All categories

3 years ago

Hello:) is hydrogen chloride and hydrochloride acid the same??

Chemistry

2 answers:

Ray Of Light [21]3 years ago

6 0

Answer:

No because hydrogen chloride is gas and hydrochloric acid is (as stated in the name) an acid. The formula for hydrogen chloride is HCI(g), the little g states that the compound is a gas and the formula for hydrochloric acid is HCl(aq) and the little aq means aqueous.

grin007 [14]3 years ago

6 0

Answer:

no. they are not the same

Explanation: Although, they both do have the same chemical formula HCL, they differ in their states of matter. Hydrogen chloride is a gas whereas hydrochloric acid is a solution of hydrogen chloride gas and water( creates an aquaeous solution). Hydrochloric acid is highly corrossive.

You might be interested in

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

hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

jenyasd209 [6]

Answer:

hiiiiiiiiiiiiiiiiiiiiiiii

6 0

4 years ago

Read 2 more answers

Bleach contains the active ingredient NaClO. Analysis of bleach involves two sequential redox reactions: First, bleach is reacte

Alisiya [41]

Answer:

0,31%

Explanation:

For the reaction:

I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻

0,043 L × 0,117 M of sodium tiosulfate = 5,031x10⁻³ moles of S₂O₃²⁻

5,031x10⁻³ moles of S₂O₃²⁻ × $\frac{1 I_2 mol}{2 S_{2}O_3 mol^{2-}}$ = 2,5156x10⁻³ moles of I₂

These moles of I₂ were produced from:

ClO⁻⁻ + 2 H⁺ + 2 I⁻ → I₂ + Cl⁻ + H2O

2,5156x10⁻³ moles of I₂ ≡ moles of NaClO

2,5156x10⁻³ moles of NaClO × $\frac{74,44 g}{1mol}$ = 0,187 g of NaClO

Thus, percentage composition by mass is:

$\frac{0,187 g of NaClO}{60 g Of Bleach} x 100$ = 0,31%

I hope it helps!

5 0

3 years ago

How many moles are in 25 grams of dihydrogen monoxide

timurjin [86]

If you'd like the full working, here it is:

I calculated this by using the formula triangle.

Mass

Number Formula
Of moles Mass

To calculate the number if moles in a substance, you need to divide the Mass by the Formula mass. You get the formula mass by adding the atomic masses of the elements in the compound together. In this situation, H2O, it would be two hydrogen molecules plus one oxygen molecule which is 2 + 16. This is because the atomic mass of Hydrogen is 1 and the atomic mass of Oxygen is 16.

Now that we have the Formula mass we can go ahead and do the calculation since we already have the Mass. You do as follows:

Mass divided by Formula mass which is in this case - 25 divided by 18

By doing this calculation you will get the answer which is 1.38 moles which can be rounded to 1.4

Hope this helps :)

6 0

3 years ago

Read 2 more answers

Iron (III) oxide - what does the III mean?

Aleonysh [2.5K]

Answer:

The iron is in the +3 oxidation state, which is what the III means.

7 0

3 years ago