What is the mass of 3.00 -1023 particles of water? Water has a molar mass of 18.0 g/mol.
3.00 grams
18.0 grams
6.00 grams
9.00 grams

Answer:

c

Explanation:

ok than not c than b maybe

Answer:

% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

% Free space in ice  = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

Explanation:

As given ,

Density for ice at 0⁰C = 0.917 g/ml

Density for water at 0⁰C = 0.999 g/ml

Radii of H atoms = 37 pm

Radii of O atoms = 66 pm

Now,

Consider 1 ml of water = 1 cm²

As , we know that mass of water in 1 cm² = 0.999 g

Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]

Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²

Now,

Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 5.48×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

Now,

Consider 1 ml of ice  = 1 cm²

S.I unit of ice = 1×[tex]10^{-6}[/tex] m²

As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g

Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]

Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012

Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]

Now,

Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 1.17×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

Answer:

light

Explanation:

light is plasma, which is a state of matter

Answer:

0.696 atoms of oxygen

Explanation:

We'll begin by calculating the number of mole in 50 g of scheelite CaWO₄. This can be obtained as follow:

Mass of CaWO₄ = 50 g

Molar mass of CaWO₄ = 40 + 184 + (4×16)

= 40 + 184 + 64

= 288 g/mol

Mole of CaWO₄ =?

Mole = mass / Molar mass

Mole of CaWO₄ = 50 / 288

Mole of CaWO₄ = 0.174 mole

Finally, we shall determine the number of oxygen atom in 50 g (i.e 0.174 mole) of CaWO₄. This can be obtained as follow:

1 mole of CaWO₄ contains 4 atoms of oxygen.

Therefore, 0.174 mole of CaWO₄ will contain = 0.696 atoms of oxygen.

Thus, 50 g (i.e 0.174 mole) of CaWO₄ contains 0.696 atoms of oxygen.

Answer:

[tex]10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].

Explanation:

Identify the elements with oxidation state changes:

Oxidation states of iron, [tex]\rm Fe[/tex]:

Oxidation state of manganese, [tex]\rm Mn[/tex]:

The change in the oxidation state of [tex]\rm Mn[/tex] is five times the opposite of the change to the oxidation state of [tex]\rm Fe[/tex]. If there are one mole of [tex]\rm Mn\![/tex] atoms in each mole of this reaction, there would be five times as many [tex]\rm Fe\![/tex] atoms per mole reaction. In other words:

[tex]\displaystyle 5\, \overset{+2}{\rm Fe}\rm SO_4 + 1\, \rm K \overset{+7}{Mn} O_4 + ?\, H_2SO_4\\ \to \frac{5}{2}\, \overset{+3}{Fe} (SO_4)_3 + ?\, K_2SO_4 + 1\, \overset{+2}{Mn}SO_4 + ?\, H_2O[/tex].

(Notice that each mole of this reaction would include five times as many [tex]\rm Fe[/tex] atoms as [tex]\rm Mn[/tex] atoms.)

Multiply the coefficients by [tex]2[/tex] to eliminate the fraction:

[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + ?\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].

Find the unknown coefficients using the conservation of atoms.

Reactants:

Therefore, among the products:

[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].

Products:

Reactants:

[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].

Products:

Therefore, among the products:

[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].

Answer:

Explanation:

The average kinetic energy for an ideal gas is directly proportional to the temperature. The average kinetic energy of the gas is a measure of the temperature of the gas molecule

Also, the average speed is usually proportional to the square root of temperature.

Similarly, there is a noticeable increase in K.E and speed in regard to temperature but sometimes it is not usually proportional.

However, provided that there is more temperature in Xe as compared to He, then after the mixture of both takes place at equilibrium; the temperature tends to fluctuate between (25 - 50)°C

Thus, since there is a decrease in temperature in Xe, both the average kinetic energy as well as the speed too will also decrease.

Answer:

See explanation

Explanation:

The valence electrons are electrons found on the valence (outermost) shell of an atom.

When an atoms form compounds, there is an exchange of valence electrons between the atoms of one element and the atoms of another element.

Let us consider a typical example, sodium has one valence electron and chlorine has seven valence electrons. This means that chlorine needs one electron to complete its octet while sodium needs to release one electron in order to attain the octet structure.

So, sodium gives out its one electron and becomes a stable sodium ion and chlorine accepts that electron and becomes a stable chloride ion. This is how the compound sodium chloride is formed.

Answer:

Answer:

Reversible

Explanation:

Changes of state are physical changes in matter. Common changes of the state include melting, freezing, sublimation, deposition, condensation, and vaporization.

Joseph Priestly is frequently credited with the discovery of oxygen, and was reported to have produced molecular oxygen from the decomposition reaction of mercury(II) oxide, which is the reverse of the synthesis of HgO depicted in the following equation. [tex]4Hg(l)+2O_2(g)\rightarrow 4 HgO(s) [/tex]Determine the value of [tex]\Delta ΔH°rxn[/tex] for the synthesis, given that [tex]\Delta H_f^0[/tex] for HgO is -90.7 kJ/mol.

Answer: The enthalpy change for this reaction is, -362.8 kJ

Explanation:

The balanced chemical reaction is,

[tex]4Hg(l)+2O_2(g)\rightarrow 4HgO(s)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{HgO}\times \Delta H_{HgO})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{Hg}\times \Delta H_{Hg})][/tex]

where,

n = number of moles

[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero

[tex]\Delta H_{Hg}=0[/tex] (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

[tex]\Delta H=[(4\times -90.7)]-[(2\times 0)+(4\times 0)][/tex]

[tex]\Delta H=-362.8kJ[/tex]

Therefore, the enthalpy change for this reaction is, -362.8 kJ

Answer:

5.27 g of NaCl

Explanation:

The balanced equation for the reaction is given below:

Na₂CO₃ + CaCl₂ —> 2NaCl + CaCO₃

Next, we shall determine the mass of CaCl₂ that reacted and the mass of NaCl produced from the balanced equation. This can be obtained as follow:

Molar mass of CaCl₂ = 40 + (35.5×2)

= 40 + 71

= 111 g/mol

Mass of CaCl₂ from the balanced equation = 1 × 111 = 111 g

Molar mass of NaCl = 23 + 35.5

= 58.5 g/mol

Mass of NaCl from the balanced equation = 2 × 58.5 = 117 g

Summary:

From the balanced equation above,

111 g of CaCl₂ reacted to produce 117 g of NaCl.

Finally, we shall determine the theoretical yield of NaCl. This can be obtained as follow:

From the balanced equation above,

111 g of CaCl₂ reacted to produce 117 g of NaCl.

Therefore, 5 g of CaCl₂ will react to produce = (5 × 117)/111 = 5.27 g of NaCl.

Thus, the theoretical yield of NaCl is 5.27 g.

Answer:

Ideal Gas

The ideal gas is extremely small and the mass is almost zero and no volume Ideal gas is also considered as a point mass.

Real Gas

The molecules of real gas occupy space though they are small particles and also have volume.

anation:

The differences between an ideal gas and a real gas are that the ideal gas follows the gas laws perfectly under all conditions. Whereas a real gas deviates from ideal gas behaviors.

The ideal gas law, also known as the general gas equation, is a fundamental principle in thermodynamics and relates the pressure, volume, temperature, and number of moles of an ideal gas.

An ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions of temperature and pressure. It is assumed to have no volume, no intermolecular forces, and elastic collisions between its particles. An ideal gas also obeys the ideal gas law.

On the other hand, a real gas is a gas that does not follow the gas laws perfectly under all conditions of temperature and pressure. Real gases have volume and intermolecular forces that affect their behavior. These forces cause deviations from ideal gas behavior, especially at high pressures and low temperatures.

In summary, while an ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions, a real gas is a gas that deviates from ideal gas behavior due to its volume, intermolecular forces, and non-elastic collisions between its particles.

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Answer:

Explanation:

From the information given;

Consider using Lande's Interval rule which can be expressed as:

[tex]\Delta E = E_{j+1} - E_jj \ = \alpha (j+1)[/tex]

here;

[tex]j+1[/tex]  = highest level of j

and

[tex]\dfrac{\Delta E_1}{\Delta E_2} = \dfrac{(j+2)}{(j+1)}[/tex]

[tex]\dfrac{5}{3} = \dfrac{(j+2)}{(j+1)}[/tex]

[tex]5(j+1) = 3(j+2)[/tex]

[tex]5j+5 = 3j+6[/tex]

[tex]2j = 1\\ \\ j = \dfrac{1}{2}[/tex]

recall that:

[tex]j = |S-L| \ \to \ |S+L |[/tex]

So;

[tex]S-L = \dfrac{1}{2} --- (1)[/tex]; &

[tex]S+L = \dfrac{5}{2} --- (1)[/tex]

Using the elimination method, we have:

[tex]2S = \dfrac{6}{2}[/tex]

[tex]S = \dfrac{3}{2}[/tex]

Since [tex]S = \dfrac{3}{2}[/tex]; then from (1)

[tex]\dfrac{3}{2} -L = \dfrac{1}{2}[/tex]

[tex]L = \dfrac{2}{2}[/tex]

[tex]L = 1[/tex]

Answer:

25.8 g/dL

Explanation:

A chemist prepares a solution of aluminum sulfate by weighing out 116.0 g of aluminum sulfate into a 450. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in g/dL of the chemist's aluminum sulfate solution. Be sure your answer has the correct number of significant digits.

Step 1: Given data

Step 2: Convert "V" to dL

We will use the following conversion factors.

450. mL × 1 L/1000 mL × 10 dL/1 L = 4.50 dL

Step 3: Calculate the concentration (C) of aluminum sulfate if g/dL

We will use the following expression.

C = m/V = 116.0 g/4.50 dL = 25.8 g/dL

Answer:

B

Explanation:

Answer: 7.77 g/ml

Explanation:

Volume of cylinder with only water = 3.28 mL

Volume of cylinder with water and metal = 8.72 mL

Volume of metal = (Volume of cylinder with water and metal ) -(Volume of cylinder with only water)

=8.72-3.28

=5.44 ml

Mass of metal = 42.26 g

Formula of Density =  [tex]\dfrac{\text{Mass}}{\text{Volume}}[/tex]

i.e. the density of the metal = [tex]\dfrac{42.26}{5.44}\approx7.77\text{ g/ml}[/tex]

Hence, the density of metal = 7.77 g/ml

Answer:

1.6734 g\ml..hope it helps

Answer:

700g

Explanation:

Given parameters:

Density of gasoline  = 0.7g/cm³

Volume of gasoline  = 1L  = 1000cm³  

Unknown:

Mass of the gasoline  = ?

Solution:

Density is the mass per unit volume of a substance. It can be expressed as;

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

 So;

      Mass  = density x volume

    Mass  = 0.7 x 1000  = 700g

Answer:

Lead

Explanation:

The subatomic particles within an atom can be used to know the atom or element given.

Of particular interest is the number of protons within the atom.

The periodic table is based on the atomic number of atoms. This atomic number is the number of protons within an atomic space.

So; If we know the number of protons within an atom, we can know the element.

The number of protons given is 82, the element is  therefore lead.

Answer:

The atomic number of polonium is 84. The atomic number lead is 82.

Explanation:

Answer:

In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M

Explanation:

In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10

In the first rinse the concentration must be of 0.9M  10 = 0.09M

2nd = 0.009M

3rd = 0.0009M

4th = 0.00009M

5th = 0.000009M →

Answer:

See explanation

Explanation:

Chemical reactions are also referred to as chemical change. A chemical change often leads to the formation of new substances and is not easily reversible.

A chemical reaction may be accompanied by the emission of heat and light, formation of a precipitate, evolution of gas, or a color change.

These observable physical effects may tell us weather a chemical reaction has taken place or not so we have to observe the system closely for any of these effects stated above.

Answer:

Molarity: 0.21M

Molality: 0.20m

Explanation:

...dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL...

To solve this question, we need to find the moles of aniline in 3.9g using its molar mass. Then, we need to find the kg and Liters of solution in order to find molarity (Moles/L solution) and molality (Moles/kg of solvent):

Moles aniline:

Molar mass:

6C: 6* 12.01g/mol = 72.06g/mol

7H: 7*1.008g/mol = 7.056g/mol

N: 1*14.007g/mol = 14.007g/mol

72.06g/mol+7.056g/mol+14.007g/mol = 93.123g/mol

Moles of 3.9g: 3.9g * (1mol / 93.123g) = 0.04188moles

Liters solution:

200mL * (1L / 1000mL) = 0.200L

kg solvent:

200mL * (1.05g/mL) * (1kg/1000g) = 0.210L

Molarity:

0.04188mol / 0.200L = 0.21M

Molality:

0.04188mol / 0.210L =0.20m

Answer:

Explanation:

From the given information:

The accuracy depends on the internal diameter of the cylinder. The cylinder with the least internal diameter is obviously more precise.

Let's assume 1% is the error of measurement.

Then, to measure 70 mL from 10 mL cylinder

The error = [tex]10 \times \dfrac{1}{100} \times 7[/tex]

= 0.7 mL

However; for a 100 mL cylinder, the error = 1 mL

Now,

The total volume for 10 mL = (70 + 0.7) = 70.7 mL

The total volume for 100 mL = (70 + 1 ) = 71 mL

Suppose the density (d) is same for both

Then;

the mass of 10 mL = ( d × 70.7) g

the mass pf 100 mL = (d × 71) g

Thus, the mass of 100 mL is greater than that of 10 mL.

The ocean is not a part of Earth's layers.

Answer:

Ocean

Explanation:

Explanation:

H2SO3 is more acid than H2TeO3. Since S is more electronegative than Te is. In H2SO3, thus, dissociation of H+ would be smoother.

So, H2SO3's got high Ka.

HCH3SO2 is more acid than HCH3CO2. Since S is more electronegative than C. So, HCH3SO2 is a high Ka.

HClO2 is more acid than HClO. Since in HClO2, after the donation of H+ ion, the negative charge is set by two oxygen atoms, while in HClO, only one oxygen atom stabilizes the negative charge.

So, HClO2 is a high Ka

Answer:

See explanation

Explanation:

The atoms of metals have fewer valence electrons than the atoms of metals and metalloids.

Atoms of metals have only very few valence electrons in their outermost shells hence they donate electrons during bonding. However, atoms of nonmetals have more electrons in their outermost shells and rather accept electrons during bonding. The atoms of metalloids just have a number of valence electrons that are intermediate between those of metals and nonmetals and mostly share electrons in covalent bonds.

Similarly, atoms of metallic elements differ from each other in the number of valence electrons present in the valence shell of the atom of each element. For instance, sodium has one electron in the valence shell of its atom while aluminium has three electrons in the valence shell of its atom.

The atoms of metallic elements are different from the atoms of non metals or metalloids base on the outer electron/ valency electrons and the its bonding pattern.

The atoms of different metals varies in it ability to bond quickly.

The atoms of metallic elements are different from the atoms of non metals or metalloids base on the outer electron/ valency electrons and how it bonds.

Metallic atoms have very few electrons in the outermost shell. The valency electrons of this metallic atoms are few and are easily lost during bonding. They have the ability to release there valency electrons easily. Example of this metals are sodium, potassium , calcium etc.

On the other hand non metallic elements have numerous electron in the outermost shell and easily receive electron during bonding. Example are chlorine, fluorine, oxygen etc.

The metalloid atoms like silicon and germanium have an average number of electron in their outermost shell. They are in between.

The atoms of different metals varies in it ability to bond quickly. For example the group 1 metals are very reactive than the group 2 metals. This simply means the group 1 metals(alkali metals) goes into bonding more easily than the group 2 metals(alkali earth metals).    

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Answer:

3

Explanation:

For numbers with decimals, count the number after the decimal.

Answer:

60.0mL of the diluted solution are needed

6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.

Explanation:

As in each titration we need to use 20.0mL of the diluted 0.100M solution. As there are 3 titration, the volume must be:

3 * 20.0mL = 60.0mL of the diluted solution are needed

Now, to prepare a 0.100M NaOH solution from a 1.00M NaOH stock solution the dilution must be of:

1.00M / 0.100M = 10 times must be diluted the solution.

As we need at least 60.0mL, the minimum volume of the stock solution must be:

60.0mL / 10 times =

Answer:

[tex]\%N=26.2\%\\\\\%H=7.5\%\\\\\%Cl=66.3\%[/tex]

Explanation:

Hello!

In this case, since the calculation of the percent composition of an element in a chemical compound is computing considering its atomic mass, subscript in the formula and molecular mass of the compound it is; for nitrogen, hydrogen and chlorine we have that ammonium chloride has a molar mass of 53.49 g/mol so the percent compositions are:

[tex]\%N=\frac{14.01*1}{53.49}*100\% =26.2\%\\\\\%H=\frac{1.01*4}{53.49}*100\% =7.5\%\\\\\%Cl=\frac{35.45*1}{53.49}*100\% =66.3\%[/tex]

Best regards!

Answer:

No

Explanation:

This does not mean that vinegar is insoluble totally. In fact, vinegar is soluble in water because water is a polar solvent.

For a substance to be soluble in another, it must obey the rule of solubility.

The rule states that "like dissolves like"

It implies that polar solvent will only dissolve polar solute.

Also, non-polar solvent will only dissolve non-polar solute.

Therefore, the answer is no, vinegar will dissolve in water.