the measured mass of a sample would have which units

Answers

Answer 1

Answer:

[tex]\longmapsto \: \frak[ hello \: mate ][/tex]

The measured mass of a sample would have unit as "Kilogram "


Related Questions

Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute.
a. 0.100 m K2S
b. 21.5 g of CuCl2 in 4.50 * 102 g water
c. 5.5% NaNO3 by mass (in water)​

Answers

The freezing points of each of the solutions are as follows;

0.100 m K2S -   - 0.558oC

21.5 g of CuCl2 in 4.50 * 102 g water -   -2oC

5.5% NaNO3 by mass (in water)​ -     - 2.6oC

What is freezing point?

The freezing point is the point at which liquid changes to solid. Let us now look at the freezing point of each solution.

a)

Since;

ΔT = K m i

K = 1.86 oC m-1

m = 0.100 m

i = 3

ΔT =   1.86 oC m-1 * 0.100 m * 3 = 0.558oC

Freezing point = 0oC - 0.558oC =  - 0.558oC

b) Number of moles of CuCl2 = 21.5 g/134.45 g/mol = 0.16 moles

molality = 0.16 moles/0.45 Kg = 0.36 m

ΔT = K m i

ΔT =  1.86 oC m-1 * 0.36 m * 3 = 2oC

Freezing point = 0oC - 2 = -2oC

c) Number of moles of NaNO3 = 5.5g/85 g/mol = 0.065 moles

molality of the solution =  0.065 moles/0.0945 Kg = 0.69 m

ΔT =  1.86 oC m-1 * 0.69 m * 2 = 2.6oC

Freezing point = 0oC - 2.6oC = - 2.6oC

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Please Help! Chemistry questions below!

Answers

2H_2S+3O_2-->2SO_2+2H_2O

So

Moles of Oxygen:-

44.7/321.396mol

3 mol of O_2 produces 2 mol SO_2

1 mol of ao_2 produces 2/3=0.6mol So_2

1.396 mol produces 0.8376mol SO_2

Mass of SO_2

0.8376(64)53.6g

Which of the following UV rays has the highest energy that damages your skin the most?

A. UVA

B. UVB

C. X-ray

D. Infrared (IR)

Answers

The answer is b) UVB

Decomposition of hydrogen peroxide: 2H2O2 --> O2(g) + 2H2O(l) How many molecules of water are produced from the decomposition of 3.4g of hydrogen peroxide, H2O2?

Answers

Answer:

Explanation:

You have the equation. Now change the 3.4 g H2 to moles. moles = grams/molar mass

3.4 g/2.016 = 1.686 moles.

Now using the coefficients in the balanced equation, convert moles H2O2 to moles H2O.

1.686 moles H2 x (2 moles H2O/2 moles H2O2) = 1.686 x (2/2) = 1.686 x (1/1) = 1.686 moles H2O.

Now you know that 1 mole of water is composed of 6.022 x 10^23 molecules. So

1.686 moles H2O x (6.022 x 10^23 molecules H2O/1 mole H2O) = ?? molecules.

H2S what species are present at 10-6 mol/L or greater when dissolved in water

Answers

Answer:

the answer has been given below have a good day

Explanation:

mark as brainiest

The species are present at 10-6 mole/L or greater when dissolved in water is when in solutions, weak acids and bases partially ionize.

What is ionization?

Ionization is defined as anything that causes electrically neutral atoms or molecules to gain or lose electrons in order to become electrically charged atoms or molecules (ions).  One of the main mechanisms by which radiation, including charged particles and X-rays, transmits their energy to matter is ionization. An anion is created when an atom or molecule picks up an electron; a cation is created when they lose an electron.

Weak Acids and Weak Bases Ionizing. Numerous acids and bases are weak, meaning they do not completely ionize in aqueous solution. The nonionized acid, hydronium ion, and conjugate base of a weak acid are all mixed together to form a solution in water, with the nonionized acid being present in the highest concentration.

Thus, the species are present at 10-6 mole/L or greater when dissolved in water is when in solutions, weak acids and bases partially ionize.

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A sample of methane gas contains 3.62x10^29 atoms of hydrogen. What is the mass of the sample?

Answers

The mass of the sample to methane that contains 3.62×10²⁹ atoms of hydrogen is 4810631.24 g

Avogadro's hypothesis

6.02×10²³ atoms  = 1 mole of Hydrogen

How to determine the mass of Hydrogen

1 mole of Hydrogen = 2 g

Thus,

6.02×10²³ atoms  = 2 g of Hydrogen

Therefore,

3.62×10²⁹ atoms =  (3.62×10²⁹ × 2) / 6.02×10²³

3.62×10²⁹ atoms = 1202657.81 g of Hydrogen

How to determine the mass of methane 1 mole of methane, CH₄ = 12 + (4×1) = 16 gMass of H in 16 g of CH₄ = 4 × 1 = 4 g

Thus,

4 g of Hydrogen is present in 16 g of methane.

Therefore,

1202657.81 g of Hydrogen will be present in = (1202657.81 × 16) / 4 = 4810631.24 g of methane

Thus, we can conclude that the mass of the sample of methane is 4810631.24 g

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If a watershed suddenly receives a great deal of rainfall, what will happen to its wetlands and rivers?

Answers

Answer:

Depending on the actual amount of water, it's going to overfill.

Explanation:

If your watershed is already one that has a lot of water and is always filled, it's likely that your wetlands and rivers will start to overfill, causing a form of flooding.

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I hope this helps!

-No one

What state of matter has the most kinetic energy?
A gas
B liquid
C solid

Answers

Answer:

gaseous state.

Explanation:

that's the answer

A sample of calcium fluoride was decomposed into the constituent elements. If the sample produced 625 mg of calcium, how many grams of fluorine were formed?

Answers

The mass of fluorine that were formed is 592 mg

Stoichiometry

From the question, we are to determine the mass of fluorine formed

First, we will write the balanced chemical equation for the decomposition reaction

The balanced chemical equation for the reaction is

CaF₂ → Ca + F₂

This means

1 mole of calcium fluoride decomposes to give 1 mole of calcium and 1 mole of fluorine

Now, we will determine the number of moles of calcium produced

From the given information,

Mass of calcium produced = 625 mg = 0.625 g

Using the formula,

[tex]Number\ of \ moles =\frac{Mass}{Atomic\ mass}[/tex]

Atomic mass of calcium = 40.078 g/mol

Then,

Number of moles of calcium produced = [tex]\frac{0.625}{40.078}[/tex]

Number of moles of calcium produced = 0.01559 mole

Since

1 mole of calcium fluoride decomposes to give 1 mole of calcium and 1 mole of fluorine

Then,

0.01559 mole of calcium fluoride will decompose to give 0.01559 mole of calcium and 0.01559 mole of fluorine

∴ Number of mole of fluorine formed was 0.01559 mole

Now, for the mass of fluorine formed

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of fluorine = 38 g/mol

Then,

Mass of fluorine formed = 0.01559 × 38

Mass of fluorine formed = 0.59242 g

Mass of fluorine formed = 592.42 mg

Mass of fluorine formed ≅ 592 mg

Hence, the mass of fluorine that were formed is 592 mg.

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During photosynthesis, a green plant produces 122 mL of oxygen gas at STP. What mass of glucose
(C6H12O6) is produced during this reaction?

6CO2+ 6H2O > C6H12O6+ 602

Show all steps (Answer: 0.164g)

Answers

Answer:

About 0.164 g of glucose.

Explanation:

We can determine the mass of glucose produced given the volume of oxygen gas produced with stoichiometry.

Recall that at STP, a mole of any gas occupies a volume of 22.4 L.

From the reaction, six moles of oxygen gas is produced for every one mole of glucose.

Lastly, the molecular weight of glucose is 180.18 g/mol.

Therefore:
[tex]\displaystyle \begin{aligned} 122\text{ mL O$_2$} & \cdot \frac{1\text{ L O$_2$}}{1000\text{ mL O$_2$}}\cdot \frac{1\text{ mol O$_2$}}{22.4\text{ L O$_2$}} \cdot \frac{1\text{ mol C$_6$H$_{12}$O$_6$}}{6\text{ mol O$_2$}} \cdot \frac{180.18\text{ g C$_6$H$_{12}$O$_6$}}{1\text{ mol C$_6$H$_{12}$O$_6$}} \\ \\ & = 0.164\text{ g C$_6$H$_{12}$O$_6$}} \end{aligned}[/tex]

Therefore, about 0.164 g of glucose is produced.

Can you think of an specific adaptations plants have made to survive in unique conditions?

Answers

Answer:

Plant Adaptations is a unique feature a plant has that allows it to live and survive in its own particular habitat (the place that it lives). Desert Plant Adaptations – Plant Adaptation is really a unique have a plant has that enables it to reside and survive in the own particular habitat (the area it lives).

Explanation:

21.10g of NaOH and Ba3(OH)2 mixture is dissolved water to prepare 1.0dm³ Solution. To neutralize 25.OO mL of this solution needs 0.5 moldm-³ HCl 15.00mL. calculate the percentage of NaOH by mass in the mixture.​

Answers

From the equation of te reaction, we know that the mass percent  of NaOH in the mixture is 1.4%.

What is neutralization?

Neutralization is a reaction that occurs between an acid and a base to yield salt and water only.

In tis case, the reaction of the NaOH and HCl occurs as follows; NaOH + HCl ----> NaCl + H2O

Number of moles of HCl reacted = 15/1000 * 0.5 moldm-³ = 0.0075 moles

Since the reaction is 1:1, 0.0075 moles of NaOH reacted.

Mass of NaOH =  0.0075 moles of NaOH * 40 g/mol = 0.3 g

Percent of NaOH = 0.3 g/21.10g * 100/1 = 1.4%

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An unidentified covalent molecular compound contains only carbon, hydrogen, and oxygen. When 7.40 mg of this compound is burned, 17.80 mg of CO2 and 2.08 mg of H2O are produced.
The freezing point of camphor is lowered by 26.4°C when 3.044 g of the compound is dissolved in 18.00 g of camphor (Kf = 40.0°C kg/mol).

What is the molecular formula of the unidentified compound?

Choose appropriate coefficients in the molecular formula below.
C H O

Answers

The molecular formular shows all the atoms in the compound. The molecular formula of the compound is C15H10O5.

What is molecular formula?

The molecular formula is the formula of the compound that shows the total number of atoms in a compound.

We know that;

ΔT = K m i

ΔT = 26.4°C

K = 40.0°C kg/mol

m = ?

i = 1

26.4°C = 40.0°C kg/mol * m * 1

m = 26.4°C/40.0°C kg/mol  = 0.66 m

We know that

m = mass/molar mass ÷ mass of solvent(in Kg)

Let the molar mass be MM

m = 3.044 g/MM ÷ (18 * 10^-3 Kg)

0.66 m =  3.044 g/0.018MM

0.66 m  * 0.018MM =  3.044 g

MM = 3.044 g/0.66 m  * 0.018

MM = 256 g/mol

Now;

Mass of C =  17.80 * 10^-3/44 * 12 = 0.00485 g

Moles of C = 0.00485 g/12 g/mol = 0.0004 moles

Mass of H = 2.08 * 10^-3/18  * 2 = 0.0002 g

Moles of H =  0.0002 g/1 g/mol = 0.0002 moles

Mass of O = [7.40 * 10^-3 - (0.00485 + 0.0002)] = 0.00235 g

Moles of O = 0.00235 g/16 g/mol = 0.000147 moles

Divide through by the lowest number of moles;

C - 0.0004/0.000147  H - 0.0002/0.000147  O - 0.000147 /0.000147

C - 3   H  -  2  O -  1

The empirical formula is C3H2O

So;

[3(12) + 2(1) + 1(16)]n = 256

[36 + 2 + 16]n = 256

n = 5

Molecular formula = C15H10O5

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How many grams of aluminum will be deposited by 0.1F? (Al=27) a.0.3g b. 0.9. c. 9.0g. d. 2.7g

Answers

Based on the charge on the aluminium ion, 0.9 g of aluminium are deposited by 0.1 F of electricity.

What is electrolysis?

Electrolysis is the decomposition of a substance known as an electrolyte when electric current is passed through it.

The mass and hence moles an electrolyte deposited when current is passed through it depends on the charge on the ion.

Aluminium ion has a charge of +3 and requires 3F of electricity to deposit 1 mole or 27 g of aluminium

0.1 F will discharge = 0.1/3 × 27 g of aluminium

mass of aluminium deposited = 0.9 g of aluminium.

Therefore, 0.9 g of aluminium are deposited by 0.1 F of electricity.

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sodium carbonate (+heat) →

Answers

Answer:

heat carbunate

Explanation:

thats true im good in here like and heart byebye

A heterogeneous mixture is:

A. composed of atoms bonded together.
B. composed of different parts that are indistinguishable in their color, texture, or appearance.
C. composed of only one type of element.
D. composed of different parts that are distinguishable in their color, texture, or appearance.

Answers

Answer:

B

Explanation:

A mixture in which the composition is not uniform throughout.

Same thing letter B says.

Calculate the specific heat in J/(g·ºC) of an unknown substance if a 2.50-g sample releases 12.0 cal as its temperature changes from 25.0ºC to 20.0ºC. ________J/(g·°C)

Answers

The specific heat of an unknown substance is 0.96 J/(g·ºC).

What is the specific heat capacity?

The specific heat capacity of a substance can be described as the quantity of heat needed to raise the temperature in one unit of substance by 1 degree Celcius.

Whenever the heat is lost or absorbed there is a change in the temperature of the substance:

Q = mCΔT

Given, the amount of energy released, Q = 12 cal

The initial temperature of the substance = 25.0ºC

The final temperature of the substance = 20.0ºC

The change in the temperature of a substance, ΔT = 25 -20 = 5°C

The mass of the substance, m = 2.50 g

We have to find the specific heat capacity of an unknown substance.

12 cal = 2.50 g × C × 5°C

C = 0.96 J/(g·ºC)

Therefore, the specific heat capacity of an unknown substance is equal to 0.96 J/(g·ºC).

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The amount of heat needed to raise a substance's temperature by one degree Celsius per gram is known as specific heat. The specific heat in J/(gºC) of an unknown substance is 0.96 J/gºC.

The amount of heat needed to increase the temperature of the entire substance by one degree is the heat capacity of a substance. Specific heat capacity is what is used when the substance's mass is equal to one.

The hydrogen bonds in water, which account for its high specific heat.

The expression used to calculate the specific heat capacity is:

Q = mcΔT

Here Q = 12 Cal

m = 2.50 g

T₁ = 25.0ºC

T₂ = 20.0ºC

ΔT = 5°C

12 cal = 2.50 g × c × 5°C

c = 0.96 J/gºC

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What is the specific heat of a mystery substance that requires 79 joules of energy to heat 4.52
grams of substance from 23 degrees Celsius to 54 degrees Celsius?

Answers

The specific heat of a mystery substance that requires 79 joules of energy to heat 4.52 grams of substance from 23 degrees Celsius to 54 degrees Celsius is 0.56 J/g°C.

How to calculate specific heat capacity?

The specific heat capacity of a substance can be calculated using the following formula:

Q = mc∆T

Where;

Q = quantity of heat absorbed or released (J)m = mass of substancec = specific heat capacity∆T = change in temperature (°C)

c = Q/m∆T

c = 79 ÷ 4.52 × (54 - 23)

c = 79 ÷ 140.12

c = 0.56 J/g°C

Therefore, the specific heat of a mystery substance that requires 79 joules of energy to heat 4.52 grams of substance from 23 degrees Celsius to 54 degrees Celsius is 0.56 J/g°C.

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Select all the correct answers.
Which two generalizations can be made based on what you know about cycles of matter in a closed system?
o New matter is added, and old matter is destroyed.
O Matter changes its physical form, allowing it to return to its original state.
O The amount of matter within the system remains the same.
O Matter and energy can cross the boundaries of the system.
The cycle has a well-defined starting and stopping point.


PLSS HELP

Answers

Answer:

3) The amount of matter within the system remains the same

5) The cycle has a well-defined starting and stopping point.

Explanation:

A closed system is a system that the boundaries of a system cannot be passed by either matter or energy.

Hence, generalisations can be made based on a closed system and the law of conservation of mass which states that matter can neither be created nor destroyed but can only change form.

The correct IUPAC name for 2-isopropylbutane is: a) 2-ethylpentane b) 2,3-dimethylpentane c) 2-methylhexane d) 3-methylhexane

Answers

Answer:

a

Explanation:

On drawing the structure, the most Carbon atoms in a row is 5

⇒ pentane

The remaining two carbon atoms are connected to the 2nd carbon atom

2-ethyl

IUPAC name is :

2-ethyl + pentane2-ethylpentane

The blood carries nutrients to where they are needed. True False

Answers

Answer:

True

Explanation:

Answer:

true

Explanation:

The partial pressures of gases A, B, and C in a mixture are
0.75 atmosphere, 0.25 atmosphere, and 1.25 atmospheres,
respectively. What is the total pressure of the gas mixture
in millimeters of Hg?
A. 1710 mm of Hg
B.
1140 mm of Hg
C. 760.0 mm of Hg
D. 570.0 mm of Hg

Answers

Answer:

A

Explanation:

.75 + .25 + 1.25 = 2.25 atm

1 atm is 760 mm hg

2.25 * 760 = 1710 mm HG

Answer:

[tex]\huge\boxed{\sf 2.25\ atm = 1710\ mm\ of\ Hg}[/tex]

Explanation:

Partial pressure of gas A = 0.75 atm

Partial pressure of gas B = 0.25 atm

Partial pressure of gas C = 1.25 atm

Total partial pressure = 0.75 atm + 0.25 atm + 1.25 atm

= 2.25 atm

We know that:

1 atm = 760 mm of Hg

Multiply 2.25 to both sides

2.25 atm  = 760 × 2.25 mm of Hg

2.25 atm = 1710 mm of Hg

[tex]\rule[225]{225}{2}[/tex]

PROBLEM SETS: %BY MASS, % BY VOLUME, MOLARITY, MOLALITY (show your solution)
1. What is the molarity of a solution in which 0.850 grams of ammonium nitrate are dissolved in 345 mL of solution?
2. Calculate the molality of a solution of 13.5g of KF dissolved in 250. g of water.
3. Calculate the molality of a solution containing 16.5 g of dissolved naphthalene (C10H8) in 54.3 g benzene (C6H6).
4. What is the mass percent of each component in the mixture formed by adding 12 g of calcium sulfate, 18 g of sodium nitrate, and 25 g of potassium chloride to 500 g of water?
5. A solution is made by dissolving 125 g of sodium chloride in 1.5 kg of water. What is the percent by mass?
6. What is the percent by volume of a solution formed by added 15 L of acetone to 28 L of water?

Answers

The concentration of a substance can be expressed using molarity, molality or percent.

What is concentration?

The term concentration has to do with the amount of substance present in solution. Now let us solve the problems individually;

a) Number of moles =  0.850 grams/80 g/mol = 0.011 moles

molarity = 0.011 moles/345 * 10^-3 L = 0.032 M

b) Number of moles =  13.5g /58 g/mol = 0.23 moles

molality = 0.23 moles/250 * 10^-3 Kg = 0.92 m

c) Number of  moles = 16.5 g/128 g/mol = 0.13 moles

molality = 0.13 moles/54.3  * 10^-3 Kg =2.39 m

d) Total mass present =  12 g +  18 g + 25 g + 500 g = 555 g

mass percent of calcium sulfate =  12 g/555 g * 100/1 = 2.2 %

mass percent of sodium nitrate =  18 g/ 555 g  * 100/1 = 3.2 %

mass percent of potassium chloride = 25 g / 555 g  * 100/1 = 4.5%

mass percent of water =  500 g / 555 g  * 100/1 =90.1%

e) Total mass present = 125 g + 1500g = 1625 g

Mass percent of NaCl =  125 g/1625 g * 100/1 = 7.7%

f) Total volume of solution = 15 L +  28 L  = 43 L

percent by volume of acetone = 15 L/43 L * 100/1 =  34.9%

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Best way to deal with calculations in chemistry? ​

Answers

Answer: Look up a video, and they will tell/show you.


There are 7 named classes of hazardous materials.
O True
O False

Answers

False, there are 9 named classes of hazardous materials.

What are hazardous materials?

Hazardous materials are substances or chemicals that pose a health hazard, a physical hazard, or harm to the environment.

There are 9 hazardous substances symbols you need to know: flammable, oxidising, explosives, gas under pressure, toxic, serious health hazard, health hazard, corrosive and environmental hazard.

Hence, the statement is false.

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Water is a _____ molecule, which gives it many of its unique properties, including its ability as a universal solvent.

Answers

Answer:

The answer is Polar.

Explanation:

Because polar has many unique properties.

2AI(NO3)3 + 3Na2CO3 → Al2(CO3)3(s) + 6NaNO3

What is the ratio of moles of AI(NO3)3 to moles Na2CO3?​

Answers

The ratio of moles of AI(NO3)3 to moles Na2CO3 in the reaction is 2:3

In the reaction 2 moles of AI(NO3)3 reacted with 3 moles Na2CO3 so the ratio of the moles is 2 is to 3 represented as 2:3


hope it helps

Which equation shows how frequency is related to velocity and wavelength?

Answers

The equation that shows how frequency is related to velocity and wavelength is f = v / λ (Option A)

What is frequency?

This is defined as the number of oscillations made in one second. It is expressed as

Frequency (f) = 1 / period (T)

f = 1 / T

Relationships between frequency, wavelength and velocity

The velocity, frequency and wavelength of a wave are related according to the following equation

v = λf

Making f the subject, we have

f = v / λ

Thus, we can conclude that the correct answer to the question is:

f = v / λ (Option A)

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How many atoms of oxygen are present: 3Al2(SO4)3
a. 36
b. 12
c. 10
d. 24

Answers

Answer:

the answer is 12

Explanation:

In one formula unit of Al2(SO4)3 A l 2 ( S O 4 ) 3 , there would be twelve atoms of oxygen.

what is galvanization

Answers

The most common hot dipping process for industrial applications is galvanizing, which refers to the coating of zinc over iron or steels for rust proofing. It is an extremely versatile and easy means of providing corrosion protection for construction materials exposed to atmospheric conditions.
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