Ask question

Ask question

All categories

1 year ago

14

you wish to make a 0.161 m hydroiodic acid solution from a stock solution of 3.00 m hydroiodic acid. how much concentrated acid

must you add to obtain a total volume of 100 ml of the dilute solution?

1 answer:

mestny [16]1 year ago

3 0

5.367 ml of the concentrated acid must be added to obtain a total volume of 100 ml of the dilute solution.

Dilution is defined as the process in which the concentration of a sample is decreased by adding more solvent. The dilution formula is given below.

C₁V₁ = C₂V₂

where C₁ = initial concentration of sample = 3.00 m

V₁ = initial volume of sample

C₂ = final concentration after dilution = 0.161 m

V₂ = total final volume after dilution = 100 ml

Plug in the values to the formula and solve for the volume of the concentrated acid that must be added.

C₁V₁ = C₂V₂

3.00 m (V₁) = 0.161 m (100 ml)

V₁ = 5.367 ml

Learn more about dilution here: brainly.com/question/1615979

#SPJ4

You might be interested in

How many chlorine atoms are found in 8.3 moles of chlorine?

erik [133]

Answer:

5*10²⁴ chlorine atoms are found in 8.3 moles of chlorine.

Explanation:

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023*10²³ particles per mole. Avogadro's number represents a quantity without an associated physical dimension, so it is considered a pure number that allows describing a physical characteristic without an explicit dimension or unit of expression. Avogadro's number applies to any substance.

Then you can apply the following rule of three: if 1 mole of the compound contains 6.023 * 10²³ atoms, 8.3 moles of the compound how many atoms does it have?

$amount of atoms=\frac{8.3 moles*6.023*10^{23}atoms }{1 mole}$

amount of atoms≅ 5*10²⁴ atoms

<u><em>5*10²⁴ chlorine atoms are found in 8.3 moles of chlorine.</em></u>

4 0

2 years ago

Can selicone conduct electricity and heat

tatuchka [14]

If you meant the word silicon then yes, silicon is a semiconductor and its ability to conduct gets better as the temp. rises

7 0

3 years ago

Read 2 more answers

How many neutrons are in a single atom of indium-115?

anzhelika [568]

Indium has 49 protons

mass number (# of neutrons and # of protons combined) is 115

115 - 49 = 66

66 neutrons

8 0

3 years ago

Two or more compounds that have the same molecular formula, but different arrangements of their atoms, are called_____ isomers.

dusya [7]

Structural isomers, as the structure of the molecule is different

6 0

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