How many grams is 4.2X10^24 atoms of sulfur?

hard because it has calcium and magnesium

Answer:

26 mol

Explanation:

Step 1: Write the balanced equation

4 FeCl₃ + 3 O₂ ⇒ 2 Fe₂O₃ + 3 Cl₂

Step 2: Determine the appropriate molar ratio

The molar ratio of FeCl₃ to O₂ is 4:3.

Step 3: Use the determined molar ratio to calculate the moles of oxygen required to completely react with 35 moles of ferric chloride

[tex]35molFeCl_3 \times \frac{3molO_2}{4molFeCl_3} = 26molO_2[/tex]

Answer:

[tex]n_{O_2}=26.25molO_2[/tex]

Explanation:

Hello,

In this case, given the reaction:

[tex]4FeCl_3 + 3O_2 \rightarrow 2Fe_2O_3 + 6Cl_2[/tex]

Since oxygen and iron (III) chloride are in a 4:3 molar ratio, he required moles of oxygen to completely react with 35 moles of iron (III) chloride result:

[tex]n_{O_2}=35molFeCl_3*\frac{3molO_2}{4molFeCl_3} \\\\n_{O_2}=26.25molO_2[/tex]

Best regards.

Answer:

C. cooler than both the crust and the core

Explanation:

It is observed that at the mantle, temperatures range from estimatedly 200 °C (392 °F) around the upper boundary with the crust to approximately 4,000 °C (7,230 °F) at the core-mantle boundary.

So we can say the mantle is cooler than both the crust and the core.

Answer:

The mass of zinc sulfide produced is  [tex]M_{ZnS} = 0.76 \ g[/tex]

Explanation:

From the question we are told that

   The mass of zinc is  [tex]m_z = 0.750 \ g[/tex]

    The mass of sulfur is  [tex]m_s = 0.250 \ g[/tex]

The molar mass   of  [tex]Zn_{(s)}[/tex]  is a constant with value  65.39 g /mol

The molar mass of [tex]S_{(s)}[/tex]  is a constant with value  32.01 g/mol

The molar mass of  [tex]ZnS_{(s)}[/tex] is a constant with value 97.46  g/mol

The reaction is  

        [tex]Zn_{(s)} + S_{(s)} ------> ZnS_{(s)}[/tex]

   So from the reaction

       1 mole of  [tex]Zn_{(s)}[/tex] react with 1 mole of  [tex]S_{(s)}[/tex] to produce 1 mole of [tex]ZnS_{(s)}[/tex]

This implies that

65.39 g /mol of  [tex]Zn_{(s)}[/tex] react with 32.01 g/mol of  [tex]S_{(s)}[/tex] to produce   97.46  g/mol  of [tex]ZnS_{(s)}[/tex]

From the values given we can deduce that the limiting reactant is sulfur cause  of the smaller mass

 So  

    0.250 g of  [tex]Zn_{(s)}[/tex] react with 0.250 of  [tex]S_{(s)}[/tex] to produce [tex]x \ g[/tex] of  [tex]ZnS_{(s)}[/tex]

So

      [tex]x = \frac{97.46 * 0.250}{32.01}[/tex]

       [tex]x = 0.76 \ g[/tex]

Thus the mass of the mass of zinc sulfide produced is

    [tex]M_{ZnS} = 0.76 \ g[/tex]

Answer:

In the case of Mount Pinatubo, the result was a measurable cooling of the Earth's surface for a period of almost two years. Because they scatter and absorb incoming sunlight, aerosol particles exert a cooling effect on the Earth's surface.

Explanation:

Temperature measures the average kinetic energy of particles of the substances. Therefore, the Mt. Pinatubo eruption result in global temperatures dropping almost two degrees.

Temperature is used to measure degree or intensity of heat of a particular substance. Temperature is measured by an instrument called thermometer.

Temperature can be measured in degree Celsius °c, Kelvin k or in Fahrenheit. Temperature is a physical quantity. Heat always flow from higher temperature source to lower temperature source.

We can convert these units of temperature into one another. The relationship between degree Celsius and Fahrenheit can be expressed as:

°C={5(°F-32)}÷9

The Mt. Pinatubo eruption result in global temperatures dropping almost two degrees because they scatter and absorb incoming sunlight, aerosol particles exert a cooling effect on the Earth's temperature.

Therefore, the Mt. Pinatubo eruption result in global temperatures dropping almost two degrees.

To know more about temperature, here:

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Answer:

.00352mol/L

Explanation:

molarity (concentration) = number of moles / volume of solvent (in L)

M = .99g / 56.108‬g/mol / .5L

M = .0176mol / .5L

M = .00352mol/L

Answer:

Endothermic, positive

Explanation:

Explanation:

MnO4−+ SO32− → Mn2++ SO42−

Splitting into half equations;

MnO4−  → Mn2+

SO32− → SO42−

Balancing the electrons;

2 MnO4−  + 10 e- → 2Mn2+

5SO32− → 5SO42− + 10 e-

In an acidic medium, it becomes;

2 MnO4−  +  8 H+ → 2 Mn2+ + 4 H2O

5 SO32−  + H2O → 5 SO42−  +  2 H+

Final equation is;

2 MnO4- + 5 SO32- + 6 H+ → 2 (Mn)2+ + 5 SO42- + 3 H2O

Coefficient of H+ = 6

Coefficient of H2O = 3

Coefficient of MnO4- = 2

Coefficient of SO32- = 5

Coefficient of (Mn)2+- = 2

Coefficient of SO42- = 5

Answer:

[tex]5SO_3^{2-}+2MnO_4^{-}+6H^+ \rightarrow 5SO_4^{2-}+ 2Mn^{2+}+3H_2O[/tex]

Explanation:

Hello,

In this case, given the reaction:

[tex]MnO_4^{-}+ SO_3^{2-} \rightarrow Mn^{2+}+ SO_4^{2-}[/tex]

We first identify the oxidation state of both manganese and sulfur at each side:

[tex]Mn^{7+}O_4^{-}+ S^{4+}O_3^{2-} \rightarrow Mn^{2+}+ S^{6+}O_4^{2-}[/tex]

So we have the oxidation and reduction half-reactions below, including the addition of water and hydronium as it is in acidic media:

[tex]S^{4+}O_3^{2-}+H_2O \rightarrow S^{6+}O_4^{2-}+2H^++2e^-[/tex]

[tex]Mn^{7+}O_4^{-}+8H^++5e^- \rightarrow Mn^{2+}+4H_2O[/tex]

Next, we exchange the transferred electrons:

[tex]5*(S^{4+}O_3^{2-}+H_2O \rightarrow S^{6+}O_4^{2-}+2H^++2e^-)\\2*(Mn^{7+}O_4^{-}+8H^++5e^- \rightarrow Mn^{2+}+4H_2O)\\\\5S^{4+}O_3^{2-}+5H_2O \rightarrow 5S^{6+}O_4^{2-}+10H^++10e^-\\2Mn^{7+}O_4^{-}+16H^++10e^- \rightarrow 2Mn^{2+}+8H_2O[/tex]

Then we add the resulting half-reactions and simplify the transferred electrons:

[tex]5S^{4+}O_3^{2-}+5H_2O+2Mn^{7+}O_4^{-}+16H^+ \rightarrow 5S^{6+}O_4^{2-}+10H^++ 2Mn^{2+}+8H_2O[/tex]

We rearrange the terms in order to simplify water and hydronium molecules:

[tex]5S^{4+}O_3^{2-}+2Mn^{7+}O_4^{-}+16H^+-10H^+ \rightarrow 5S^{6+}O_4^{2-}+ 2Mn^{2+}+8H_2O-5H_2O\\\\5S^{4+}O_3^{2-}+2Mn^{7+}O_4^{-}+6H^+ \rightarrow 5S^{6+}O_4^{2-}+ 2Mn^{2+}+3H_2O[/tex]

Finally we write the balanced reaction in acidic media:

[tex]5SO_3^{2-}+2MnO_4^{-}+6H^+ \rightarrow 5SO_4^{2-}+ 2Mn^{2+}+3H_2O[/tex]

Best regards.

Answer:

[tex]pH=10.5[/tex]

Explanation:

Hello,

In this case, since we define the pH as:

[tex]pH=-log([H^+])[/tex]

We can directly compute it by using the given concentration of H⁺:

[tex]pH=-log(3.5x10^{-11})\\\\pH=10.5[/tex]

For which we conclude it is a basic solution as the pH is greater than 7.

Best regards.

Answer:

[tex]m_{PbI_2}=18.2gPbI_2[/tex]

Explanation:

Hello,

In this case, we write the reaction again:

[tex]Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)[/tex]

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

[tex]n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI[/tex]

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

[tex]0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI[/tex]

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

[tex]m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2[/tex]

Best regards.

Answer:

Mass PbI2 = 18.19 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.19 grams

Answer:

Magma

Explanation:

Magma is the hot liquid under earths surface

Answer:

See explanation below

Explanation:

To do aflow diagram for extraction of these two components of a sample, we need to analize both reagents so we can make a great diagram and separate both reagents.

First, let's see the acid. The 2-chlorobenzoic acid is a relativel strong acid, so, in order to separate this from a sample we need to use a base to do so. However, it's very important the use of the base here, we cannot use any base to do it, for the main reason that the sample has other component, and this component may react too with the base and the separation will not be succesfull. So, as the chlorobenzoic acid is a relatively strong acid, if we use a strong base such as NaOH, this will react with the acid, but it will also react with the 1,4-dichlorobenzene forming a Sn2 product and a salt like this.

C₆H₄Cl₂ + NaOH ---------> C₆H₅OCl + NaCl

This is the reason which we cannot use NaOH, because it's a strong base that may react with other compounds in the sample.

To solve this, we can use a weak base such as NaHCO₃. This weak base has the strength enough to react with the benzoic acid, but not strong enough to react with the dichlorobenzene.

So, the first step is dissolve the sample in an organic solvent like ether. The next step is mixing the sample with NaHCO₃. This will cause the layer to divide into two layers. One aqueous layer that will have the product of the acid with the base, and an organic layer with the dichlobenzene. Afterward, we just need to use a strong acid like HCl, but dilluted in the aqueous layer to regenerate the acid, and in the organic layer, just heat the solution til the whole solvent evaporates completely.

The flow diagram is below in the attached picture.

Answer:

[tex]P_2=50kPa[/tex]

Explanation:

Hello,

In this case, we can apply the Boyle's law in order to understand the pressure-volume relationship as an inversely proportional relationship relating the initial and the final volume:

[tex]P_1V_1=P_2V_2[/tex]

Next, we compute the final pressure P2:

[tex]P_2=\frac{P_1V_1}{V_2}=\frac{100.00kPa*500.0mL}{1000.0mL} \\\\P_2=50kPa[/tex]

Thus we validate, the higher the volume the lower the pressure.

Best regards.

Answer:

Explanation:

Adding sugar

Due to prokaryote

Explanation:

A crystal of table salt (NaCl) is dissolved in water. Which of the following statements explains why the dissolved salt does not recrystallize as long as the temperature and the amount of water stay constant?

Na+ and Cl- ions lose their charges in the water.

Water molecules surround the Na+ and Cl- ions.

Na+ and Cl- ions leave the water through vaporization.

Water molecules chemically react with the Na+ and Cl- ions.

Your answer: -

Answer: B - Water molecules surround the Na+ and Cl- ions.

Answer:

pH = 9.69

Explanation:

When pyridine (C₅H₅N) is added to water, the equilibrium that occurs is:

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) Kb = 1.7x10⁻⁹

Where Kb is defined as:

Kb = 1.7x10⁻⁹ = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]

If you have a solution of 1.4M C₅H₅N, the equilibrium concentration of each specie is:

[C₅H₅N] = 1.4 - X

[C₅H₅NH⁺] = X

[OH⁻] = X

Where X represents the reaction coordinate

Replacing in Kb expression:

1.7x10⁻⁹ = [X] [X] / [1.4 - X]

2.38x10⁻⁹ - 1.7x10⁻⁹X =  X²

0 = X² + 1.7x10⁻⁹X - 2.38x10⁻⁹

Solving for X:

X = -0.0000488M → False answer, there is no negative concentrations

X = 0.0000488M → Right answer

Thus, [OH⁻] = 0.0000488M. As pOH = -log [OH⁻]

pOH = 4.31

Knowing pH = 14 - pOH

pH = 9.69

The pH of a 1.4 M pyridine solution is 9.69. When pyridine (C₅H₅N) is added to water, the equilibrium occurs. The rate of forward reaction is equals to the rate of backward reaction.

When pyridine (C₅H₅N) is added to water, the equilibrium that occurs is:

C₅H₅N(aq) + H₂O(l) ⇌ C₅H₅NH⁺(aq) + OH⁻(aq) Kb = 1.7x10⁻⁹

Where Kb is defined as:

Kb = 1.7x10⁻⁹

Kb= [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]

If you have a solution of 1.4M C₅H₅N, the equilibrium concentration of each specie is:

[C₅H₅N] = 1.4 - x

[C₅H₅NH⁺] = x

[OH⁻] = x

Where x represents the reaction coordinate

Replacing in Kb expression:

1.7*10⁻⁹ = [x] [x] / [1.4 -x]

2.38*10⁻⁹ - 1.7x10⁻⁹x =  x²

0 = x² + 1.7*10⁻⁹x - 2.38*10⁻⁹

Solving for x:

x = 0.0000488M

Thus, [OH⁻] = 0.0000488M.

As pOH = -log [OH⁻]

pOH = 4.31

Knowing pH = 14 - pOH

pH = 9.69

Find more information about Equilibrium constant here:

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Answer:

2 HCl

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer: kg= 0.37

Explanation:

Use the molality formula.

M= m/kg

Answer:

Explanation:

∆H positive means heat is required or absorbed -- endothermic

∆H negative means heat is produced or removed -- exothermic

Answer:

It would be applied force.

Explanation:

Hope it helps:)

Answer:carbon, hydrogen and oxygen

Explanation:

Alcohol functional group is composed of carbon, hydrogen and oxygen

Answer:

400 kPa

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

Answer:

400 kPa

The answer is right since its from ck12

Answer:

.01

Explanation:

H30+=10^-pH

- Hope that helps! Please let me know if you need further explanation.

Answer:

400 mL

Explanation:

Boyle's Law: [tex]P_1*V_1 = P_2*V_2[/tex]

Let x = the resulting volume

350 (200) = 700 (x)

x = 400 mL

Answer:

By reacting carbon monoxide and hydrogen the formation of methanol takes place, the reaction is,

CO(g)+2H₂(g)⇔CH₃OH (g)

Based on the given reaction, one mole of methanol is obtained by reacting one mole of carbon monoxide (CO) with the two moles of hydrogen (H₂). It is mentioned in the question that for the reaction 500 mol of carbon monoxide and 750 moles of hydrogen are present.

Therefore for 500 moles of carbon monoxide, there is a requirement of 2 × 500 moles of hydrogen, which is equivalent to 1000 moles of hydrogen (H₂). However, only 750 moles of hydrogen is present. Therefore, the limiting reactant in the given case is H₂. The present moles of H₂ will react with 0.5 × 750 moles of CO = 375 mole of CO

The additional or excess concentration of CO, which is the excess reactant will be, 500-375 = 125 moles.

Answer:

Sulfur Trioxide

Explanation:

Sulfur trioxide is a gas that reacts with liquid water to produce aqueous sulfuric acid,or acid rain.

Answer:

The correct option is 1, since by changing the partial pressures the gas pressures change, the gases go from the zones of higher partial pressure to the zones of lower partial pressure, an example of this is the homeostasis of the human pulmonary alveolus in gas exchange with CO2 and O2.

Explanation:

In the first it increases, in the second the volume is maintained, and in the third reaction it decreases.

Answer:

1. Increase volume.

2. No change.

3. No change.

Explanation:

Hello,

In this case, if we want to shift the reaction rightwards, based on the Le Chatelier's principle we would have to:

1. For this reaction:

[tex]CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)[/tex]

- Increase the volume or decrease the pressure, since there are more gaseous moles at the products.

2. For this reaction:

[tex]S(s) + 3F_2(g) \rightleftharpoons SF_6(g)[/tex]

- Do nothing since it is not possible to achieve it as we have the same number of gaseous moles at both reactants and products.

3. For this reaction:

[tex]Cl_2(g) + I_2(g)\rightleftharpoons 2ICl(g)[/tex]

- Do nothing since it is not possible to achieve it as we have the same number of gaseous moles at both reactants and products.

Regards.