I need help on this!!!!!

Answer:

Detrimental- Toxic chemicals leach out of plastic and are found in the blood and tissue of nearly all of us. Exposure to them is linked to cancers, birth defects, impaired immunity, endocrine disruption and other ailments

Beneficial- Plastic packaging helps protect and preserve goods, while reducing weight in transportation,

Answer:

No matter where that object is in the vast universe, it will have the same mass. Weight, on the other hand, is a measure of how much gravitational force is exerted on an object.

Answer:

iconic compounds is your anwser

Answer:

Iconic Compounds

Explanation:

I watched a video about it.

Have a nice Friday!

Answer:

2 sig figs.

Explanation:

Sig Fig Rules:

Any non-zero digit is a significant figure.

Any zeros between 2 non-zero digits are significant figures.

Trailing zeros after the decimal are significant figures.

Answer:

well ,if I have done my sum rights there are

14 atom in the one formula unit of aluminium carbonate.

Answer:

14 atoms

Explanation:

Answer:

Divide by 100

Explanation:

Answer:

MM/n will be less than the theoretical value

Explanation:

Provided that

The plating efficiency is proportional to the element weight

i.e.

[tex]\frac{Zn^2}{(aq)}[/tex]

That represents [tex]\frac{MM}{n\ value}[/tex]

So,

MM by N value is

= 65.39 g /mol ÷ 2

= 32.70 g/mol

Also, the experimental plating value is 0.55 that represents that it is much lesser than the normal value

So it would be concluded that the MM by N would be lower than the theoretical value

Answer:

Explanation:

In organic chemistry, the reaction between 2-butanol with TsCl and Et3N is known as the tosylation of the alcohol hydroxyl group. Alcohol is being changed to tosylate by the use of tosyl chloride under the influence of a base. Tosylation of alcohol is an example of a nucleophilic substitution reaction. From the image attached below, we will see how the reaction between 2-butanol proceed into the product by using tosyl chloride and a base(Et3N).

Answer:

220000 in/min

Explanation:

Step 1: Write out conversions

1 in ≈ 2.54 cm

60 sec = 1 min

Step 2: Set up dimensional analysis

[tex]\frac{933cm}{s} (\frac{60s}{1min} )(\frac{1in}{2.54cm} )[/tex]

Step 3: Evaluate

220039.4 in/min

Step 4: Sig figs

There are 3 sig figs

220039.4 in/min ≈ 220000 in/min

Answer:

the substance is element.

because an element is pure substance which is made of only one kind of atom.

Answer:

✔ Labor Relations Specialist

✔ Compensation and Benefits Manager

✔ Human Resources Assistant

✔ Human Resources Manager

B

D

A

C

Answer:

Helps employees and management negotiate:

✔ Labor Relations Specialist or B.)

Oversees payment of employees:

✔ Compensation and Benefits Manager or D.)

Maintains employee records, such as contact information:

✔ Human Resources Assistant or A.)

Oversees the Human Resources department of a company:

✔ Human Resources Manager or C.)

Answer:

oxygen and Carbon dioxide

Answer:

26150 years

Explanation:

First we need to determine the number of half lives -- not years -- it'll take for 100g (of any element) to decay to 3.125 g

First Half Life 100/2 = 50

Seconf Half Life 50/2 = 25

Third Half Life 25/2 = 12.5

Fourth Half Life 12.5/2 = 6.25

Fifth Half Life 6.25/2 = 3.125

So it'll take 5 half lives for 100g to decay to 3.125g.

If the Half Life of Carbon-14 is 5230 and it will take 5 half lives, to determine the number of years we just need to multiply 5*5230

5*5230 = 26150

This means that it will take 26150 years for 100g Carbon-14 to decay to 3.125g.

Hope this helps!

Answer:

1 cubic centimeter (cu cm) = 0.001 kilogram (kg)

110cm³ =110*0.001=0.11kg

Explanation:

The volume has been the space occupied by the object and substances and uses the derived units. The mass of the iron ball with a volume of 110 cubic centimeters is 0.11kg.

Volume has been the three-dimensional space that an object has been using in space. The volume can be measured differently for the various states of matter in various units. Volume is the physical parameter to measure the states of matter.

The estimation of the volume can be given by density or mass and has been expressed by the various units like liters, milliliters, cubic meters, cubic centimeters, kiloliters, cubic decimeters, gallons, etc.

The unit cubic centimeter can be converted into kilogram (kg) as:

1 cubic centimeter = 0.001 kilogram

Given,

The volume of a cannonball made of iron = 110 cubic centimeters

The unit conversion from cubic centimeter to kg is shown as:

1 cubic centimeter = 0.001 kg

110 cubic centimeters = 0.11 Kg

Therefore, 0.11 kilogram is the mass of a cannonball made of iron.

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Answer:

s, p

Explanation:

The four parts, or orbitals, on the periodic table are s, p, d, and f.

Answer:

Benzene must be kept away from flames.

Explanation:

Hello,

In this case since MSDS tables show the potential risks of chemical compounds, for benzene which which has a substantial vapor pressure and a fire rating of 3, which means that is highly flammable (very likely to get on fire), we conclude then that benzene must be kept away from flames due to its high risk. The other options are incorrect since it has 0 reactivity, it is nor fragile so it is not affected by shaking and it could be storage at standard pressures since it is a liquid at standard conditions as well.

Regards.

The graphs which use axes both labeled with quantitative values and show individual points of data include:

Thus is defined as a pictorial representation of data or variables in an organized or definite manner.

The graphs mentioned above always have each point matching with one value on the x-axis and one value on the y-axis.

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Answer:0.005

Explanation:kilo means 1000 do you divide 5 by 1000 and you get.005

Answer: 0.005 kilolitres

Explanation:

5 liters = 0.005 kilolitres

Why? Because liters - kilolitres:  ÷ 1000

Answer:

36.66%

Explanation:

Step 1: Given data

Step 2: Calculate the mass of salt

The mass of the sample is equal to the sum of the masses of the components.

m(sample) = m(iron) + m(sand) + m(salt)

m(salt) = m(sample) - m(iron) - m(sand)

m(salt) = 2.875 g - 0.660 g - 1.161 g

m(salt) = 1.054 g

Step 3: Calculate the percent of salt in the sample

We will use the following expression.

%(salt) = m(salt) / m(sample) × 100%

%(salt) = 1.054 g / 2.875 g × 100% = 36.66%

The percent of salt in the original sample containing only iron, sand, and salt is 36.66%.

Salt is a mineral made up of primarily sodium chloride.

Seawater contains a vast amount of salt.

Given,

The total mass of the sample is 2.875 g

Mass of iron is  0.660 g

Mass of sand is 1.161 g

Step 1: To find the mass of salt

The total mass of the sample = the sum of the masses of the compounds

m(sample) = m(iron) + m(sand) + m(salt)

2.875 g =  0.660 g + 1.161 g + m(salt)

m(salt) = 2.875 g - 0.660 g - 1.161 g

m(salt) = 1.054 g

Step 2: Calculate the percent of salt in the sample

[tex]\%(salt) = \dfrac{m(salt) }{m(sample)} \times 100\%\\\\\\\%(salt) = \dfrac{1.054 g }{2.875 g} \times 100\% = 36.66\%[/tex]

Thus, the percentage of salt in the sample is 36.66%.

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Answer:

The concentration of fructose-6-phosphate is 1.50x10⁻³M

Explanation:

Gibbs Free energy, ΔG°, is defined as:

ΔG° = -RTln K

Where ΔG° = 1670J/mol

R is gas constant = 8.314J/molK

T is absolute temperature (25°C + 273.15 = 298.15K)

Thus, equilibrium constant, k, is:

1670J/mol = -8.314J/molK* 298.15K ln K

-0.6737 = ln K

0.5098 = K

K of the reaction:

G6P ⇄ F6P

is:

K = 0.5098 = [F6P] / [G6P]

As [G6P] is 2.95x10⁻³M:

0.5098 = [F6P] / [2.95x10⁻³M]

[F6P] = 1.50x10⁻³M

Answer:

Copper(I) chloride, commonly called cuprous chloride, is the lower chloride of copper, with the formula CuCl. The substance is a white solid sparingly soluble in water, but very soluble in concentrated hydrochloric acid. Impure samples appear green due to the presence of copper(II) chloride.

Explanation:

Please mark me brainliest

Answer:

B. Fluids

I hope this helps!

Answer:

B

Explanation:

Answer:

The strength of a bond between two atoms increases as the number of electron pairs in the bond increases.

Explanation:

Answer:

Explanation:

% mass of Ethylene glycol =(  mass of ethylene glycol/ mass of solution )x100%

                        =  62.1 g/ (62.1  +  250 )   x 100%

                       = 24.24%

Mass percentage is a term that represents the concentration, basically mass by mass percentage. It tells about the percentage of a component in a whole component. Therefore, the mass % of ethylene glycol is  24.24%.

Mass percentage represents the the percentage of each element that is making a particular compound. Mass percentage does not have any unit as numerator and denominator contains the same unit.

Mathematically,

Percentage of mass = (component’s mass ÷ total mass) x 100%

% mass of Ethylene glycol =(  mass of ethylene glycol/ mass of solution)x100%

% mass of Ethylene glycol =  62.1 g/ (62.1  +  250 )   x 100

                                             = 24.24%

Therefore, the mass % of ethylene glycol is  24.24%.

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Answer:

Approximately [tex]0.291\; \rm M[/tex] (rounded to two significant figures.)

Explanation:

The unit of concentration [tex]\rm M[/tex] is the same as [tex]\rm mol \cdot L^{-1}[/tex] (moles per liter.) On the other hand, the volume of both the [tex]\rm NaOH[/tex] solution and the original [tex]\rm HCl[/tex] solution here are in milliliters. Convert these two volumes to liters:

Calculate the number of moles of [tex]\rm NaOH[/tex] in that [tex]0.0182\; \rm L[/tex] of [tex]0.160\; \rm M[/tex] solution:

[tex]\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}[/tex].

[tex]\rm HCl[/tex] reacts with [tex]\rm NaOH[/tex] at a one-to-one ratio:

[tex]\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)[/tex].

Coefficient ratio:

[tex]\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1[/tex].

In other words, one mole of [tex]\rm NaOH[/tex] would neutralize exactly one mole of [tex]\rm HCl[/tex]. In this titration, [tex]0.291\; \rm mol[/tex] of [tex]\rm NaOH\![/tex] was required. Therefore, the same amount of [tex]\rm HC[/tex] should be present in the original solution:

[tex]\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}[/tex].

Calculate the concentration of the original [tex]\rm HCl[/tex] solution:

[tex]\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M[/tex].

Answer:

They are more spread out.

Explanation:

The atoms and molecules in gases are much more spread out than in solids or liquids. They vibrate and move freely at high speeds. A gas will fill any container, but if the container is not sealed, the gas will escape. Gas can be compressed much more easily than a liquid or solid.

Answer:

Explanation:

Explanation:

Gas – In a gas, particles are in continual straight-line motion. The kinetic energy of the molecule is greater than the attractive force between them, thus they are much farther apart and move freely of each other. In most cases, there are essentially no attractive forces between particles

Answer:

Dissolves a chemical sample well at high temperatures.

Does not dissolve a chemical sample well at low temperatures.

Explanation:

Recrystallization involves the dissolution of the solute in a solvent at high temperature. As the solution cools, it becomes saturated and the solute crystallizes out of the solution.

A good recrystallization solvent must only dissolve the solute at high temperature and must not dissolve it at low temperature. Hence, these are the two conditions required for any good recrystallization solvent.

Answer:

Concentration of H3O⁺  [H3O⁺] = 0.864 M

Explanation:

Given that:

The mass concentration of MO = 43 g/L

The volume of MO = 7.9 mL = 7.9 × 10⁻³ L

Recall that

The mass number of MO = Mass concentration of MO × Volume of MO

The mass number of MO = (43 g/L) * (7.9 × 10⁻³ L)

The mass number of MO =  0.3397 g

number of  moles of MO = (mass number of MO) / (molar mass of MO)

number of  moles of MO = (0.3397 g) / (327.33 g/mol)

moles of MO = 0.00104 mol

The total volume = 7.9 mL + 3.13 mL + 5.49 mL + 3.43 mL

The total volume = 19.95 mL = 19.95 × 10⁻³ L

Concentration of MO [MO} =(number of moles of MO) / (total volume)

[MO] = 0.00104 mol  /  19.95 × 10⁻³ L

[MO] = 5.2130 × 10⁻⁸ M

the number of moles of H3O⁺ = molarity of HCl in the solution × the volume of HCl in solution

the number of moles of H3O⁺ = [(2.0 M) * (3.13 mL)] + [(2.0 M) * (5.49 mL)]

the number of moles of H3O⁺ = 17.24 mmol

Concentration of H3O⁺  [H3O⁺] = (the number of moles of H3O⁺) / (total volume)

Concentration of H3O⁺  [H3O⁺] = (17.24 mmol) / (19.95 mL)

Concentration of H3O⁺  [H3O⁺] = 0.864 M

The Concentration of H3O⁺  [H3O⁺] is 0.864 M

But before that following calculations should be needed.

The mass number of MO = Mass concentration of MO × Volume of MO

= (43 g/L) * (7.9 × 10⁻³ L)

=  0.3397 g

Now

number of  moles of MO = (mass number of MO) / (molar mass of MO)

= (0.3397 g) / (327.33 g/mol)

= 0.00104 mol

Now

The total volume = 7.9 mL + 3.13 mL + 5.49 mL + 3.43 mL

= 19.95 mL

= 19.95 × 10⁻³ L

Now

Concentration of MO [MO} =(number of moles of MO) / (total volume)

= 0.00104 mol  /  19.95 × 10⁻³ L

5.2130 × 10⁻⁸ M

Now

the number of moles of H3O⁺ = molarity of HCl in the solution × the volume of HCl in solution

= [(2.0 M) * (3.13 mL)] + [(2.0 M) * (5.49 mL)]

= 17.24 mmol

Now

Concentration of H3O⁺  [H3O⁺] = (the number of moles of H3O⁺) / (total volume)

= (17.24 mmol) / (19.95 mL)

= 0.864 M

Hence, The Concentration of H3O⁺  [H3O⁺] is 0.864 M

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Answer:

The full form of TLC is Thin layer chromatography which is used in separating non volatile mixtures.

Explanation:

In the context, the Rf values could not be read properly or determined properly and accurately if the solvent front flows to the top of the plate. If the solvent reaches the plate's top, then the Rf values reading will be different from the TLC plate where the solvent did not reach the plate's top.

If the solvent front reaches the top, then it is not possible to accurately determine draw the line and to determine how far the spot went. And thus it would be difficult to solve the ratio as the plate would keep advancing or developing and also the compound spot keeps rising. As a result, spots reaches higher or rises than the normal rise but the solvent front does not change, it remains same. Thus, the Rf value would be completely off.