Hypochlorous acid decays in the presence of ultraviolet radiation. Assume that degradation occurs accord- ing to first-order kinetics and the rate of degradation was measured to be 0.12 day−1 (at a particular sun- light intensity and temperature). Given this, how long does it take for the concentration of hypochlorous acid to reach nondetectable levels (0.05 mg · L−1) if the initial concentration were 3.65 mg · L−1?

Answer:

D. the minerals they contain

Hope this answer is right!!

Answer:

0.189 g

Explanation:

Step 1: Write the balanced equation

NCl₃ + 3 H₂O ⇒ NH₃ + 3 HCIO

Step 2: Calculate the moles corresponding to 1.33 g of NCl₃

The molar mass of NCl₃ is 120.36 g/mol.

1.33 g × 1 mol/120.36 g = 0.0111 mol

Step 3: Calculate the moles of NH₃ produced from 0.0111 moles of NCl₃

The molar ratio of NCl₃ to NH₃ is 1:1. The moles of NH₃ produced are 1/1 × 0.0111 mol = 0.0111 mol.

Step 4: Calculate the mass corresponding to 0.0111 moles of NH₃

The molar mass of NH₃ is 17.03 g/mol.

0.0111 mol × 17.03 g/mol = 0.189 g

Answer:

Group 1A: alkali metals, or lithium family.

Group 2A: alkaline earth metals, or beryllium family.

Group 7A: the manganese family.

Group 8A: the iron family.

Explanation:

Answer:

1A: Alkali Metals

2A: Alkaline Earth Metals

7A: Halogens

8A: Noble Gases

Answer:

SI

Explanation:

I would say silicon because it is also another metalloid. Boron is a metalloid.

Answer:

Chlorine gas reacts with potassium bromide to form potassium chloride in solution and liquid bromine.

Answer:

DDDDD

Explanation:

Answer: The mass of [tex]CO_2[/tex] produced is 12.32 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.  The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of ethane = 4.21 g

Molar mass of ethane = 30 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of ethane}=\frac{4.21g}{30g/mol}=0.140mol[/tex]

Given mass of oxygen gas = 31.9 g

Molar mass of oxygen gas= 32 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of oxygen gas}=\frac{31.9g}{32g/mol}=0.997mol[/tex]

The chemical equation for the combustion of ethane follows:

[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]

By stoichiometry of the reaction:

If 2 moles of ethane reacts with 7 moles of oxygen gas  

So, 0.140 moles of ethane will react with = [tex]\frac{7}{2}\times 0.140=0.49mol[/tex] of oxygen gas

As the given amount of oxygen gas is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, ethane is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 2 moles of ethane produces 4 moles of [tex]CO_2[/tex]

So, 0.140 moles of ethane will produce = [tex]\frac{4}{2}\times 0.140=0.28mol[/tex] of [tex]CO_2[/tex]

We know, molar mass of [tex]CO_2[/tex] = 44 g/mol

Putting values in above equation, we get:

[tex]\text{Mass of }CO_2=(0.28mol\times 44g/mol)=12.32g[/tex]

Hence, the mass of [tex]CO_2[/tex] produced is 12.32 g

Solution :

The equation is :

[tex]$HA (aq) + NaOH(aq) \rightleftharpoons NaA(aq) + H_2O(l)$[/tex]

The number of the moles of HA os 0.00285, and the volume is 25 mL.

15 mL of the 0.0950 M NaOH is added.

The total volume of a solution is V = 25 mL + 15  mL = 40 mL

The pH of the solution is 6.50

Calculating the [tex]K_a[/tex] of HA

[tex]$HA(aq) \rightleftharpoons A^-(aq)+H^+$[/tex]

[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]

Let s calculate the concentration of HA and NaOH

[tex]$[HA] = \frac{^nH_A}{V}$[/tex]

        [tex]$=\frac{0.00285 \ mol}{0.04 \ L}$[/tex]

       = 0.07125 M

[tex]$[NaOH]= \frac{0.015L \times 0.0950 M}{V}$[/tex]

            [tex]$=\frac{0.001425 mol}{0.04L}$[/tex]

           = 0.0356 M

                                      [tex]$HA(aq) \ \ + \ \ NaOH(aq) \ \ \rightleftharpoons NaA(aq) \\ + \ \ H_2O(aq)$[/tex]

Initial conc. (M)            0.07125 M       0.0356 M            0 M

Change in conc. (M)   -0.0356 M       -0.0356 M        + 0.0356 M

Equilibrium conc. (M)   0.03565 M        0 M                0.0356 M

Therefore, the concentration of HA and the NaA at the equilibrium are [HA] = 0.03565 M and [NaA]= 0.0356 M

0.0356 M of NaA dissociates completely into 0.0356 M [tex]Na^+[/tex] and 0.0356 M [tex]A^-[/tex]

Now for [tex][H^+][/tex]

[tex]$[H^+] = 10^{-pH}$[/tex]

       [tex]$=10^{-6.5}$[/tex]

       [tex]$=3.16 \times 10^{-7}$[/tex]

Calculating the value of [tex]K_a[/tex],

[tex]K_a=\frac{[A^-].[H^+]}{[HA]}[/tex]

     [tex]$=\frac{0.0356 \times 3.16 \times 10^{-7}}{0.03565}$[/tex]

     [tex]$=3.16\times 10^{-7}$[/tex]

Therefore the the value of [tex]K_a[/tex] for the unknown acid is [tex]$3.16\times 10^{-7}$[/tex].

Answer: 26 moles of [tex]PH_3[/tex] are required for the reaction.

Explanation:

We are given:

Moles of [tex]P_4O_{10}[/tex] = 6.5 moles

The given chemical reaction follows:

[tex]4PH_3(g)+8O_2(g)\rightarrow P_4O_{10}(s)+6H_2O(g)[/tex]

By the stoichiometry of the reaction:

If 1 mole of [tex]P_4O_{10}[/tex] is produced by 4 moles of [tex]PH_3[/tex]

So, 6.5 moles of [tex]P_4O_{10}[/tex] will be produced by = [tex]\frac{4}{1}\times 6.5=26mol[/tex] of [tex]PH_3[/tex]

Hence, 26 moles of [tex]PH_3[/tex] are required for the reaction.

Answer:

B

Explanation:

So, a pot of boliling is hot right? of course, since it is hot thermal energy will be transferred from one place to another. I don't know if this is correct but I just wanted to give it a try.

Answer:

3.59x10⁻⁴ mol

Explanation:

Assuming ideal behaviour we can solve this problem by using the PV=nRT formula, where:

We input the data given by the problem:

And solve for n:

Answer:

36g

Explanation:

you need to know the equation mass=moles*mr (in this case mr of carbon which is 12)

so 3*12=36g

hope this helps :)

Answer:

I think A should be the answer because oxygen is the chemical change of carbon.

Answer: c potassium.

Explanation:

potassium is the most reactive metal among the given options.

D is absolutely wrong.

Answer:

[tex]v_{solute}=160mL[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the volume of methyl alcohol solute by using the definition of by-volume percentage:

[tex]\%v=\frac{v_{solute}}{v_{solution}} *100\%[/tex]

Whereas we solve for the volume of the solute as shown below:

[tex]v_{solute}=\frac{\%v*v_{solution}}{100\%} \\\\v_{solute}=\frac{80\%*200mL}{100\%}\\\\ v_{solute}=160mL[/tex]

Regards!

Answer:

the answer is b.CH3NO2 I guess I'm correct

Answer:

Sound travels through the air in approximately 332 metres per second?

Explanation:

is this what you are looking for>

Answer:

The maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water = 94.9%

Note: The question is incomplete. A similar but complete question is given below:

The solubility of benzoic acid in water is 6.80g per 100mL at 100 degrees C and 0.34 g per 100mL at 25 degrees C.

Calculate the maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water, assuming the solution is filtered at 25 degrees C.

Explanation:

Solubility of benzoic acid in water at 100 degrees C = 6.80g per 100mL

Solubility of benzoic acid in water  at 25 degrees C =  0.34 g per 100mL

Mass of benzoic acid to be theoretically recovered from 100 mL of water = 6.80 g - 0.34 g = 6.46 g

At 25 degrees;

0.34 g of benzoic acid is present in 100 mL of water

x g of  benzoic acid will be present in 15 mL of water

x = 0.34 × 15 / 100 = 0.051 g

Mass of benzoic acid to be theoretically recovered from 25 mL of water = 1.00 g - 0.051 g = 0.949 g

Maximum theoretical percent recovery = (mass recovered / original mass dissolved) x 100%

Maximum theoretical percent recovery =  (0.949 / 1.00) × 100% = 94.9 %

Therefore, the maximum theoretical percent recovery from the recrystallization of 1.00 g of benzoic acid from 15 mL of water = 94.9%

Answer:

The answer is C... I am almost positive.

The question is incomplete, the complete question is:

When aqueous solutions of NaCl and [tex]Pb(NO_3)_2[/tex] are mixed, a solid forms. Determine the mass of solid formed when 140.7 mL of 0.1000 M NaCl is mixed with an excess of an aqueous solution of

Answer: The mass of lead chloride produced is 1.96 g

Explanation:

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{ \text{Volume of solution (mL)}}[/tex] .....(1)

Given values:

Molarity of NaCl = 0.1000 M

Volume of the solution = 140.7 mL

Putting values in equation 1, we get:

[tex]0.1000=\frac{\text{Moles of NaCl}\times 1000}{140.7}\\\\\text{Moles of NaCl}=\frac{0.1000\times 140.7}{1000}=0.01407mol[/tex]

The chemical equation for the reaction of NaCl and lead nitrate follows:

[tex]Pb(NO_3)_2(aq)+2NaCl(aq)\rightarrow PbCl_2(s)+2NaNO_3(aq)[/tex]

By the stoichiometry of the reaction:

If 2 moles of NaCl produces 1 mole of lead chloride

So, 0.01407 moles of NaCl will produce = [tex]\frac{1}{2}\times 0.01407=0.007035mol[/tex] of lead chloride

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(2)

Molar mass of lead chloride = 278.1 g/mol

Plugging values in equation 2:

[tex]\text{Mass of lead chloride}=(0.007035mol\times 278.1g/mol)=1.96g[/tex]

Hence, the mass of lead chloride produced is 1.96 g

Answer:

Option 1. NO

Explanation:

The balanced equation for the reaction is given below below:

4NH₃ + 6NO —> 5N₂ + 6H₂O

From the balanced equation above,

4 moles of NH₃ reacted with 6 moles of NO.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

4 moles of NH₃ reacted with 6 moles of NO.

Therefore, 1.24 moles of NH₃ will react with = (1.24 × 6)/4 = 1.86 moles of NO

From the calculation made above, we can see that a higher amount of NO (i.e 1.86 moles) than what was given (i.e 1.79 moles) is needed to react completely with 1.24 moles of NH₃.

Therefore, NO is the limiting reactant and NH₃ is the excess reactant.

Thus, the 1st option gives the correct answer to the question

Answer:

1. NO .

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to identify the limiting reactant by simply calculating the moles of any product, say N2, via the moles of each reactant and including the corresponding mole ratio (4:5 and 6:5):

[tex]1.24molNH_3*\frac{5molN_2}{4molNH_3}=1.55molN_2 \\\\1.79molNO*\frac{5molN_2}{6molNO}=1.50molN_2[/tex]

Thus, since NO yields the fewest moles of N2 product, we infer it is the limiting reactant.

Regards!

Answer:

NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)

Explanation:

Sodium hydrogen sulfate is a strong electrolyte, that is, when dissolved in water it completely dissociates into the cation sodium and the anion hydrogen sulfate. The corresponding chemical equation is:

NaHSO₄(s) --H₂O--> Na⁺(aq) + HSO₄⁻(aq)

Answer:

Explanation:

578kj/mol

Answer:

[tex]\%m=66.7\%[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the concentration of the solution obtained, by knowing 20 mL of water are the same to 20 g and therefore the mass of the solution is 40g+20g=60g.

Next, we apply the following equation to obtain the required concentration:

[tex]\%m=\frac{40g}{60g} *100\%\\\\\%m=66.7\%[/tex]

Regards!

Answer:

95.9 kg

Explanation:

First we convert 15.0 mi² to m²:

Then we convert 27.0 ft to m:

Now we calculate the total volume of the lake:

Converting 3.20x10⁸ m³ to L:

Now we calculate the total mass of mercury in the lake, using the given concentration:

Finally we convert μg to kg:

Four quantities must be known for a complete physical description of a sample of a gas: temperature, volume, amount, and pressure. ... The height of the mercury column is proportional to barometric pressure, which is often reported in units of millimeters of mercury (mmHg), also called torr.

Answer:

Consider the reaction between an alcohol and tosyl chloride, followed by a nucleophile. Write the condensed formula of the expected main organic product.

CH3CH2CH2OH---------- 2.CI 1.TsCl,pyridine__________

Explanation:

Given alcohol is propanol.

When it reacts with TsCl, the hydrogen in -OH group is replaced with tosyl group.

Pyridine is a weak base and it neutralizes the HCl (acid) formed during the reaction.

The reaction is shown below:

Answer:

See explanation

Explanation:

Esterification is a reaction that involves the combination of an alkanoic acid and an alkanol. The product is always a sweet smelling substance.

Sulphuric acid acts as a catalyst in this reaction because it is a dehydrating agent thereby pushing the equilibrium position towards the right by the removal of water molecules.

The test tubs is heated in a water bath and not directly moved the flame because the alcohol is flammable. Also heating in a water bath helps to separate the reaction mixture from the newly formed ester.

Esters are used in industries that produces soaps and perfumes. There is a great need for the use of fragrances which are ester compounds in these industries.

The wet paper towel in the opening of the test tube cools the top of the test tube. It usually serves as a kind of condenser preventing an excess loss of vapour from the reaction mixture.

The reaction of pentanol and ethanoic acid yields pentyl ethanoate according to IUPAC nomenclature.

Answer:

option b is your right answer