Which of the following elements is in the same family as fluorine?
a. silicon
b. antimony
O c. iodine
O d. arsenic
e. None of these.

Answers

Answer 1

Answer:

c iodine

Explanation:

fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride


Related Questions

Find the percentage composition of each element in the compound having 9.8 grams of nitrogen,0.7 grams of hydrogen and 33.6 grams of oxygen

Answers

Answer: The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.

Explanation:

Percentage composition is defined as the ratio of mass of substance to the total mass in terms of percentage.

Percentage composition=[tex]\frac{\text {mass of the element}}{\text {Total mass of the substance}}\times 100\%[/tex]

a) [tex]{\text {percentage composition of nitrogen}}=\frac{\text {mass of nitrogen}}{\text {Total mass}}\times 100\%[/tex]

[tex]{\text {percentage composition of nitrogen}}=\frac{9.8g}{9.8+0.7+33.6}\times 100\%=22.2\%[/tex]

b) [tex]{\text {percentage composition of hydrogen}}=\frac{\text {mass of hydrogen}}{\text {Total mass}}\times 100\%[/tex]

[tex]{\text {percentage composition of hydrogen}}=\frac{0.7}{9.8+0.7+33.6}\times 100\%=1.59\%[/tex]

c) [tex]{\text {percentage composition of oxygen}}=\frac{\text {mass of oxygen}}{\text {Total mass}}\times 100\%[/tex]

[tex]{\text {percentage composition of oxygen}}=\frac{33.6}{9.8+0.7+33.6}\times 100\%=76.2\%[/tex]

The percentage composition of nitrogen , hydrogen and oxygen is 22.2 % , 1.59 % and 76.2% respectively.

What's the difference between velocity time graph and distance time graph​

Answers

Explanation:

Hi there!

I attached a photo of a unit summary that states the difference between s-t and v-t graph.

Hope this helps ;) ❤❤❤

What may be expected when K < 1.0? Choose the THREE correct statements. The concentration of one or more of the reactants is small. The concentration of one or more of the products is small. The reaction will not proceed very far to the right. The reaction will generally form more reactants than products.

Answers

Answer:

The concentration of one or more of the products is small.

The reaction will not proceed very far to the right.

The reaction will generally form more reactants than products  

Explanation:

We often write

K =[Products]/[Reactants]

Thus, if K is small

We have fewer products than reactants We have more reactants than products The position of equilibrium lies to the left

A. is wrong. Usually, if K < 1, the concentration of reactants is greater than that of the products.

Which Group is in the second column of the periodic table?
A. Noble gases
B. Halogens
C. Alkali metals
D. Alkaline earth metals

Answers

Answer:

Hey there!

That would be the alkaline earth metals.

Hope this helps :)

Answer: alkaline earth metals

Explanation:

A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?

Answers

Answer:

V2 = 17371.43ml

Explanation:

We use Boyles laws

since temperature is constant

P1V1=P2V2

760 x 400 = 17.5 x V2

304000 = 17.5 x V2

V2 = 304000/17.5

V2 = 17371.43ml

The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at  760 torrs will be 18 ml.

What is vapor pressure?

The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.

The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -

P1 V1 / T1 = P2 V2 / T1

here, P = pressure

       T = temperature

        V = volume

substituting the value in the equation,

400 ×760 / 20 = 17.5× V / 20

V = 400× 760 / 20 × 17.5 / 20

V = 18 ml

Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.

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some students believe that teachers are full of hot air. If I inhale 3.5 liters of gas at a temperature of 19 degrees Celsius and it heats to a temperature of 58 degrees celsius in my lungs. what is the new volume of the gas?

Answers

Answer:

3.97 L

Explanation:

Data obtained from the question include the following:

Initial volume (V1) = 3.5 L

Initial temperature (T1) = 19 °C

Final temperature (T2) = 58 °C

Final volume (V2) =..?

Next, we shall convert celsius temperature to Kelvin temperature. This can be done as shown below:

Temperature (K) = temperature (°C) + 273

T (K) = T (°C) + 273

Initial temperature (T1) = 19 °C

Initial temperature (T1) = 19 °C + 273 = 292 K

Final temperature (T2) = 58 °C

Final temperature (T2) = 58 °C + 273 = 331 K

Finally, we shall determine the new volume of the gas by using Charles' law equation as shown below:

Initial volume (V1) = 3.5 L

Initial temperature (T1) = 292 K

Final temperature (T2) = 331 K

Final volume (V2) =..?

V1 /T1 = V2 /T2

3.5 /292 = V2 /331

Cross multiply

292 x V2 = 3.5 x 331

Divide both side by 292

V2 = (3.5 x 331) / 292

V2 = 3.97 L

Therefore, the new volume of the gas is 3.97 L.

what is the pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4

Answers

Answer:

2

Explanation:

First, find the hydronium ion concentration of the solution with a pH of 4.

[H₃O⁺] = 10^-pH

[H₃O⁺] = 10⁻⁴

[H₃O⁺] = 1 × 10⁻⁴

Next, multiple the hydronium ion concentration by 100 to find the hydronium ion concentration of the new solution.

[H₃O⁺] = 1.0 × 10⁻⁴ × 100 = 0.01

Lastly, find the pH.

pH = -log [H₃O⁺]

pH = -log (0.01)

pH = 2

The pH of a solution that has a hydronium ion concentration 100 times greater than a solution with a pH of 4 is 2.

Hope this helps.

Which of the following reagents should be used to convert to Question 26 options: A) H2, Lindlar's catalyst B) Na, liquid NH3 C) H2 / Nickel D) H2SO4, H2O

Answers

Answer: The question is is not complete...here is the complete question.

Which of the following reagents should be used to convert 3-Hexyne to E-3-hexene

Option B.

Na, liquid NH3.

Explanation:

3-Hexyne to E-3-hexene can be converted with by using the reagent of Na, NH3 (birch reduction) and this can be done by hdrogenation of H2.

The reagents NaNH3 convert 3-Hexyne to E-3-hexene because it is a reducing agents that convert or has the ability to reduce alkynes to trans alkenes.

3 Hexyne is an alkynes and it is converted to E- 3 hexene Na and NH3.

Daniel has a sample of pure copper.its mass is 89.6 grams (g),and its volume is 10 cubic centimeters (cm3). whats the destiny of the sample.

Answers

Answer:

8.96g \ cm3

Explanation:

(89.6\ 10) (g\ cm3) = 8.96g\cm3

Resonance Structures are ways to represent the bonding in a molecule or ion when a single Lewis structure fails to describe accurately the actual electronic structure. Equivalent resonance structures occur when there are identical patterns of bonding within the molecule or ion. The actual structure is a composite, or resonance hybrid, of the equivalent contributing structures. Draw Lewis structures for thecarbonate ion and for phosphine in which the central atom obeys the octet rule. ... How many equivalent Lewis structures are necessary to describe the bonding in CO32-

Answers

Answer:

See explanation

Explanation:

A Lewis structure is also called a dot electron structure. A Lewis structure represents all the valence electrons on atoms in a molecule as dots. Lewis structures can be used to represent molecules in which the central atom obeys the octet rule as well as molecules whose central atom does not obey the octet rule.

Sometimes, one Lewis structure does not suffice in explaining the observed properties of a given chemical specie. In this case, we evoke the idea that the actual structure of the chemical specie lies somewhere between a limited number of bonding extremes called resonance or canonical structures.

The canonical structure of the carbonate ion as well as the lewis structure of phosphine is shown in the image attached to this answer.

The heat of vaporization delta Hv of dichloromethane (Ch2CL2) is 28.0 kJ/mol . Calculate the change in entropy delta S when 473 g of dichloromethane boils at 39.8 degree.

Answers

Answer:

16 J/K.mol

Explanation:

From the question,

ΔS = ΔH/T............... Equation 1

Where ΔH = Heat change, T = Temperature

But,

ΔH = n(Hv).................. Equation 2

Where n = number of  mole, Hv = heat of vaporization.

Given: Hv = 28.0 kJ/mol, n = 473/85 = 5.59 mole.

Substitute these values into equation 2

ΔH = 28/5.59

ΔH = 5.01 kJ.

Also: T = 273+39.8 = 312.8 J

Substitute into equation 1

ΔS = 5.01/312.8

ΔS = 0.016 kJ/K

ΔS = 16 J/K.mol

1L of bleach has a mass of 1,100 grams, 7.25% of the mass of bleach is NaClO, 1 mol of NaClO has a mass of 74.44 grams. What is the molarity (mol/L) of NaClO in the bleach? A.0.097 B.0.93 C. 1.07 D.79.75

Answers

Answer:

C. 1.07 M.

Explanation:

Hello,

In this case, we can define the molarity of the bleach as shown below:

[tex]M=\frac{moles_{NaClO}}{V_{solution}}[/tex]

In such a way, given the mass of bleach in a 1-L solution, we can compute the density:

[tex]\rho = \frac{1100g}{1L}=1100g/L =1.1 kg/L=1.1g/mL[/tex]

In such a way, we can use the previously computed density to compute the volume of the solution, assuming a 100-g solution given the by-mass percent:

[tex]V_{solution}=100g*\frac{1mL}{1.1g} *\frac{1L}{1000mL} =0.091L[/tex]

Afterwards, using the by-mass percent of bleach we compute the mass:

[tex]m_{NaClO}=100g*0.0725=7.25g[/tex]

And the moles:

[tex]n_{NaClO}=7.25g*\frac{1mol}{74.44g} =0.097mol[/tex]

Therefore, the molarity turns out:

[tex]M=\frac{0.097mol}{0.091L}\\ \\M=1.07M[/tex]

Thus, answer is C. 1.07 M.

Regards.

A sample of an unknown gas effuses in 11.1 min. An equal volume of H2 in the same apparatus at the same temperature and pressure effuses in 2.42 min. What is the molar mass of the unknown gas

Answers

Answer:

Molar mass of the gas is 0.0961 g/mol

Explanation:

The effusion rate of an unknown gas = 11.1 min

rate of [tex]H_{2}[/tex] effusion = 2.42 min

molar mass of hydrogen = 1 x 2 = 2 g/m

molar mas of unknown gas = ?

From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.

from

[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]

where

[tex]R_{h}[/tex] = rate of effusion of hydrogen gas

[tex]R_{g}[/tex] = rate of effusion of unknown gas

[tex]M_{h}[/tex] = molar mass of H2 gas

[tex]M_{g}[/tex] = molar mass of unknown gas

substituting values, we have

[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]

[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587

[tex]\sqrt{M_{g} }[/tex] = 0.31

[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol

The molar mass of the unknown gas will be "0.0961 g/mol".

Given:

Effusion rate of unknown gas,

[tex]R_g = 11.1 \ min[/tex]

Effusion rate of [tex]H_2[/tex],

[tex]R_h = 2.42 \ min[/tex]

Molar mass of hydrogen,

[tex]M_h = 1\times 2[/tex]

              [tex]= 2 \ g/m[/tex]

According to the Graham's law, we get

→    [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]

By substituting the values, we get

→   [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]

→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]

→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]

   [tex]\sqrt{M_g} = 0.31[/tex]

       [tex]M_g = 0.0961 \ g/mol[/tex]

Thus the above solution is right.          

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A 40.80 gram sample of copper is heated in the presence of excess sulfur. A metal sulfide is formed with a mass of 51.09 g. Determine the empirical formula of the metal sulfide.

Answers

Answer:

Cu₂S

Explanation:

From the question,

                     Cu                                 S

Mass:            40.80 g                      51.09-40.80 = 10.29 g

Mole ratio:     40.80/63.5                 10.29/32.1

                         0.64          :                0.32

Divide by the smallest,

                         0.64/0.32   :                0.32/0.32

                           2                :                  1

   Therefore,

Empirical formula = Cu₂S.

Which of the following is true regarding a voltaic (or galvanic) cell? a) It can only be used with hydrogen. b) It produces electrical current spontaneously. c) It consumes electrical current to drive a nonspontaneous chemical reaction.

Answers

Answer:

b) It produces electrical current spontaneously.

Explanation:

Cells capable of converting chemical energy to electrical energy and vice versa are termed Electrochemical cells. There are two types of electrochemical cells viz: Galvanic or Voltaic cells and Electrolytic cells. Voltaic cell is an elctrochemical cell capable of generating electrical energy from the chemical reaction occuring in it.

The voltaic cell uses spontaneous reduction-oxidation (redox) reactions to generate ions in a half cell that causes electric currents to flow. An half cell is a part of the galvanic cell where either oxidation or reduction reaction is taking place. Hence, the spontaneous production of electric currents is true about Voltaic/Galvanic cells.

Name the following alkanes, please need answer for f,g,h?!

Answers

Answer:

f is =2,2-dimethyl butane

g is = 2,2-dimethyl propane

h is = 3,3-diethyl pentane

Explanation:

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Determine the volume occupied by 10 mol of helium at
27 ° C and 82 atm

Answers

Answer:

3.00 L

Explanation:

PV = nRT

(82 atm × 101325 Pa/atm) V = (10 mol) (8.314 J/mol/K) (27 + 273) K

V = 0.00300 m³

V = 3.00 L

There are parts of a standing wave that do not move at all. These parts are called ___________.

Answers

Answer:

They are called nodes.

Explanation:

Answer:

i guess this is the ans nodes

hope this helps

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A solution of malonic acid, H2C3H2O4, was standardized by titration with 0.0990 M NaOH solution. If 20.52 mL mL of the NaOH solution is required to neutralize completely 11.13 mL of the malonic acid solution, what is the molarity of the malonic acid solution

Answers

Answer:

0.0913 M

Explanation:

We'll begin by writing the balanced equation for the reaction.

This is given below:

H2C3H2O4 + 2NaOH —> C3H2Na2O4 + 2H2O

From the balanced equation above, we obtained the following:

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 2

Data obtained from the question include:

Molarity of base, NaOH (Mb) = 0.0990 M

Volume of base, NaOH (Vb) = 20.52 mL

Volume of acid, H2C3H2O4 (Va) = 11.13 mL

Molarity of acid, H2C3H2O4 (Ma) =..?

The molarity of the acid, H2C3H2O4 can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 11.13 / 0.0990 x 20.52 = 1/2

Cross multiply

Ma x 11.13 x 2 = 0.0990 x 20.52 x 1

Divide both side by 11.13 x 2

Ma = (0.0990 x 20.52)/ (11.13 x 2)

Ma = 0.0913 M

Therefore, the molarity of malonic acid, H2C3H2O4 solution is 0.0913 M

If phosphorus (P) has 4 naturally occurring isotopes, phosphorus-29(32.7.%), phosphorus-30(48.03%), phosphorus-31(18.4%), and phosphorus-33 (0.87%), what is its average r.a.m.?

Answers

The Average atomic mass of phosphorus is 29.9.

What is Average atomic mass ?

The average atomic mass (sometimes called atomic weight) of an element is the weighted average mass of the atoms in a naturally occurring sample of the element.

Average masses are generally expressed in unified atomic mass units (u), where 1 u is equal to exactly one-twelfth the mass of a neutral atom of carbon-12.

An element can have differing numbers of neutrons in its nucleus, but it always has the same number of protons.

The versions of an element with different neutrons have different masses and are called isotopes.

The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on Earth i.e,

Average atomic mass of P = ∑(Isotope mass) (its abundance)

∴ Average atomic mass of P = (P-29 mass) (its abundance) + (P-30 mass)(its abundance) + (P-31 mass) (its abundance) + (P-33 mass) (its abundance)

Abundance of isotope = % of the isotope / 100.

∴ Average atomic mass of P = (29)(0.327) + (30)(0.4803) + (31)(0.184) + (33)(0.0087) = 29.88 a.m.u ≅ 29.9 a.m.u.

Hence , The Average atomic mass of phosphorus is 29.9.

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The ionization constant of lactic acid ch3ch(oh) co2h am acid found in the blood after strenuous exercise is 1.36×10^-4 If 20.0g of latic acid is used to make a solution with a volume of 1.00l what is the concentration of hydronium ion in the solution

Answers

Answer:

Explanation:

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

ionisation constant = 1.36 x 10⁻⁴ .

molecular weight of lactic acid = 90 g

moles of acid used = 20 / 90

= .2222

it is dissolved in one litre so molar concentration of lactic acid formed

C = .2222M

Let n be the fraction of moles ionised  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionisation constant Ka

Ka = nC x nC / C - nC

= n²C ( neglecting n in the denominator )

n² x .2222 = 1.36 x 10⁻⁴

n = 2.47  x 10⁻²

nC = 2.47  x 10⁻² x .2222

= 5.5 x 10⁻³

So concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per litre .

The concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter .

Ionization of lactic acid can be represented as:

CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻  + H⁺

Given:

ionization constant = 1.36 x 10⁻⁴

mass= 20.0 g

Now, Molecular weight of lactic acid = 90 g

[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]

It is dissolved in 1.00L so molar concentration of lactic acid formed will be

C = 0.22M

Consider "n" to be the fraction of moles ionized  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionization constant Ka

[tex]K_a =\frac{nC*nC}{C-nC}[/tex]

[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )

On substituting the values we will get:

[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]

To find the concentration of hydronium ion in the solution,

[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]

So, concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter.  

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Draw the major product(s) obtained when the following compounds are treated with bromine in the presence of iron tribromide.

a. Bromobenzene
b. ortho-Xylene
c. Benzene sulfonic acid
d. Benzaldehyde
e. meta-Nitrotoluene
f, para-Dibromobenzene
g. Nitrobenzene tert-Butylbenzene
h. Benzoic acid
i. Dibromobenzene

Answers

Answer:

The halogens are the ortho and para directing groups. Whenever they react with other benzene compounds they will attach to the ortho or para positions of the benzene ring.

Major products which are obtained by reacting these given compounds are given in attached pictures with complete reactions.      

HBr will always be the side product of the bromine reactions along with the major compound.                                      

Explanation:

What is it’s molecular formula for C5H4 if it’s molar mass is 128.17g/mol

Answers

Answer :QUESTION ①)

✔ C5H4 has a molecular molar mass of :

M(C5H4) = 5 x M(C) + 4 x M(H)

M(C5H4) = 5 x 12 + 4 x 1 M(C5H4) = 60 + 4 M(C5H4) = 64 g/mol

✔ The molecular mass of C5H4 is therefore 64 g/mol.

But, 128/64 = 2 This is double the molar mass of C5H4, this molecule has the formula 2C5H4.

Answer:

C10H8

Explanation:

I clicked on that answer, and it is correct.

Phosphorus pentafluoride, PF5, acts as a __________ during the formation of the anion PF−6. Select the correct answer below: A. Lewis acid B. Lewis base C. catalyst D. drying agent

Answers

Answer:

Lewis acid

Explanation:

In chemistry, a Lewis acid is any chemical specie that accepts a lone pair of electrons while a Lewis base is any chemical specie that donates a lone pair of electrons.

If we look at the formation of PF6^-, the process is as follows;

PF5 + F^- -----> PF6^-

We can see that PF5 accepted a lone pair of electrons from F^- making PF5 a lewis acid according to our definition above.

Hence in the formation of PF6^-, PF5 acts a Lewis acid.

Solid iron(II) oxide reacts with oxygen gas to produce solid iron(III) oxide. Balance the equation for this reaction (in lowest multiple integers). Write the unbalanced equation for this reaction.

Answers

Answer

Hello

I think the reaction is like this FeO+OFe²O³

And the balance reaction is 2Fe+OFe²O³

Explanation:

At first we should find sth that has more atoms than the other then for example we realized that we have two atoms of Fe in Fe²O³ then put 2 before FeO and now we have 2 atoms of Fe in right side and 2 atoms of Fe in left then Oxygen in FeO change to 2 atoms of Oxygen and we have an other one in right side that they become 3 atoms of Oxygen and now we have 3 atoms of Oxygen in both right and left side.

Finally our reaction balanced.

Good luck

Describe the process of scientific inquiry ?

Answers

Answer:

It usually consists of six steps: question, observation or investigation, hypothesis, experiment, analysis of data (reviewing what happened during the experiment), and conclusion. Scientific inquiry, on the other hand, is non-linear, which means it does not follow a consistent step-by-step process.

Explanation:

Hope it helps

if your acetic acid buret was still wet inside with deionized water when you filled it with acetic acid?

Answers

Answer:

The water would act as a base and would produce an undesired product of ethanol (CH3OH) through a dissociation reaction. If doing a titration reaction, it will likely yield inaccurate results.

Bayer Villiger Provide a balanced chemical equation of the reaction performed in this experiment. Use structures and compound names to show ALL reactants and products involved. Baeyer-Villiger Reaction of Acetophenone Data Results
• Moles of acetophenone used: (Show calculations) 0.020 moles (2.40g/120.151 g mol-1 =0.0199 moles)
• Moles of mCPBA used: (Show calculations) 0.036 moles_(6.25 grams/ 172.56 g.mol-1)
• Expected mass of the product: (Show calculation. Clearly show the limiting and excess reactants)

Answers

Answer:

See the explanations

Explanation:

In the Baeyer-Villiger reaction, we will produce an ester from a ketone (see the first reaction). In our case, the ketone is Acetophenone therefore phenyl acetate would be produced.

Now, for the mass calculation, we have to keep in mind that we have a reaction with a 1:1 ratio. So, if we have 0.02 moles of acetophenone and 0.036 moles of m-CPBA the limiting reagent would be the smallest value in this case acetophenone.

Additionally, if we have a 1:1 ratio and the limiting reagent is 0.02 moles of acetophenone we will have as product 0.02 of phenyl acetate, if we take into account the molar mass of phenyl acetate (136.05 g/mol), we can do the final calculation:

[tex]0.02~mol~acetophenone\frac{1~mol~phenyl acetate}{1~mol~acetophenone}\frac{136.05~g~phenyl acetate}{1~mol~phenyl acetate}=2.72~g~phenyl acetate[/tex]

I hope it helps!

Which of the following metals has a low melting point?
2 A. Rubidium
B. Potassium
C. Calcium
D. Sodium​

Answers

Answer:

Rubidium

Explanation:

Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 26.34 mL sample of 0.0100 M K I O 3 with a solution of N a 2 S 2 O 3 of unknown concentration. The endpoint was observed to occur at 15.51 mL . How many moles of K I O 3 were titrated

Answers

Answer:

0.1 M

Explanation:

The overall balanced reaction equation for the process is;

IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)

Generally, we must note that;

1 mol of IO3^- require 6 moles of S2O3^2-

Thus;

n (iodate) = n(thiosulfate)/6

C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6

Concentration of iodate C(iodate)= 0.0100 M

Volume of iodate= V(iodate)= 26.34 ml

Concentration of thiosulphate= C(thiosulfate)= the unknown

Volume of thiosulphate=V(thiosulfate)= 15.51 ml

Hence;

C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)

0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M

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