what is work and state the law of conservation mass?

Answers

Answer 1

Answer:

Work is a force causing the movement or displacement of an object

law of conservation mass:

1. Atoms cannot be created or destroyed in a chemical reaction.

2. Molecules cannot be created or destroyed in a chemical reaction.

3. Compounds cannot be created or destroyed in a chemical reaction.

4. Heat cannot be created or destroyed in a chemical reaction.


Related Questions

Use the drop-down menus to classify each of the following as an addition, substitution, elimination, or
condensation reaction.
CH3CHO+H2O → CH3OCH3
CH,CICH CI + Zn → C2H4 + ZnCl2
CH3CH3Br + OH – CH3CH3OH + Br
2CH2COOH
>>
(CH3CO)20 + H20

Answers

Answer:

CH3CHO+H2O → CH3OCH3 - addition

CH,CICH CI + Zn → C2H4 + ZnCl2 - elimination

CH3CH3Br + OH – CH3CH3OH + Br - substitution

2CH2COOH >>(CH3CO)20 + H20 - condensation

Explanation:

An addition reaction is a reaction in which a specie is added across the double bond as we can see in CH3CHO+H2O → CH3OCH3.

In an elimination reaction, a small molecule is lost from a saturated compound to form the corresponding unsaturated compound as in CH,CICH CI + Zn → C2H4 + ZnCl2

In a substitution reaction, a chemical moiety replaces another in a molecule as in; CH3CH3Br + OH – CH3CH3OH + Br .

A condensation reaction is in which two molecules are joined together to form a bigger molecule as in; 2CH2COOH >>(CH3CO)20 + H20.

Answer:

addition

elimination

substitution

condensation

Explain why each of the following names is incorrect
(a) 2,2-Dimethyl-6-ethylheptane
(b) 4-Ethyl-5,5-dimethylpentane
(c) 3-Ethyl-4,4-dimethylhexane
(d) 5,5,6-Trimethyloctane
(e) 2-Isopropyl-4-methylheptane

Answers

Explanation:

Alkanes are the hydrocarbons in which single bonds are present between the carbon atoms

The suffix for alkane hydrocarbons used is '-ane'. The rules to write the nomenclature of alkanes follows:

Select the longest possible continuous carbon chain and it will be the parent chainNumbering is done as such that the alkyl substituents are given the lowest numberIf different alkyl groups are present, they are written in an alphabetical order irrespective of their position in the carbon chainIf two or more similar alkyl groups are present, the words, di, tri, tetra, and so on are used to specify the number of times alkyl groups appearIf two or more alkyl groups are present and the branching occurs, the numbering is done which gives the minimum possible number to all the substituents

For the given options:

(a) 2,2-Dimethyl-6-ethylheptane

The longest possible carbon chain has 8 carbon atoms (prefix used is 'oct-') and not 7(prefix used is 'hept-').

Thus, the correct IUPAC name will be 2,2,6-trimethyloctane.

(b) 4-Ethyl-5,5-dimethylpentane

The longest possible carbon chain has 6 carbon atoms (prefix used is 'hex-') and not 5(prefix used is 'pent-').

Thus, the correct IUPAC name will be 4-isopropylhexane.

(c) 3-Ethyl-4,4-dimethylhexane

The substituents are not given the minimum possible number. Total sum of the number in original compound = [4 + 4 + 3] = 11

But, if both the methyl groups are placed at 3rd position, then total sum of the number = [3 + 3 + 4] = 10

Thus, the correct IUPAC name will be 4-ethyl-3,3-dimethylhexane

(d) 5,5,6-Trimethyloctane

The substituents are not given the minimum possible number. Total sum of the number in original compound = [5 + 5 + 6] = 16

But, if both the methyl groups are placed at 3rd position and 4th position, then total sum of the number = [3 + 3 + 4] = 10

Thus, the correct IUPAC name will be 3,3,4-trimethyloctane

(e) 2-Isopropyl-4-methylheptane

The longest possible carbon chain has 8 carbon atoms (prefix used is 'oct-') and not 7 (prefix used is 'hept-').

Thus, the correct IUPAC name will be 2,3,5-trimethyloctane

The structures of the compounds are attached below.

1. Express in conventional notation (no exponents) in the space provided within the
parentheses, state how many significant figures are in the number (apis, cach)
a) 3.2 X 102
b) 2.366 X 104
C) 7.30 x 101
d) 5.325 x 102

Answers

Answer:

a) 320: two significant figures.

b) 2,366: four significant figures.

c) 73.0: three significant figures.

d. 532.5: four significant figures.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to write each number by knowing we move the decimal places to the right as much as the exponent is, and also, we count every figure, even zeros, because they are to the right of the first nonzero digit:

a) 320: two significant figures because the rightmost zero is not preceded o followed by a decimal place.

b) 2,366: four significant figures.

c) 73.0: three significant figures, because the zero is followed by the decimal place.

d. 532.5: four significant figures.

Regards!

HELP PLS THIS IS SO HARD AHHHHHHH

Answers

1. The box like figure in the given image is the [BATTERY SOURCE] from where the current drawn into the circuit.

2. A string connecting positive terminal of battery to the bulb is an [ELECTRIC WIRE] through which current flows in the circuit.

3. A bubble like object in the circuit is a [BULB] which lights up when current moves through the circuit.

4. A component connected to the negative terminal of batter source is a [SWITCH].

The answer are uppercase.

I hope this helps c:

What happens to an electroscope when a negatively charged rod is brought close to the metal sphere at the top?

Answers

Answer:

When the negatively-charged rod is brought close to the electroscope, positive charges are attracted to it and negative charges are repelled away from it. The electroscope has a net neutral charge and the rubber rod has a net negative charge. If they are brought into contact, they will both take a net negative charge.

Explanation:

I looked it up

A certain liquid has a normal freezing point of and a freezing point depression constant . A solution is prepared by dissolving some glycine () in of . This solution freezes at . Calculate the mass of that was dissolved.

Answers

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of [tex]-6.4^oC[/tex] and a molal freezing point depression constant [tex]K_f= 3.96^oC.kg/mol[/tex]. A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at [tex]-13.6^oC[/tex]. Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

Answer: 129.66 g of glycine will be dissolved.

Explanation:

Depression in the freezing point is the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m[/tex]

                                 OR

[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times \frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent (g)}}[/tex]        ....(1)

where,

i = Van't Hoff factor = 1 (for non-electrolytes)

Freezing point of pure solvent = [tex]-6.4^oC[/tex]

Freezing point of solution = [tex]-13.6^oC[/tex]

[tex]K_f[/tex] = freezing point depression constant  = [tex]3.96^oC/m[/tex]

[tex]M_{solute}[/tex] = Molar mass of solute (glycine) = 75.07 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 950 g

Plugging values in equation 1:

[tex]-6.4-(-13.6)=1\times 3.96\times \frac{\text{Given mass of glycine}\times 1000}{75.07\times 950}\\\\\text{Given mass of glycine}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\\text{Given mass of glycine}=129.66g[/tex]

Hence, 129.66 g of glycine will be dissolved.

Calculate the amount of heat associated with cooling a 350.0 g aluminum bar from 70.0 oC to 25.0 oC. The specific heat of aluminum is 0.897 J/g oC. (–14,127.75 J)

Answers

Explanation:

[tex]q = 350 \times0.897 \times (70 - 25) \\ q = 14127.75[/tex]

electron affinity of lithium is -60 whereas of cesium is -45.this trend is due to... plz give me accurate answer

Answers

Electron affinity is defined as the change in energy (in kJ/mole) of a neutral atom (in the gaseous phase) when an electron is added to the atom to form a negative ion. In other words, the neutral atom's likelihood of gaining an electron.


Electron Affinity of Lithium is 59.6 kJ/mol.

Electron Affinity of Caesium is 45.5 kJ/mol.

Electron Affinity of Lithium is 59.6 kJ/mol. Electronegativity of Lithium is 0.98. ... Electron affinities are more difficult to measure than ionization energies. An atom of Lithium in the gas phase, for example, gives off energy when it gains an electron to form an ion of Lithium.

Trends

The ionization energy of the elements within a period generally increases from left to right. This is due to valence shell stability.

The ionization energy of the elements within a group generally decreases from top to bottom. This is due to electron shielding.

The noble gases possess very high ionisation energies because of their full valence shells as indicated in the graph. Note that helium has the highest ionization energy of all the elements.

2. Which of the following best represents the nucleus in a model of an atom?
a positively charged nucleus consisting of protons and neutrons
O a positively charged nucleus consisting of electrons and neutrons
a neutral nucleus consisting of protons and neutrons
a neutral nucleus consisting of electrons and neutrons

Answers

Answer:

A positively charged nucleus consisting of protons and neutrons

Explanation:

The atomic nucleus is a positively charged region located at the core of an atom that consists of positively charged protons and neutral neutrons while the negatively charged electrons make up the outer cloud (electrons are therefore not contained in the nucleus).

A 66.4 gram sample of Ba(ClO4)2 3 H2O was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, how many grams of the anhydrous compound remained?

Answers

Answer:

57.2 g

Explanation:

First we convert 66.4 grams of Ba(ClO₄)₂·3H₂O into moles, using its molar mass:

Molar mass of Ba(ClO₄)₂·3H₂O = Molar mass of Ba(ClO₄)₂ + (Molar Mass of H₂O)*3Molar mass of Ba(ClO₄)₂·3H₂O = 390.23 g/mol66.4 g ÷ 390.23 g/mol = 0.170 mol Ba(ClO₄)₂·3H₂O

0.170 moles of Ba(ClO₄)₂·3H₂O would produce 0.170 moles of 0.170 moles of Ba(ClO₄)₂. Meaning we now convert 0.170 moles of Ba(ClO₄)₂ into grams, using the molar mass of Ba(ClO₄)₂:

0.170 mol * 336.23 g/mol = 57.2 g

1 gallon =3.785 L how many liters of gasoline will fill a 10.00 tank

Answers

Answer:

37.85 L

Explanation:

3.785 x 10.00 = 37.85 L

it would take 37.85 L  to fill a 10.00 tank

(sorry if im wrong pls dont report)

(hope this helps can i plz have brainlist :D hehe)

A 1.0 g sample of propane, C3H8, was burned in calorimeter. The temperature rose from 28.5 0C to 32.0 0C and heat of combustion 10.5 kJ/g. Calculate the heat capacity of the calorimeter apparatus in kJ/0C

Answers

Answer:

A 1.0 g sample of propane, C3H8, was burned in the calorimeter.

The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.

Calculate the heat capacity of the calorimeter apparatus in kJ/0C

Explanation:

[tex]Heat of combustion = heat capacity of calorimeter * deltaT\\[/tex]

Given,

The heat of combustion = 10.5kJ/g.

[tex]deltaT = (32.0-28.5)^oC\\deltaT = 3.5^oC[/tex]

Substitute these values in the above formula to get the value of heat capacity of the calorimeter.

[tex]deltaT =heat capacity of calorimeter * (change in temperature)\\10.5kJ/g = heat capacity of calorimeter * (3.5^oC)\\\\=>heat capacity of calorimeter = \frac{10.5kJ/g}{3.5^oC} \\=>heat capacity of calorimeter = 3.0 kJ/g.^oC[/tex]

Answer:

The heat capacity of the calorimeter is [tex]3.0kJ/g.^oC.[/tex]

4-A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% X, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.

Answers

Answer:

C₁₅N₃H₅

Explanation:

Let's assume we have 240 g of the dye (1 mol), in that case we'd have:

240 g * 75.95/100 =  182.28 g of C240 g * 17.72/100 =  42.53 g of N240 g * 6.33/100 =  15.19 g of H

Now we convert the masses of each element into moles, using their respective molar masses:

182.28 g C ÷ 12 g/mol = 15.19 mol C ≅ 1542.53 g N ÷ 14 g/mol = 3.04 mol N ≅ 3 15.19 g C ÷ 1 g/mol = 15.19 mol H ≅ 15

Thus the molecular formula is C₁₅N₃H₅.

A typical dollar bill is 15.50 cm by 6.50 cm.
Calculate the surface area in square meters, square centimeters and square nanometers

Answers

Answer:

0.010075 m²100.75 cm²1.0075x10¹⁶ nm²

Explanation:

As the measurements are given to us in centimeters, let's start by calculating the surface area in square centimeters:

Area = 15.50 cm * 6.50 cm = 100.75 cm²

Now we convert 100.75 cm² to m², as follows:

100.75 cm² * [tex](\frac{1m}{100cm}) ^2[/tex] = 0.010075 m²

Finally we convert 0.010075 m² to nm², as follows:

0.010075 m² * [tex](\frac{1nm}{1x10^{-9}m}) ^2[/tex] = 1.0075x10¹⁶ nm²

You combine 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4 what is the pH of the solution

Answers

Answer:

pH = 3.68

Explanation:

We can solve this problem by using Henderson-Hasselbach's equation:

pH = pKa + log[tex]\frac{[Formate]}{[Formic Acid]}[/tex]Where pKa = -log(Ka)pKa = -log(1.8x10⁻⁴) = 3.74

Assuming we have 1 L of the buffer solution then the molar concentrations of formate and formic acid would be:

[Formate] = 0.75 mol / 1 L = 0.75 M[Formic Acid] = 0.85 mol / 1 L = 0.85 M

We now have all required data to calculate the pH:

pH = 3.74 + log[tex]\frac{0.75}{0.85}[/tex]pH = 3.68


A 200. gram sample of a salt solution contains 0.050 grams of NaCl. What is the concentration of the
solution in parts per million (ppm)?

Answers

Answer:

2.5 × 10² ppm

Explanation:

Step 1: Given data

Mass of NaCl: 0.050 gMass of the sample: 200. g

Step 2: Convert 0.050 g to μg

We will use the conversion factor 1 g = 10⁶ μg.

0.050 g × 10⁶ μg/1 g = 5.0 × 10⁴ μg

Step 3: Calculate the concentration of NaCl in ppm

The concentration of NaCl in ppm is equal to the micrograms of NaCl per gram of the sample.

5.0 × 10⁴ μg NaCl/200. g = 2.5 × 10² ppm

Answer:250 ppm

Explanation:

1 pc
8. Some chemicals, most often inorganic salts, in the laboratory have the nasty tendency to absorb water from the
atmosphere. This property is called hygroscopicity. Anhydrous (water-free) Cupric Perchlorate [Cu(C104)2] weighs
262.447g/mol, but after sitting on out on the benchtop absorbs water from the air and now weighs 370.540g/mol. How
many water molecules did our Cupric Perchlorate absorb?

Answers

I JATW THAT I NEED TO ANSWER TO ASK

2) Calculate the density of a rod of metal (in g/cm) with a mass of 196.41g, a diameter of 3.10
cm and a height of 7.55 cm. Volume cylinder = rh. Remember the radius is half the length
of the diameter.
Volume =
Density

Answers

Answer:56.98496

Explanation:

half if diameter is radius or 1.55 and double the radius is 3.10 or the length of the diameter. you have a height given, so use the radius and heaight to plug it in the circular cylinder calculator and you get 56.98496

Gold is a precious metal that is chemically unreactive. It is used mainly in jewelry, dentistry, and electronic devices. A piece of gold ingot with a mass of 301 g has a volume of 15.6 cm3. Calculate the density of gold.

Answers

Answer:

Density = 19.3 g/cm³

Explanation:

In order to answer this question we need to keep in mind the following definition of density:

Density = Mass / Volume

As both the mass and the volume are given by the problem, we can proceed to calculate the density of gold:

Density = 301 g / 15.6 cm³Density = 19.3 g/cm³

TIME REMAINING
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Scientists use different types of microscopes to observe objects that are not visible to the naked eye. A scientist is
reviewing various samples of pond water to determine what species of microorganisms live in the pond. The scientist
wishes to make drawings of the structure of each microorganism and study each one's method of movement. Which of
the following microscopes would be best for the scientist to use?
transmission electron microscope
b. scanning electron microscope
c. compound light microscope
d. dissecting microscope
a.
Please select the best answer from the choices provided
ОА
ОВ
D
Nox
Submit
Save and Exit
Mark this and retum
Sono

Answers

Answer:

compound light microscope

briefly describe the action of hardwater with soap​

Answers

Answer:

The correct answer is - hard water reacts to form the calcium or magnesium salt of the organic acid of the soap.

Explanation:

Soaps are made up of fatty acids or oils by treating with strong alkali and are salts of sodium and potassium. Hard water, on the other hand, has a high concentration of minerals in comparison to soft water. When hard water and soap are mixed together the salts of the minerals like Ca2+ and Mg2+ ions present in hard water react with fatty acids of the soap.

The sodium salts are changed to the salts of calcium and magnesium which are precipitated to an insoluble substance. The insoluble salts of the calcium or magnesium dirt stick on the clothes thus cleaning ability of soap is reduced.

If the pH is 9 what is the concentration of hydroxide ion [OH]? (hint: find the pOH first)

Answers

realize that pH + pOH = 14
so, 9 + pOH = 14 -> pOH = 5
pOH = -log[OH-]
5 = -log[OH-]
plug it into a calculator and you get 1.0 x 10^-5
alternatively, use [OH-] = 10^-pOH to get the same answer
[OH-] = 1.0 x 10^-5
the answer to this is 7

PLEASE TELL ME THE AWNSERS ITS A DOC FILE SO OPEN IT I WILL GIVE BRAINLIEST PLS HURRY

Answers

ummmmmm we cant open it

What the correct answer

Answers

Answer:

[Ar] 4s²3d³

Explanation:

Vanadium has atomic number of 23. The electronic configuration of vanadium can be written as:

V (23) => 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d³

NOTE: After the 18th electron, 4s will be filled before 3d.

We can also write the electronic configuration of an element in its condensed form by writing the symbol of the noble before the desired element in a squared bracket followed by the remaining electrons to complete the electronic configuration of the element.

The electronic configuration of vanadium in its condensed form is given below:

The noble gas before vanadium is Argon (Ar) with atomic number of 18. Thus, the electronic configuration of vanadium becomes:

V (23) => [Ar] 4s²3d³


What is the Name of molecule and smiles strings ?

Answers

Explanation:

Aromatic nitrogen bonded to hydrogen, as found in pyrrole must be represented as [nH] ; thus imidazole is written in SMILES notation as n1c[nH]cc1 . When aromatic atoms are singly bonded to each other, such as in biphenyl, a single bond must be shown explicitly: c1ccccc1-c2ccccc2 .

PLS HELP The average atomic mass of carbon is 12.01 amu. Based on the atomic
masses of the two isotopes of carbon, how do the relative abundances of the
isotopes compare?

Answers

Answer:

B. There is a very large percentage of C-12.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize that, since the average atomic mass is 12.01 amu, then the C-12, with an atomic mass of 12.000 am prevails over C-13 with an atomic mass of 13.003 amu as long as the average is nearer to the former.

In such a way, the answer will be B. There is a very large percentage of C-12.

Regards!

Iodide ion catalyzes the decomposition of hydrogen peroxide. The reaction is first-order in H2O2. What is the value of the rate constant, k, if the initial rate is 0.00842 mol/(L·s) and the initial concentration of H2O2 is 0.500 mol/L.

Answers

Answer:

0.01684 s⁻¹

Explanation:

In a first-order reaction, the rate is proportional to the concentration of only one reactant (raised by 1). In this case, the reactant is H₂O₂. Thus, the rate law is the following:

rate = k [H₂O₂]

We have the following data for the initial rate:

rate = 0.00842 mol/(L·s)

[H₂O₂] = 0.500 mol/L

So, we introduce the data in the expression for the rate law to calculate k:

k = rate/[H₂O₂] = (0.00842 mol/L·s)/0.500 mol/L = 0.01684 s⁻¹

g There are two substances, 1 and 2, that diffuse across identical surface areas. The substances have diffusion constants D1 and D2, and D1 > D2. The substances have identical concentration gradients. Which substance will diffuse at a faster rate

Answers

Answer:

Substance 1 will diffuse at a faster rate.

Explanation:

We can solve this problem by keeping in mind Fick's law, which states:

J = -D * (dc/dx)

Where:

J is the fluxD is the diffusion constant(dc/dx) is the concentration gradients

As (dc/dx) is equal for both substances, as stated by the problem, the substance with the higher diffusion constant will diffuse at a faster rate.

Thus the answer is substance 1.


What is the overall charge of an ion that has 31 protons, 36 electrons, and 30 neutrons?

Answers

Answer:

(d) the ion with 24 electrons, 30 neutrons, and a 3+ charge ... (b) the ion with 36 electrons, 35 protons, and 45 neutrons.

for each 20 grams of glucose made by the plant, calculate the mass of water used​

Answers

Answer: 6

Explanation:

You would need 6

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