What is the major organic product obtained from the reaction of 2,2-dimethyl-1-propanol aqueous HBr at reflux

Answer:

The polyps get most of their food from the algae.

Explanation:

Answer:

The answer is barium phosphate

Answer: 0.0450 moles of [tex]CaCl_2[/tex]

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

To calculate the moles, we use the equation:

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

[tex]\text{Number of moles of} CaCl_2=\frac{5.00g}{110.98g/mol}=0.0450moles[/tex]

Thus there are 0.0450 moles of [tex]CaCl_2[/tex]

Answer:

i dunno

Explanation:

Balanced Equation: 2 KCIO3 ⇒ 2 KCl + 3 O2

3 mol KClO3 x [tex]\frac{3 mol O2}{2 mol KClO3}[/tex] = 4.5 mol O2

Answer:

option B is correct answer of this question

it is. Evaporation

Answer:

It's filtration!

Explanation:

Filtration is a physical chemical operation that separates solid matter and fluid from a mixture!

Answer:

I hope it help

your welcome

Answer:

344.1g

Explanation:

Answer: The empirical formula is [tex]C_{17}H_{19}O_3N[/tex]

Explanation:

Mass of C= 17.900 g

Mass of H = 1.680 g

Mass of O = 4.225 g

Mass of N = 1.228 g

Step 1 : convert given masses into moles.

Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{17.990g}{12g/mole}=1.5moles[/tex]

Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.680g}{1g/mole}=1.680moles[/tex]

Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{4.225g}{16g/mole}=0.264moles[/tex]

Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{1.228g}{14g/mole}=0.087moles[/tex]

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

For C = [tex]\frac{1.5}{0.087}=17[/tex]

For H = [tex]\frac{1.680}{0.087}=19[/tex]

For O =[tex]\frac{0.264}{0.087}=3[/tex]

For N = [tex]\frac{0.087}{0.087}=1[/tex]

The ratio of C : H: O: N = 17: 19: 3: 1

Hence the empirical formula is [tex]C_{17}H_{19}O_3N[/tex]

Answer:

Frogs

Explanation:

Answer:

Hello there! I'm assuming frog and or dog :)

Answer:

Complex II

Explanation:

Complex II does not directly pump protons but rather sends two protons on to Complex III in the form of the reduced UQ known as ubiquinol.

Thus, the correct answer is Complex II

Answer:

42.9 h

Explanation:

From the question given above, the following data were obtained:

Original amount (N₀) = 100%

Amount remaining (N) = 25%

Time (t) = 85.8 h

Half-life (t½) =?

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Original amount (N₀) = 100%

Amount remaining (N) = 25%

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

25 = 1/2ⁿ × 100

Cross multiply

25 × 2ⁿ = 100

Divide both side by 25

2ⁿ = 100/25

2ⁿ = 4

Express 4 index form with 2 as the base

2ⁿ = 2²

n = 2

Thus, two half-lives has elapsed.

Finally, we shall determine the half-life of the of the isotope. This can be obtained as follow:

Time (t) = 85.8 h

Number of half-lives (n) = 2

Half-life (t½) =?

n = t / t½

2 = 85.8 / t½

Cross multiply

2 × t½ = 85.8

Divide both side by 2

t½ = 85.8 / 2

t½ = 42.9 h

Thus, the half-life of the of the isotope is 42.9 h

Answer:

1 qualiative physical  

2 qualiative   chemical

3 quantitative physcial

4 quantiative chemical

5qualiative   physcial

6quatiative    chemcial

7quanatiative physcial

8 qualiative   chemical

9quatliate  physcial

Mole measure the number of elementary entities of a given substance that are present in a given sample. Therefore, 1.98×10²²molecules of aspartame are present in 10.00 grams of aspartame.

The SI unit of amount of substance in chemistry is mole. The mole is used to measure the quantity or amount of substance. We know one mole of any element contains 6.022×10²³ atoms which is also called Avogadro number.

Mathematically,

mole =given mass ÷ molar mass

       =10.00  ÷ 294.30

       =0.033moles

number of molecules= Avogadro's number × number of moles

                                   =6.022×10²³× 0.033

                                 = 1.98×10²²molecules

Therefore, 1.98×10²²molecules of aspartame are present in 10.00 grams of aspartame.

To know more about mole, here:

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Answer:

Step 1 should be convert atoms to moles (n). Step 2 should be convert moles (n) to mass (m).

Step 1

Use dimensional analysis to convert the number of atoms to moles.

1 mole atoms = 6.022 × 10²³ atoms

n(Ag) = 2.3 × 10²⁴ Ag atoms × (1 mol Ag/6.022 × 10²³ Ag atoms) = 3.8193 mol Ag

Step 2

Convert the moles of Ag to mass.

mass (m) = moles (n) × molar mass (M)

n(Ag) = 3.8193 mol Ag

M(Ag) = atomic weight on the periodic table in g/mol = 107.868 g Ag/mol Ag

m(Ag) = 3.8193 mol × 107.868 g/mol = 412 g Ag = 410 g Ag rounded to two significant figures

The mass of 2.3 × 10²⁴ Ag atoms is approximately 410 g.

Explanation:

The number of moles that are present is 2.3 × 10²⁴ atoms of silver are 3.8193 moles that can be determined using Avogadro's law.

One mole:

A mole is defined as 6.022 × 10²³ of some chemical unit, be it atoms, molecules, ions, or others.

1 mole = 6.022 × 10²³ atoms

n(Ag) = 2.3 × 10²⁴ Ag atoms × (1 mol Ag/6.022 × 10²³ Ag atoms)

n(Ag) = 3.8193 mol Ag

Thus, the number of moles is 3.8193 moles.

Find more information about one mole here:

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Answer and Explanation:

When using simulations, it is important that the researcher knows some limitations. This is because the simulations do not behave exactly like what they represent, in addition to not presenting very important details for analysis and without approximations of what really happens with what they are representing. When modeling the synthesis of Ammonia, for example, these limitations can cause inaccuracies that will compromise the entire analysis.

When using simulations, it is important that the researcher knows some limitations. This is because the simulations do not behave exactly like what they represent, in addition to not presenting very important details for analysis and without approximations of what really happens with what they are representing. When modeling the synthesis of Ammonia, for example, these limitations can cause inaccuracies that will compromise the entire analysis.

Answer:

12430mL of water must be added

Explanation:

To solve this question we need to convert the 20%w of CuSO₄ to molarity. Then, using the diulution factor we can find the amount of water required:

20g CuSO₄ / 100mL * (1mol / 159.609g CuSO₄) = 0.1253 moles / 100mL = 0.1253mol / 0.1L =

1.25M is the concentration of CuSO₄. To dilute this concentration to 0.01M, the dilution factor must be of:

1.25M / 0.01M = 125 times must be diluted the solution.

As the volume of the concentrated solution is 100mL, the total volume of the solution to have a 0.01M solution must be of:

100mL * 125 times = 12530mL is the final volume of the solution. That means the amount of water added must be of:

12530mL - 100mL =

Answer:

D

Explanation:

Answer:9.012

Explanation:

I used the periodic table, the mass is 9.012

Answer:

Pretty sure it would be 69(nice).

Explanation:

You add the number of protons and neutrons to find the mass

Answer and Explanation:

The introduced species is, literally, a species that was introduced into an environment by human action. In other words, an introduced species is one that is not native to a region, does not occur naturally, but has been taken by humans to that region.

An invasive species, on the other hand, is one that was introduced naturally in an environment, but multiplied in a harmful way, causing a strong imbalance in the region.

Answer:

the de Broglie wavelength of one of the electrons is 5.3787 × 10⁻¹¹ m

Explanation:

Given the data in the question;

we know that; ( electron volt ) 1 eV = 1.602 × 10⁻¹⁹ J

so given that Kinetic Energy KE = 520 eV

KE = 520 × 1.602 × 10⁻¹⁹ = 8.3304 × 10⁻¹⁷ J

we know that;

Kinetic Energy = 1/2 × mv²

where m is the electron mass

and mass of electron is  9.109 × 10⁻³¹ kilograms

so

8.3304 × 10⁻¹⁷ = 1/2 × 9.109 × 10⁻¹⁷× v²

v = √[(8.3304 × 10⁻¹⁷) / (1/2 × 9.109 × 10⁻³¹)]

v = 1.3524 × 10⁷ m/s

speed of the electron v = 1.3524 × 10⁷ m/s

now, de Broglie equation;

wavelength λ = h / mv

where h is Planck constant ( 6.626 × 10⁻³⁴ m² kg / s )

so we substitute

wavelength λ = 6.626 × 10⁻³⁴ / ( 9.109 × 10⁻³¹ × 1.3524 × 10⁷)

wavelength λ = 5.3787 × 10⁻¹¹ m

Therefore, the de Broglie wavelength of one of the electrons is 5.3787 × 10⁻¹¹ m

Answer:

Implement rainwater harvesting systems to collect and store rainwater for drinking or recharging underground aquifers. Build wells to extract groundwater from underground aquifers. Provide home water-treatment capability through the use of filters, solar disinfection, or flocculants, to make drinking water safe.

Explanation:

the government of the United States preserves water sources by making waste water treatment and preventing water pollution

Answer: [tex]3Ag^+(aq)+3Cl^-(aq)\rightarrow 3AgCl(s)[/tex]

Explanation:

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolytes and  are represented in the form of ions.

The complete ionic equation will be:

[tex]3Ag^+(aq)+3NO_3^-(aq)+Fe^{3+}(aq)+3Cl^-(aq)\rightarrow 3AgCl(s)+Fe^{3+}(aq)+3NO_3^-(aq)[/tex]

Net ionic equation : In the net ionic equations, we do not not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

  In this equation, [tex]Fe^{3+}[/tex] and [tex]NO_3^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be:

[tex]3Ag^+(aq)+3Cl^-(aq)\rightarrow 3AgCl(s)[/tex]

The question is incomplete, the complete question is;

Calculate the volume of concentrated HCL (12 M) needed to convert 1.5 g of sodium benzoate back to benzoic acid.

Answer:

8.33 * 10^-4 L

Explanation:

Equation of the reaction;

C6H5COONa + HCl -------->>>> C6H5COOH + NaCl

Number of moles of  sodium benzoate = mass/molar mass

molar mass of sodium benzoate =144.11 g/mol

mass of sodium benzoate = 1.5 g

Number of moles of  sodium benzoate = 1.5g/144.11 g/mol

Number of moles of  sodium benzoate = 0.01 moles

Since the reaction is 1:1,  0.01 moles of HCl reacted.

but;

n = CV

n = number of moles

C = concentration

V = volume

V = n/C

V =  0.01 moles/12 M

V = 8.33 * 10^-4 L

Answer:

New volume V2 = 2.565 L

Explanation:

Given:

Old volume V1 = 10.8 L

Old pressure P1 = 0.38 atm

New presuure P2 = 1.6 atm

Find:

New volume V2

Computation:

P1V1 = P2V2

So,

10.8 x 0.38 = V2 x 1.6

New volume V2 = 2.565 L

Answer:

0.012 M

Explanation:

Step 1: Given and required data

Step 2: Calculate the moles (n) of solute (KCl)

We will use the following expression.

n = m/M = 0.64 g / (74.553 g/mol) = 8.6 × 10⁻³ mol

Step 3: Calculate the molar concentration (C) of the solution

The molarity is equal to the moles of solute divided by the liters of solution.

C = 8.6 × 10⁻³ mol/0.725 L = 0.012 M

Answer:  153

Explanation:

californium is a synthetic radioactive element found on the bottom row of the periodic table, along with other actinides.

It was first discovered in 1950 at the University of California Radiation Laboratory in Berkeley by bombarding curium with alpha particles (helium nuclei). Pure californium metal has a silvery-white appearance and is soft enough to be easily cut with razor blade.

In initial experiments with californium in its concentrated form began in 1958.

Since then, 20 isotopes of californium have been characterize, with mass numbers ranging from 237 to 256. One of these isotopes is californium-252, which was first isolated from a neutron-irradiated sample of plutonium-239. As an exclusively synthetically made element (it cannot be found in nature),

californium-252 is very rare an is produced via neutron bombardment in high flux isotope reactor or particle accelerators.

On this,page we walk through some of the properties  and uses of his unique isotope and provide additional resources for your reference.

Hope this helps! :)