How many moles of Ba(NO3)2 are there in 0.25 L of a 2.00 M Ba(NO3)2 solution?
Use Molarity equals StartFraction moles of solute over liters of solution EndFraction..
0.13 mol
0.50 mol
2.25 mol
8.0 mol

Answers

Answer 1

Answer:

0.5moles

Explanation:

Given parameters:

Volume of solution= 0.25L

Molarity of solution = 2.00M

Unknown:

Number of moles  = ?

Solution:

To solve this problem;

  Molarity is the number of moles of solute in a given volume of solution, so

 Number of moles  = molarity x volume

 Now insert the parameters and solve;

  Number of moles  = 0.25 x 2  = 0.5moles

Answer 2

Answer:

b

Explanation:

got it right on test


Related Questions

What makes the Bactria move

Answers

The answer is B ^_^

Answer:

b

Explanation:

What is the boiling point (in C) of a 0.743 m
aqueous solution of KCI?
Enter your rounded answer with
3 decimal places.

Answers

The boiling point (in C) of a 0.743 m  aqueous solution of KCI : 100.7608°C

Further explanation

Solutions from volatile substances have a higher boiling point and lower freezing points than the solvent  

ΔTb = Tb solution - Tb solvent  

ΔTb = boiling point elevation  

[tex]\large {\boxed {\boxed {\bold {\Delta Tb \: = \: Kb.m}}}[/tex]

For electrolyte solutions there is a van't Hoff factor = i  

i = 1 + (n-1) α  

n = number of ions from the electrolyte  

α = degree of ionization, strong electrolyte α = 1, for non electrolytes i = 1  

KCl⇒K⁺+Cl⁻⇒ electrolyte  solution(2 ions K⁺ and Cl⁻),  strong electrolyte α = 1

ΔTb=Kb.m.i

i = 1 + (n-1) α  

i=1+(2-1).1=2

Kb for water (solvent) : 0.512 °C kg/mol

molal KCl = 0.743 m

The boiling point of solution :

[tex]\tt Tb-100=0.512\times 0.743\times 2\\\\Tb-100=0.7608\\\\Tb=100.7608^OC[/tex]

The correct answer is 100.761.

In the current periodic table, how are the elements arranged?

a. the bonding power with oxygen
b. the number of neutrons
c. atomic mass
d. atomic number

Answers

Answer:

The answer would be

C. The number of neutrons

You're welcome <3

1.) a copper penny has a mass of 3.1 g and a volume of 0.35 cm^3. what is the dencity of the copper penny?

2.) A block of wood has the dimensions 7.0 cm x 10.0 cm x 5.0 cm. What is the
volume of the block of wood?

3.) A graduated cylinder has a volume of 50.0 ml of water. After a rubber cork is
placed in the graduated cylinder, the volume of the water rises to 53.0 ml. If
the mass of the rubber cork is 4.3 g, what is its density?

4.) Calculate the volume of a book with a density of 9.9 g/cm3 and a mass of 58.5 g.

5.) Find the mass of iron if its density is 6.7 g/ml and it has a volume of 5 ml.

Answers

1.)

density is found by [tex]\rho=\frac{m}{V}[/tex]

[tex]\rho=\frac{3.1}{0.35}\\\\ \rho=8.857 g/cm^3[/tex]

2.)

Volume is [tex]lwh[/tex]

7*10*5

V=350[tex]cm^3[/tex]

3.)

using the density formula

[tex]\rho=\frac{m}{V}[/tex]

[tex]\rho=\frac{4.3}{3}[/tex]

[tex]\rho=1.43g/ml[/tex]

4.)

rearrange the density formula for V

[tex]V=\frac{m}{\rho}[/tex]

[tex]V=\frac{58.5}{9.9}\\\\ V=5.909cm^3[/tex]

5.)

rearrange the density formula for [tex]m[/tex]

[tex]m=\rho V\\\\m=(6.7)(5)\\\\m=33.5g[/tex]

Select the answer with the correct number of significant figures for each calculation. 31.580 + 4.26 = 35.8 35.84 35.840

Answers

Answer:

35.8.540 its korrect now

An atom has a diameter of 2.00 Å and the nucleus of that atom has a diameter of 9.50×10−5 Å . Determine the fraction of the volume of the atom that is taken up by the nucleus. Assume the atom and the nucleus are a sphere.

Answers

fraction of the volume of the atom that is taken up by the nucleus is 1.07x10⁻¹³Å³

Sphere volume is calculated by formula =

V=4/3(πr³) , V is volume , r is radius
Radius = diameter/2

Radius of atom = 2/2=1

Radius of nucleus = (9.50x10⁻⁵)/2 = 4.75x10⁻⁵


now we may calculate the volume for the atom and the nucleus

atom volume = (4/3) × 3.14 × (1 )³ = 4.186 ų

nucleus volume = (4/3) × 3.14 × (4.75× 10⁻⁵)³ = 4.48 × 10⁻¹³Å³

now the fraction of volume of the atom that is taken up by the nucleus:

fraction = nucleus volume / atom volume 

fraction =4.48 × 10⁻¹³/4.186 = 1.07x10⁻¹³Å³


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What is the importance of the fundamental laws of chemistry?

Answers

Answer:

Law of Conservation of mass: Mass cannot be created nor destroyed. Reference: Antoine Lavoisier – By carefully weighing the reactants & products of chemical reactions. The laws of chemical combination describe the basic principles obeyed by interacting atoms and molecules, interactions that can include many different combinations that happen in many different ways. This amazing diversity of interactions allows for an astounding variety of chemical reactions and compounds. The most fundamental concept in chemistry is the law of conservation of mass, which states that there is no detectable change in the quantity of matter during an ordinary chemical reaction. Modern chemistry is based on several fundamental laws, including:

The law of multiple proportions.The law of definite proportions.The law of conservation of mass.

b. onsia, t. conard, s. de gendt, m. heyns, i. hoflijk, p. mertens, m. meuris, g. raskin, s. sioncke, i. teerlinck, a. theuwis, j. van steenbergen, c. vinckier, "a study of the influence of typical wet chemical treatments on the germanium wafer surface," solid state phenom. 103-104, 19-22 (2005)

Answers

The influence of various wet chemical treatments on the germanium wafer surface has been examined. The GeO2 part of the native oxide is readily dissolved in water. The solubility of the GeO2 in aqueous solutions and the fast oxidation of germanium makes commonly used silicon treatments nonapplicable due to the relatively high etch rates. In various etching chemistries, the underlying suboxide is not attacked. A hydroxylated suboxide might be an ideal starting surface for ALD, because of its low contribution to electrical thickness. HBr removes even the suboxide and yields a completely oxide-free germanium surface which makes it a candidate surface preparation technique for epitaxial growth. The same result is obtained using a HI dip followed by a DIW rinse. Oxide is growing in CR-air, while in N 2 ambient the surface remains stable.

An oxide is a category of chemical compound that has one or more oxygen atoms as well as another element in its composition such as Li2O, CO2, H2O, etc.

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CHOOSE ONE AND I WILK GIEV YOU BRAINLIEST AND DONT GUESS PLEASE!!!..

Answers

6272.2737362282736363737373733.282726363

Answer:

c

Explanation:

85miles/hours south is an example of? this is science btw

Answers

Answer:

I believe it's an example of velocity

Explanation:

This is because 85miles/hour south is giving both magnitude (85miles/hour) and direction (south)

why law of conservation of mass should better be called as law of conservation of mass and energy​

Answers

Answer:

Energy can neither be created not be destroyed

Mass too can neither be created nor be destroyed

Both of them remain conserved during any chemical or physical reaction

Some part of them gets transformed to undesired useless form

Hence law of conservation of mass should better be called the law of conservation of mass and energy

I hope helps you and if you consider and want you can give me a Brainly crown

How many orbitals would mercury (Hg) have?

Answers

Answer:

6 orbitals

Explanation:

An atom of mercury will have 6 - orbitals because it can be found on the 6-period in the periodic table of elements.

Orbitals or shells are energy levels in which electrons are placed within an atom.

Let us write the shell notation of the element:

    Hg has 80 electrons:

            K     L    M   N     O   P

            2      8   18    32    18  2

So, we have six orbitals

Substances that can either take up hydrogen ions or release hydroxide ions into water are called.

Answers

Bases are substances that can release hydroxide ions or take up hydrogen ions into the water.

Bases are those substances that have a pH above 7. They are bitter. When bases are added to water, they release hydroxide ions. And takes up hydrogen ions. Bases that fall at the pH of 8 are considered weak bases whereas those bases that fall at the pH of 14 are considered strong bases. A few examples of bases are sodium hydroxide, calcium hydroxide, potassium hydroxide, etc.

[tex]NaOH + H _{2}O→ Na ^{ + } + OH ^{ - } [/tex]

Substances that fall within the pH of 7 are considered neutral substances.

Acids are substances that have a pH below 7. They are sour. When acids are added to water they release hydrogen ions. Acids that fall at the pH of 6 are considered weak acids whereas those bases that fall at the pH of 1 are considered strong acids. A few examples of acids are acetic acid, hydrochloric acid, sulphuric acid, nitric acid, etc.

[tex]CH _{3}COOH + H _{2} O→COO ^{ - } + H ^{ + } [/tex]

Therefore, the substances that release hydroxide ions or take up hydrogen ions into the water are known as bases.

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HELP ME ASAP PLEASE LOTS OF POINTS
Determine the average atomic mass of Sulfur, given the following:

There are 3 isotopes of the element Sulfur the mass numbers and abundances are as follows:

S-32: 94.99%; S-33: .75%; S-34: 4.26%

Answers

Answer: 32.06 atomic mass units

Explanation:

In a solid, the particles..

A. Vibrate in place

B. Overcome the strong attraction

C. Slide past one another

D. Move freely

Answers

answer: a
explanation

Please help me to do this assignment

Answers

Answer:

1. Objective

2. Objective

3. Opinion

4. Objective

5. Opinion

6. Opinion

7. Opinion (I think.)

8. Opinion (I think.)

Explanation:

How hot would the water be if all the light reaching the two beakers had been transmitted

Answers

the water would be cooling down if the light is gone

If the light completely falls on the two beakers, the degree of hotness increases with respect to time, while if the light is transmitted, the water would gradually cool down.

What is the effect of transmitted light on temperature?

The effect of transmitted light on temperature is understood by the fact that the intensity of transmitted light became maximum during the process of cooling. While it gets the opposite when the water warms significantly.

According to the context of this question, if all the light reaching the two beakers had been transmitted, the degree of hotness of the water gradually declines. Due to this, the water in the two beakers cool down.

Therefore, the water cools slowly if all the light reaching the two beakers had been transmitted.

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whoever answers this first gets brainly

Answers

Answer:

D. lysosome

Explanation:

hope it helps :D

sorry if its wrong :/

You need 150g of pure lithium for an experiment you're doing. You have 675g of lithium oxide (Li2O). Can you extract all the lithium you need from the amount of compound you have? Show your reasoning.

Answers

Answer: Yes we can extract all the lithium from the amount of compound we have.

Explanation:

To calculate the moles :

[tex]moles=\frac{\text {given mass}}{\text {Molar mass}}[/tex]

[tex]\text{Moles of} Li=\frac{150g}{7g/mol}=21.4moles[/tex]

[tex]\text{Moles of} Li_2O=\frac{675g}{30g/mol}=22.5moles[/tex]

[tex]2Li_2O\rightarrow 4Li+O_2[/tex]

According to stoichiometry :

2 moles of [tex]Li_2O[/tex] produce 4 moles of [tex]Li[/tex]

Thus 22.5 moles of [tex]Li_2O[/tex] will produce=[tex]\frac{4}{2}\times 22.5=45moles[/tex]  of Li

As 21.4 moles is a lesser quantity than 45 moles, thus it can be produced

Yes, you can extract all the lithium you need from the compound

We'll begin by calculating the mass of lithium, Li in 1 mole of Li₂O. This can be obtained as described below:

1 mole of Li₂O = (2×7) + 16

1 mole of Li₂O = 14 + 16 = 30 g

SUMMARY

30 g of Li₂O contains 14 g of Li.

With the above information in mind, we can determine the mass of Li in 675 g of lithium oxide, Li₂O. This can be obtained as follow:

30 g of Li₂O contains 14 g of Li.

Therefore,

675 g of Li₂O Will contain = (675 × 14)/30 = 315 g of Li.

From the calculation made above, we can see that there are 315 g of Li in 675 g of Li₂O. Thus, we can conclude that 150 g of Li can be extracted from the compound (i.e 675 g of Li₂O)

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How many grams of CO2 are contained in 4.77x10^24 molecules of of CO2?

Answers

Answer:

3.5x1024 3.5 x 10 24 carbon dioxide molecules weighs 256 g.

a 657-ml sample of unknown HCL solution reacts completely with Na2CO3 to form 11.1 g CO2. What was the concentration of the HCI solution?

Answers

The concentration of the HCI solution : 0.767 M

Further explanation

Reaction

Na₂CO₃ (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO₂ (g) + H₂O (l)

mass of CO₂ = 11.1 g

mol of CO₂ (MW= 44,01 g/mol) :

[tex]\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{11.1}{ 44,01}\\\\mol=0.252[/tex]

From the equation above, mol ratio of HCl : CO₂ = 2 : 1, so mol HCl :

[tex]\tt mol~HCl=\dfrac{2}{1}\times 0.252=0.504[/tex]

Molarity shows the number of moles of solute in every 1 liter of solution.

[tex]\large{\boxed {\bold {M ~ = ~ \frac {n} {V}}}[/tex]

The molarity of unknown HCl :

mol=n=0.504

volume=V=657 ml=0.657 L

[tex]\tt M=\dfrac{0.504}{0.657}\\\\M=0.767[/tex]

what type of friction is talking

Answers

Answer:

does friction talks ? friction is a oppossing force tho

Which type of organism is a prokaryote?

OA) plant

OB) animal

OC) fungus

OD) bacterium​

Answers

Answer:

D

Explanation:

Answer:

D) Bacterium

Explanation:

An object in motion has what type of energy?

A magnetic
B kinetic
C possible
D chemical

Answers

Answer:

B) Kinetic

Explanation:

Kinetic Energy is energy relating to or resulting from motion. Therefore, the answer to this question is B.

Answer:

The answer is option B - Kinetic energy.

Explanation:

The Energy that a moving object has due to its motion is Kinetic Energy.

Juan wanted to investigate how exercise affects heart rate. a classmate sat quietly for 5 minutes before juan used a stopwatch to count her pulse for 15 seconds. a second classmate ran in place for 5 minutes before juan used the stopwatch to find his pulse rate for 15 seconds. after both measurements, juan multiplied the number by 4 to determine hear rate per minute. what did juan do wrong in this investigation? what should juan have done instead?

Answers

The wrong Juan did in this investigation is he multiplied by 4 and did not count it for 60 seconds to calculate the heart rate per minute.

What to do to know the pulse rate or heart beat and what did Juan did wrong?See to know the heart beat or pulse rate we simply do it for 60 seconds because we generally obtain the data for a minute .Here Juan is investigating the heart rate for 15 seconds .Plus the the cardinal mistake he did was he multiplied the beats by 4 of both the classmate knowing a student just rushed into the classroom and is exercised.So obviously the student who just arrived the classroom would have high heart rates comparatively.Calculating both heart rates and multiplying with 4 is the main mistake Juan is commiting.

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why elements are arranged according to atomic number rather than atomic mass in the periodic table

Answers

Answer:

hope it helps...

Explanation:

Atomic Number as the Basis for the Periodic Law

Assuming there were errors in atomic masses, Mendeleev placed certain elements not in order of increasing atomic mass so that they could fit into the proper groups (similar elements have similar properties) of his periodic table.

10. Overpopulation occurs when the number of organisms of a species
exceeds the ecosystem's carrying capacity.
True
O False

Answers

True.

However, overpopulation also increases if there is a shortage in the population of consumers. As a result population of producers increase.

Answer:

True (statement is correct)

Explanation:

Yeah, overpopulation occurs when the number of organisms of a species exceeds the ecosystem's carrying capacity. Hence, the given statement is correct (true).

Calculate the mass of 1.25 mol ammonium sulfide, (NH4)2S. *

Answers

Answer:

85g

Explanation:

To convert the moles of a substance to grams we need to know the molar mass of the substance. We, as first, must obtain the molar mass of (NH₄)₂S as follows:

There are 2 atoms of N, 8 of H and 1 of S:

N = 2*14g/mol = 28g/mol

H = 8*1g/mol = 8g/mol

S = 1*32g/mol = 32g/mol

Molar mass: 68g/mol

That means 1 mole of (NH₄)₂S has a mass of 68g.

1.25moles have a mass of:

1.25moles * (68g/mol) =

85g

Any element above 92 on the
periodic table has to be constructed
in a lab.
A.False
B.True

Answers

false
explanation

All of the elements with atomic numbers 1 to 92 can be found in nature, have stable or very long half-life isotopes, and are created as common products of the decay of uranium and thorium.

answer: it should be false

to save time you can approximate the initial volume of water to ±1 ml and the initial mass of the solid to ±1 g . for example, if you are asked to add 23 ml of water, add between 22 ml and 24 ml . which metals in each of the following sets will have equal density?

Answers

The metals in the following set which will have equal density are:

20.2 g of silver when placed in 21.6 mL of water and 12.0 g of silver when placed in 21.6 mL of water.15.2 g of copper when placed in 21.6 mL of water and 50.0 g of copper when placed in 23.4 mL of water.11.2 g of gold when placed in 21.6 mL of water and 14.9 g of gold when placed in 23.4 mL of water

So, the correct options are 2,3 and 6.

A substance or object's density can be found out by comparing its mass to volume. Density is calculated using the formula:

[tex]density = \frac{mass}{volume}[/tex]

It is an essential property that the density of the same material remains constant at varied masses and volumes. Since the metal used in both situations is the same and metal has a fixed value of density, the choice with sets of the same metals has equal densities across the possibilities provided.

The complete question is as follows:

To save time you can approximate the initial volume of water to ±1 mL and the initial mass of the solid to ±1 g. For example, if you are asked to add 23 mL of water, add between 22 mL and 24 mL. Which metals in each of the following sets will have equal density?

1. 20.2 g gold placed in 21.6 mL of water and 12.0 g copper placed in 21.6 mL of water.

2. 20.2 g silver placed in 21.6 mL of water and 12.0 g silver placed in 21.6 mL of water.

3. 15.2 g copper placed in 21.6 mL of water and 50.0 g copper placed in 23.4 mL of water.

4. 15.4 g gold placed in 20.0 mL of water and 15.7 g silver placed in 20.0 mL of water.

5. 20.2 g silver placed in 21.6 mL of water and 20.2 g copper placed in 21.6 mL of water.

6. 11.2 g gold placed in 21.6 mL of water and 14.9 g gold placed in 23.4 mL of water.

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