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3 years ago

Can someone help me

Chemistry

1 answer:

iogann1982 [59]3 years ago

5 0

Answer:

i cant see it or i would

Explanation:

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Consider the following equilibrium:

Ainat [17]

Answer: The equation which is wrong is $K_p=K_c(RT)^{-5}$

Explanation:

For the given reaction:

$4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)$

The expression for $K_c\text{ and }K_p$ is given by:

$K_c=\frac{1}{[O_2]^3}$

$K_p=\frac{1}{[O_2]^3}$

The concentration of solids are taken to be 1, only concentration of gases and liquid states are taken. The pressure of only gases are taken.

Relationship between $K_p\text{ and }K_c$ is given by the expression:

$K_p=K_c\times (RT)^{\Delta n_g}$

where,

$\Delta n_g$ = number of moles of gaseous products - number of moles of gaseous reactants

R = gas constant

T= temperature

For the above reaction,

$\Delta n_g$ = number of moles of gaseous products - number of moles of gaseous reactants = 0 - 3 = -3

Hence, the expression for $K_p$ is:

$K_p=K_c\times (RT)^{-3}$

Therefore, the equation which is wrong is $K_p=K_c(RT)^{-5}$

7 0

3 years ago

7. Calculate moles of H2 needed to form 2.25 moles HCl in the reaction H2 + Cl2==>2HCl

mart [117]

Answer:

1.125 moles

Explanation:

2mole of HCl produced 1mole of H2

2.25moles of HCl will produce x moles

cross multiply

2x=™1×2.25

x= 2.25÷2

x=1.125mole

8 0

3 years ago

The surface rocks on Venus are similar to ______ rocks on Earth.

alexandr1967 [171]

D. granitic

*+.*+.*+.*+.

6 0

2 years ago

What kind of bond does these have?

RUDIKE [14]

1. Tap water has a small concentration of H+ & OH- ions as well as water molecules, hence there would be permanent dipole-permanent dipole (p.d.-p.d.) forces of attraction between the water molecules (aka H-bonds) as well as ionic bonds between the H+ & OH- ions.

2. Distilled water does not have H+ & OH- ions, hence only H-bonds exist between the water molecules.

3. There are covalent bonds between the individual sugar molecules.

4. There are ionic bonds between the Na+ & Cl- ions in NaCl.

5. There are p.d.-p.d. forces of attraction between the Na+ ions and the O2- partial ions of the water molecules as well as between the Cl- ions and the H+ partial ions of the water molecules. There are also H-bonds between the individual water molecules and ionic bonds between the Na+ & Cl- ions (although these are in much lower abundance than in unsolvated solid NaCl).

6. There are i.d.-i.d. as well as p.d.-p.d. forces of attraction between the sugar molecules and the water molecules. There are also H-bonds between the individual water molecules and covalent bonds within the sugar molecules.

7 0

2 years ago

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to

Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density $\rho_{avg}$ ; we have:

$\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$

The average atomic weight is:

$A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$

So; in terms of vanadium, the Concentration of iron is:

$C_{Fe} = 100 - C_v$

From a unit cell volume $V_c$

$V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}$

where;

$N_A$ = number of Avogadro constant.

SO; replacing $V_c$ with $a^3$ ; $\rho_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$ ; $A_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$ and

$C_{Fe}$ with $100-C_v$

Then:

$a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }$

Replacing the values; we have:

$(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }$

$2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }$

$2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})$

$\mathbf{C_v = 9.1 \ wt\%}$

4 0

2 years ago

Can someone help me ​

Can someone help me