2. What organelle in plant cells makes it possible for plants to carry out photosynthesis?
a. cell wall
b. chloroplast
C. lysosome
d. nucleus

Cyclohexene is a cyclic, six-membered hydrocarbon that contains one double bond. The types of reactions that can occur in cyclohexene would be those that are typical with alkenes generally.

The pi-bonded electrons in the double bond are nucleophilic. So, electrophilic addition reactions could occur with cyclohexene. For example,

cyclohexene + HBr → bromocyclohexane

cyclohexene + H2O/H+ → cyclohexanol

cyclohexene + Br2 → trans-1,2-dibromocyclohexane (racemic)

The latter is a common test for alkenes where one adds bromine to a sample to see if there is decolorization, which would indicate the presence of nucleophilic pi bonds. Bromine, which is dark reddish-brown, will become clear as it reacts with an alkene to form a colorless haloalkane.

Cyclohexene can also be converted to the fully saturated cyclohexane by hydrogenation: cyclohexene + H2/Pd → cyclohexane.

Explanation:

copper (II) nitrate

Illustrate variety of substances of which an element can be part:

metal --> blue solution --> blue solid --> black solid --> blue solution (again) --> metal (again).When solid copper reacts with aqueous silver nitrate, the products are copper (II) nitrate and solid silver. and carbon monoxide gas produces solid iron and carbon dioxide gas.

Answer:

Partial pressure of Ar = 100.5 KPa

Explanation:

The following data were obtained from the question:

Number of mole of Ne = 1 mole

Number of mole of Ar = 2 moles

Total pressure = 150.0 KPa

Partial pressure of Ar =?

Next, we shall determine the mole fraction of Argon (Ar). This can be obtained as follow:

Number of mole of Ne = 1 mole

Number of mole of Ar = 2 moles

Total mole = 1 + 2 = 3 moles

Mole fraction of Ar = mole of Ar / total mole

Mole fraction of Ar = 2/3

Mole fraction of Ar = 0.67 mole

Finally, we shall determine the partial pressure of argon (Ar). This can be obtained as follow:

Mole fraction of Ar = 0.67 mole

Total pressure = 150.0 KPa

Partial pressure of Ar =?

Partial pressure = mole fraction of gas × total pressure

Partial pressure of Ar = 0.67 × 150

Partial pressure of Ar = 100.5 KPa

Answer:

energy is used to break bonds in reactants and energy is released when new bonds form in products.The law of conservation of energy states that matter cannot be created or destroyed. Whether a chemical reaction absorbs or releases energy there is no overall change in the amount of energy during the reaction.

Explanation:

Sorry if im wrong

Answers:

1. firstly Start numbering from that carbon atom which would be nearest to chlorine .

2. then if u observed properly , both chlorine is attached are attached to 2nd carbon atom.

3.but in first one chlorine is on the left side

whereas in the second one chlorine right side.

p.s:

1. well, were u have number carbon three in above pic take it as C1 than C2( if u observed it is were Cl is attached to C2 ) than C3 ,C4.

2. Then similarly in second molecules where u have number C3 in above pic take it as C1 than C2 (if u observed , again Cl is attached to C2 but this time it is one on opposite sides.

now imagine if u kept first molecules on second one.

matching C1 C2 C3 C4 of first molecules with

C1 C2 C3 C4 of second molecule .

u will observed that in both of them Cl is on C2 but they in exactly opposite position

Explanation:

Enantiomers are a pair of molecules that exist in two forms that are mirror images of one another but cannot be superimposed one upon the other.

Answer:

85.952 ml [tex]Zn^2^+[/tex]  needed to titrate the excess complexing reagent .

Explanation:

Lets calculate

After addition of 80 ml of EDTA the solution becomes = 20 + 70 = 90 ml

As the number of moles of [tex]CoSO_4[/tex] =[tex]\frac{Given mass }{molar mass}[/tex]

                                                       =[tex]\frac{1.817}{155}[/tex]

                                                          =0.01172

Molarity = [tex]\frac{no. of moles}{volume of solution}[/tex]

           =[tex]\frac{0.01172}{20}[/tex]

        =0.000586 moles

Excess of EDTA = concentration of EDTA - concentration of CoSO4

                            = 0.009005 - 0.000586

                           = 0.008419 M

As M1V1 ( Excess of EDTA ) = M2V2 [tex](Zn^2^+)[/tex]

           [tex]0.008419\times100ml=0.009795\times V2[/tex]

           [tex]V2=\frac{0.008419\times100}{0.009795}[/tex]

             V2 =85.952 ml

Therefore , 85.952 ml [tex]Zn^2^+[/tex] needed to titrate the excess complexing reagent .

Answer:

34 mL

Explanation:

We'll begin by calculating the number of mole in 1.25 g of sodium carbonate, Na₂CO₃. This can be obtained as follow:

Mass of Na₂CO₃ = 1.25 g

Molar mass of Na₂CO₃ = (23×2) + 12 + (16×3)

= 46 + 12 + 48

= 106 g/mol

Mole of Na₂CO₃ =?

Mole = mass /molar mass

Mole of Na₂CO₃ = 1.25 / 106

Mole of Na₂CO₃ = 0.012 mole

Next, we shall determine the number of mole HCl needed to react with 0.012 mole of Na₂CO₃.

The equation for the reaction is given below:

Na₂CO₃ + 2HCl —> H₂CO₃ + 2NaCl

From the balanced equation above,

1 mole of Na₂CO₃ reacted with 2 moles of HCl.

Therefore, 0.012 mole of Na₂CO₃ will react with = 0.012 × 2 = 0.024 mole of HCl.

Next, we shall determine the volume of HCl required for the reaction. This is illustrated:

Mole of HCl = 0.024 mole

Molarity of HCl = 0.715 M

Volume of HCl =?

Molarity = mole /Volume

0.715 = 0.024 / volume of HCl

Cross multiply

0.715 × volume of HCl = 0.024

Divide both side by 0.715

Volume of HCl = 0.024 / 0.715

Volume of HCl = 0.034 L

Finally, we shall convert 0.034 L to mL

This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.034 L = 0.034 L × 1000 mL / 1 L

0.034 L = 34 mL

Therefore, 34 mL of HCl is needed for the reaction.

The amount of HCl required for counterbalancing 1.25 g of Na2CO3(Sodium Carbonate) would be:

- [tex]34 ml[/tex]

Given that,

Mass of Na2CO3 [tex]= 1.25 g[/tex]

To find the Moles of Na2CO3, we will find the molar mass of Na2CO3,

Molar Mass of or Na2CO3 [tex]= 106 g/mol[/tex]

So,

Moles of Na2CO3 [tex]= mass /molar mass[/tex]

[tex]= 1.25/106[/tex]

[tex]= 0.012 mol[/tex]

To determine the quantity of HCl required to display the reaction with 0.012 mol of Na2CO3

[tex]Na_{2} CO_{2} + 2HCl[/tex] → [tex]H_{2}CO_{3} + 2NaCl[/tex]

While balancing the equation, we know that [tex]0.012[/tex] × [tex]2 = 0.024 mole of HCl[/tex] is necessary to process the reaction.

Now,

As we know,

HCl moles [tex]= 0.024[/tex]

Molarity of HCl [tex]= 0.715 M[/tex]

∵ Quantity of HCl required = Moles/Molarity

[tex]= 0.024 / 0.715[/tex]

[tex]= 0.034 l[/tex] [tex]or 34ml[/tex]

Thus, 34 ml is the correct answer.

Learn more about 'Molarity' here:

brainly.com/question/12127540

Answer:

Assuming that the atmospheric pressure is standard, the temperature will rise steadily until it reaches 100°C

Answer: 1. The empirical formula is [tex]PNF_2[/tex]  

2. The molecular formula is [tex]PNF_2[/tex]

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of P = 37.32 g

Mass of N = 16.88 g

Mass of F = 45.79 g

Step 1 : convert given masses into moles.

Moles of P =[tex]\frac{\text{ given mass of P}}{\text{ molar mass of P}}= \frac{37.32g}{31g/mole}=1.20moles[/tex]

Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.88g}{14g/mole}=1.20moles[/tex]

Moles of F =[tex]\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{45.79g}{19g/mole}=2.41moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For P = [tex]\frac{1.20}{1.20}=1[/tex]

For N = [tex]\frac{1.20}{1.20}=1[/tex]

For F =[tex]\frac{2.41}{1.20}=2[/tex]

The ratio of P: N: F= 1: 1: 2  

Hence the empirical formula is [tex]PNF_2[/tex]

The empirical weight of [tex]PNF_2[/tex]= 1(31)+1(14)+2(19)= 82.98 g.

The molecular weight = 82.98 g/mole

Now we have to calculate the molecular formula.

[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{82.98}{82.98}=1[/tex]

The molecular formula will be=[tex]1\times PNF_2=PNF_2[/tex]

Answer:

Helium and hydrogen......

Answer: See explanation

Explanation:

Digital technologies refers to the electronic tools, or devices that can be used to generate data, store them or process data. e.g. mobile phones, ebooks etc.

Digital learning makes use of technology. Effective use of digital learning tools can help increase the engagement of students. This csn.aksi be used by the teachers to enhance personalized learning and also improve the lesson plans.

Digital technology can enhance collaboration, and communication with the students. The use of videos, robots, virtual classrooms, augmented reality and other technologies can bring about fun and foster inquisitiveness.

Answer: It decreases

Answer: It decreases.

Explanation: a p e x :)

Answer:

NO is the limiting reagent.  

In this reaction 0.886 mole of NO2 is produced

Explanation:

The chemical equation for this reaction is  

2NO(g) + O2(g) → 2NO2(g)

In this limiting reagent reaction, 2 moles of NO reacts with one mole of O2  to produce 2 mole of 2NO2

0.886 mole of NO * (2 mole of NO2/2 mole of NO) = 0.886 mole of NO2

0.503 mole of O2 * (1 mole of NO2/1 mole of O2) = 1.01 mole of NO2

Hence, NO is the limiting reagent.  

In this reaction 0.886 mole of NO2 is produced

Answer:

a PE teacher would be in both catagories  of heath and fitness

Explanation:

Answer:

4.2 mol NH3

Explanation:

First, balance your reaction.

N2 + 3H2 --> 2NH3

Multiply 6.3 mol H2 by the mole ratio of 2 mol NH3 for every 3 mol H2 to get moles of NH3 produced.

6.3 mol H2 • (2 mol NH3 / 3 mol H2) = 4.2 mol NH3

Answer:

Here is the answer from somewhere in the internet

Answer:  Lavoisier organized a list of the known elements of his day as four categories. Newlands was the first to organize the elements and show that properties repeated in a periodic way. Mendeleev and Meyer proposed periodic tables showing a relationship between atomic mass and elemental properties. Moseley organized the elements by atomic number instead of atomic mass.

Explanation:

Answer:

Avogadro's constant says that  

1

mole of any atom contains  

6.022

⋅

10

23

atoms. In this case you have  

3.40

⋅

10

22

atoms:

3.40

⋅

10

22

atoms

6.022

⋅

10

23

atoms

mol

=  

5.65

⋅

10

−

2

m

o

l

Step 2

The atomic mass of helium (He) will give you the weight of one mole of this molecule:  

1

mol =  

4.00

gram:

4.00

g

mol

⋅

5.65

⋅

10

−

2

mol

=  

0.226

g

So the  

3.40

⋅

10

22

helium atoms weigh  

0.226

gram

.

Explanation:

Mark as brainlyiest

Answer:

Is the question not supposed to be what is the partial pressure of Nitrogen not Sodium

Answer:

The correct answer is - 4-nitroaniline.

Explanation:

It is given that all three solid compounds salicylic acid + 4-nitroaniline + naphthalene are equal in the ratio in the mixture and then 1 gram of this mixture is dissolved in the diethyl ether and run a drop of the solution on TLC plate. This plate shows three spots.

The salicylic acid and naphthalene would stay dissolved in the diethyl ether solution due to the 4-nitroaniline could be extracted by adding aqueous acid and involve in the aqueous layer and thus spot of 4-nitroaniline would be with largest Rf value.

Answer:

hydro, water

Explanation:

Answer:

5.86mol 2Fe(OH)3 * [3 mols Na SO4 / 2 mols Fe(OH)3] = 8.79 mols sodium sulfate

Explanation:

To solve this question you need to look at the ratio of sodium sulfate to iron(III) hydroxide. The ratio is 3 to 2, meaning that for every 3 moles of Na SO4 there are also 2 moles of Fe(OH)3. So, you can multiply 5.86 by (3/2) to find the mols of sodium sulfate.

Answer:

For a): The mass percent of naphthalene in the original sample is 31.00 %.

For b): The mass percent of 3-nitroaniline in the original sample is 32.03 %.

For c): The mass percent of benzoic acid in the original sample is 29.97 %.

For d): The total percent recovered is 93.00 %.

Explanation:

Percentage by mass is defined as the ratio of the mass of a substance to the mass of the solution multiplied by 100. The formula used for this is:  

[tex]\text{Percent by mass}=\frac{\text{Mass of substance}}{\text{Mass of a solution}} \times 100[/tex]               ......(1)

a):

Mass of naphthalene = 0.483 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of naphthalene}=\frac{0.483 g}{1.558}\times 100\\\\\%\text{ mass of naphthalene}=31.00 \%[/tex]

b):

Mass of 3-nitroaniline = 0.499 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of 3-nitroaniline}=\frac{0.499 g}{1.558}\times 100\\\\\%\text{ mass of 3-nitroaniline}=32.03 \%[/tex]

c):

Mass of benzoic acid = 0.467 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of benzoic acid}=\frac{0.467 g}{1.558}\times 100\\\\\%\text{ mass of benozic acid}=29.97 \%[/tex]

d):

Total mass recovered = [0.483 + 0.499 + 0.467] = 1.449 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\text{Total percent recovered}=\frac{1.449 g}{1.558}\times 100\\\\\text{Total percent recovered}=93.00\%[/tex]

Answer:

25 minutes

Explanation:

V = Volume = [tex]90\ \text{m}^3[/tex]

[tex]C_0[/tex] = Radon concentration under steady state = 1.5 Bq/L

k = Radon decay rate = [tex]2.09\times 10^{-6}\ \text{s}^{-1}[/tex]

[tex]Q[/tex] = Venting rate = [tex]0.14\ \text{m}^3/\text{s}[/tex]

[tex]C_f[/tex] = Final concentration of radon = 0.15 Bq/L

Theoretical detention time is given by

[tex]\theta=\dfrac{V}{Q}\\\Rightarrow \theta=\dfrac{90}{0.14}\\\Rightarrow \theta=642.86\ \text{s}[/tex]

We have the relation

[tex]C_f=C_0e^{-(\dfrac{1}{\theta}+k)t}\\\Rightarrow t=\dfrac{\ln\dfrac{C_f}{C_0}}{-(\dfrac{1}{\theta}+k)}\\\Rightarrow t=\dfrac{\ln\dfrac{0.15}{1.5}}{-(\dfrac{1}{642.86}+2.09\times 10^{-6})}\\\Rightarrow t=1478.25\ \text{s}=\dfrac{1478.25}{60}=24.63\approx 25\text{minutes}[/tex]

The time taken to reach the acceptable level of concentration is 25 minutes.

Answer:

The answer is agriculture.

Explanation:

Answer: Agriculture

Explanation:

I got it right on my exam

Answer:

The incorrect or the false options are -

(B) - If NH4NO3 is added, the pH becomes more basic.

(D) - If a small amount of HCl is added, the pH becomes slightly more basic.

 (E) - If NaOH is added, the OH- ions react with the [tex]NH_3[/tex].

Explanation:

The given solution the buffer basic .

Therefore , the options A ( If a small amount of NaOH is added, the pH becomes slightly more basic.) and C ( If HCl is added, the H+ ions react with the NH3.) All other B , D , E are false .

Answer:

1.75 g HCl

Explanation:

First we convert 2.00 g of PCl₅ into moles, using its molar mass:

Then we convert PCl₅ moles into HCl moles, using the stoichiometric coefficients of the equation:

Finally we convert HCl moles into grams, using its molar mass: