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3 years ago

2KCI -> K + CI2 balanced or unbalanced pls help

Chemistry

1 answer:

Mnenie [13.5K]3 years ago

6 0

Answer:

unbalanced I know you should know to

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Nonmetals rarely lose electrons in chemical reactions because?

ziro4ka [17]

The reason for this is because nonmetals, have close to fulfilling an octet and need to gain few more electrons to do this, not to lose more. Nonmetals, because of the fact they need only few more electrons to satisfy their octet they would receive or share electrons to do this.

The property that nonmetals have are that they are very electronegative, they possess a strong affinity to pull electron density closer, because they possess fewer electron shells and possess even protons this allows for this.

4 0

3 years ago

Calculate the ionization constant for the following acids or bases from the ionization constant of its conjugate base or conjuga

suter [353]

Answer:

7.41 × 10⁻⁵

Explanation:

Let's consider the basic dissociation reaction of trimethylamine (CH₃)N).

(CH₃)N + H₂O = (CH₃)NH⁺ + OH⁻

According to Brönsted-Lowry, in this reaction (CH₃)N is a base and (CH₃)NH⁺ is its conjugate acid. The pKb for (CH₃)N is 9.87. We can calculate the pKa of (CH₃)NH⁺ using the following expression.

pKa + pKb = 14

pKa = 14 - pKb = 14 - 9.87 = 4.13

Then, we can calculate the acid dissociation constant for (CH₃)NH⁺ using the following expression.

pKa = -log Ka

Ka = antilog - pKa = antilog -4.13 = 7.41 × 10⁻⁵

8 0

3 years ago

Balance<br> _C2H6() +<br> O2(g) →

nignag [31]

Answer:

2C2H6 + 7O2 → 4 CO2 + 6H20

Explanation:

3 0

3 years ago

How many elements are in C5H4O2S

elena-s [515]

Answer:

4 elements

Explanation:

The four elements would be:

C-Carbon

H-Hydrogen

O-Oxygen

S-Sulfur

Hope this helps :)

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3 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

3 years ago