values in above equation, we get:

\Delta G^o=-(8.314J/Kmol)\times 873.15 K\times \ln (0.900 )

\Delta G^o=764.85 J/mol

The ΔG° of the reaction at 764.85 J/mol is 764.85 J/mol.

Equilibrium constant at 600°C = K_1=0.900

Equilibrium constant at 1000°C = K_2=0.396

T_1=[273.15+600]K=873.15 K

T_2=[273.15+1000]K=1273.15 K

\ln \frac{K_2}{K_1}=\frac{\Delta H^o}{R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

\ln \frac{0.396}{0.900}=\frac{\Delta H^o}{8.314 J/mol K}\times [\frac{1}{873.15 K}-\frac{1}{1273.15 K}]

\Delta H^o=-18,969.30 J/mol

The ΔH° of the reaction at 600 C is -18,969.30 J/mol.

ΔG° = ΔH° - TΔS°

764.85 J/mol = -18,969.30 J/mol - 873.15 K × ΔS°

ΔS° = -22.60 J/K mol

The ΔS° of the reaction at 600 C is -22.60 J/K mol.

FeO (s) + CO(g)\rightleftharpoons Fe(s) + CO_2(g)

Partial pressure of carbon dioxide = p_1=P\times \chi_1

Partial pressure of carbon monoxide = p_2=P\times \chi_2

Where \chi_1\& \chi_2 mole fraction of carbon dioxide and carbon monoxide gas.

The expression of K_p is given by:

K_p=\frac{p_1}{p_2}=\frac{P\times \chi_1}{P\times \chi_2}

0.900=\frac{\chi_1}{\chi_2}

\chi_1=0.900\times \chi_2

\chi_1+\chi_2=1

0.9\chi_2+\chi_2=1

1.9\chi_2=1

\chi_2=\frac{1}{1.9}=0.526

\chi_1=1-\chi_2=1-0.526=0.474

Mole fraction of carbon dioxide at 600°C is 0.474.

6 0
3 years ago
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