How can a cation be formed?
Add 1 clectron
Take away 1 clectron
Add 1 proton
Take away 1 prekon

Answer:  The balanced equation for the given reaction is

[tex]2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)[/tex].

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side.

For example, [tex]SO_{2}(g) + O_{2}(g) \rightarrow SO_{3}(g)[/tex]

Here, number of atoms on reactant side are as follows.

Number of atoms on product side are as follows.

To balance this equation, multiply [tex]SO_{2}[/tex] by 2 on reactant side and multiply [tex]SO_{3}[/tex] by 2. Hence, the equation will be re-written as follows.

[tex]2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)[/tex]

Here, number of atoms on reactant side are as follows.

Number of atoms on product side are as follows.

Now, there are same number of atoms on both reactant and product side. So, this equation is balanced.

Thus, we can conclude that the balanced equation for the given reaction is [tex]2SO_{2}(g) + O_{2}(g) \rightarrow 2SO_{3}(g)[/tex].

Answer:

Explanation:

First we convert 134 lb into kg:

Then we convert 275 mg into g:

Now we can calculate the dosage in grams per kilogram of body weight, keeping in mind that two tablets are consumed:

As for the second part, first we convert 31 lb into kg:

Now we calculate how many penicillin grams should be consumed:

We convert 1.27 g of penicilin into mg:

Finally we calculate how many 150 mg tables should be taken:

Answer:

liquid, and of course

Explanation:

Answer:

1116g

Explanation:

We'll convert moles O2 -> moles of Na2O -> grams of Na2O.

Based on our balanced equation, we have 1 mole of O2 for every 2 moles of Na2O. This is our mole to mole ratio.

9 mol O2 x [tex]\frac{2 mol Na2O}{1 mol O2}[/tex] = 18 mol Na2O

We can convert mols -> grams using the molar mass of Na2O- 62g.

18 mol Na2O x [tex]\frac{62g Na2O}{1 mol}[/tex] = 1116g

Answer:

2200 L

Explanation:

There are 22.4 L in 1 mole of neon so if you have 98.2 L, then you have 98.2 mole x (22.4 L/mole) = 2200 L of neon.

Answer:

No, no precipitate is formed.

Explanation:

Hello there!

In this case, since the reaction between ammonium sulfide and aluminum nitrate is:

[tex]3(NH_4)_2S(aq)+2Al(NO_3)_3(aq)\rightarrow Al_2S_3(s)+6NH_4NO_3(aq)[/tex]

In such a way, we can calculate the concentration of aluminum and sulfide ions in the solution as shown below, and considering that the final total volume is 140.00 mL:

[tex][Al^3^+]=\frac{120.00mL*0.0082M}{140.00mL}=0.00703M[/tex]

[tex][S^2^-]=\frac{20.00mL*0.0090M}{140.00mL}=0.00129M[/tex]

In such a way, we can calculate the precipitation quotient by:

[tex]Q=[Al^3^+]^2[S^2^-]^3=(0.00703)^2(0.00129)^3=1.05x10^{-13}[/tex]

Which is smaller than Ksp and meaning that the precipitation does not occur.

Regards!

Solution :

According to Chick's law

[tex]$\frac{N_t}{N_0}=e^{-k'C^n t}$[/tex]

where, t = contact time

            c = concentration of disinfectant

            k' = lethality coefficient = 4.71

            n = dilution coefficient = 1

            4 log removal = % removal = 99.99

[tex]$\frac{N_t}{N_0}=\frac{\text{bacteria remaining}}{\text{bacteria initailly present}}$[/tex]

      = 1 - R

      = 1 - 0.9999

Now for plug flow reactor contact time,

[tex]$\tau =\frac{V}{Q} =\frac{75000}{40 \times 10^6}$[/tex]

          = 0.01875 days

          = 27 minutes

For CSTR, [tex]$\tau =\frac{V}{Q} =\frac{150000}{40 \times 10^6}$[/tex]

                             [tex]$=3.75 \times 10^{-3}$[/tex] days

                            = 5.4 minute

There are 3 reactors, hence total contact time = 3 x 5.4

                                                                            = 16.2 minute

Or [tex]$\frac{N_t}{N_0}=e^{-k'C^n t}$[/tex]

or [tex]$(1-0.9999)=e^{-4.71 \times C \times t}$[/tex]

∴ C x t = 1.955

For PFR, [tex]$t_1 = 27 $[/tex] min

∴ C [tex]$=\frac{1.955}{27}$[/tex] = 0.072 mg/L

For CSIR, [tex]$t_2=16.2$[/tex] min

[tex]$C=\frac{1.955}{16.2} = 0.1206$[/tex] mg/L

∴ Chlorine required for PFR in kg/day

[tex]$=\frac{0.072 \times 40 \times 10^6 \times 3.785}{10^6}$[/tex]      (1 gallon = 3.785 L)

= 18.25 kg/day

Therefore we should go for PFR system.

The more energy that particles have, the more they move.

Explanation:

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Answer:

+2

Explanation:

Answer:

2 electrons

Explanation:

In Iron (II) oxide, or Ferrous Oxide, or FeO, the Iron element (Fe) is bonded to the Oxygen, in the oxidation state of "2". This means that the Iron has accepted 2 electrons from the Oxygen.

Answer:25,06 kJ of energy must be added to a 75 g block of ice.

ΔHfusion(H₂O) = 6,01 kJ/mol.

T(H₂O) = 0°C.

m(H₂O) = 75 g.

n(H₂O) = m(H₂O) ÷ M(H₂O).

n(H₂O) = 75 g ÷ 18 g/mol.

n(H₂O) = 4,17 mol.

Q = ΔHfusion(H₂O) · n(H₂O)

Q = 6,01 kJ/mol · 4,17 mol

Q = 25,06 kJ.

Explanation:

Answer:

67.91 g of CuCl2; 32.09 g of Cu.

Explanation:

The two masses add to 100.0 g, the initial amount of starting material, demonstrating the law of conservation of matter.

Answer: 8%

Explanation:

Can't you just do 4/50= 8%?  If not, just remember that for empirical formula, do percent to grams, and grams to moles, divide by smallest and multiply to whole!

Answer:

The correct answer is - transparent medium.

Explanation:

A transparent substance or medium is the substance that allows light to pass through it. Light moves through these substances as they do not absorb the light and do not reflect too.

The example of such substances is glass, air or water. These substances allow light to pass through them.

Thus, The correct answer is - transparent medium.

Answer:

Explanation:

Remark

Interesting que8stion. You have to figure out how many mols are present in each reactant. Since all periodic tables are different, I'm going to use rounded numbers. If it is too close, I will go further.

NaCl

Na = 23

Cl = 35.5

1 mol = 58.5 grams

given = 50.0 grams

Mols for the reaction = 50/58.5 = 0.855

H2SO4

H2       =    2*1        2

S          =   1 * 32   32

O4       =   4*16     64

1 mol  =               98   grams

mols  present = 50/98 = 0.510

MnO2

Mn     = 1 * 55    = 55

O2   =   2*16      = 32

1 mol =                  87 grams

mols available    = 50/87 = 0.5747

Discussion

Na Cl and H2SO4 both require 2 moles for every mol of Cl2 produces.

H2SO4  has 0.51 mols available for a reaction

NaCl has  0.855 moles available for a reaction

MnO2 has 0.575 moles available for a reaction.

Given those numbers 0.510 mols of H2SO4 will only produce 0.255 mols of chlorine and the rest will be reduced in a similar manner. H2SO4 is the limiting reagent (reactant).

In other words only 0.510 moles of NaCl will be used and 0.855 - 0.510 moles will be left over on the reactants side.

only 0.575 moles of MnO2 will be used and 0.065 moles will be left over.

The oddity in the result shows up because the balance numbers in the equation give a ratio of 2 to 1 for H2SO4 and NaCl The 2 belongs to the reactants and the 1 for the chlorine.

Answer:

the answer is something  

Explanation:

Answer:

B

Explanation:

Vaccines prevent illness and disease

Answer:

c4 is an explosive..

contains RDX, DOS, DOA, and PIB.

Explanation:

Answer:

Pressure Affects the Boiling Point

Atmospheric pressure influences the boiling point of water. When atmospheric pressure increases, the boiling point becomes higher, and when atmospheric pressure decreases (as it does when elevation increases), the boiling point becomes lower.

Explanation:

i think it will help you

Answer:

155.1 L

Explanation:

Step 1: Given data

Mass of CO₃: 415.4 g

Step 2: Calculate the moles corresponding to 415.4 g of CO₃

The molar mass of CO₃ is 60.01 g/mol.

415.4 g × 1 mol/60.01 g = 6.922 mol

Step 3: Calculate the volume occupied by 6.922 moles of CO₃

The volume of a gas depends on conditions such as Temperature and Pressure. Since the conditions are not specified, we may assume that the gas is at Standard Pressure and Temperature (1 atm and 273.15 K). At STP, 1 mole of a gas occupies 22.41 L.

6.922 mol × 22.41 L/1 mol = 155.1 L

Answer:

0.500 moles

Explanation:

In order to convert grams of any given substance into moles, we need the substance's molar mass:

We can find the molar masses of each element in the periodic table:

Now we can divide the given mass by the molar mass in order to calculate the number of moles:

The temperature of a gas sample rises, the molecules quicken, and as a result, the root mean square molecular speed rises.

Degree of warmth or coolness determined by a specific scale

The temperature of a gas sample rises, the molecules quicken, and as a result, the root mean square molecular speed rises.

In a gas, the speed of the molecules is inversely related to the molar mass of the gas and inversely proportional to temperature.

The rate of diffusion is inversely related to the square root of the gas's molar mass, according Graham's law of Diffusion.

To learn more about temperature, refer to the below link:

https://brainly.com/question/13122916

# SPJ2

Answer:

a. True

Explanation:

Palmitic acid is the acid found in animals and plants. It is the saturated acid that contains fats. It is extracted from palm oil and from fat sources like butter, cheese, milk and meat. It is added in milk with low fat content to add Vitamin A in it. The fat content of palmitic acid is very high. excessive consumption of palmitic acid leads to heart diseases and life risks.

Answer:

Explanation:

2 H₂S(g) +S0₂(g) =  3 S(s) +  2H₂0(g)

2 x 34 g     64 g        3 x 32 g

68 g of  H₂S reacts with 64 g of S0₂

3.89 g of H₂S reacts with 64 x 3 .89 / 68 g of S0₂

3.89 g of H₂S reacts with 3.66  g of S0₂

S0₂ given is 4.11 g , so it is in excess .

Hence H₂S is limiting reagent .

68 g of  H₂S reacts with  S0₂ to give 96 g of Sulphur

3.89 g of  H₂S reacts with  S0₂ to give 96 x 3.89 / 68 g of Sulphur

3.89 g of  H₂S reacts with  S0₂ to give 96 x 3.89 / 68 g of Sulphur

5.49 g of Sulphur is produced .

Actual yield is 4.89

percentage yield = 4.89 x 100 / 5.49

= 89 % .  

Considering the reaction stoichiometry and the definition of percent yield:

a. H₂S will be the limiting reagent.

b. the maximum mass of sulfur that can be produced in the reaction is 5.49 grams.

c. the percent yield of the reaction is 89%.

The balanced reaction is:

2 H₂S(g) + SO₂(g) → 3 S(s) +  2 H₂O(g)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of each compound is:

So, by reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 64 grams of SO₂ reacts with 68 grams of H₂S, 4.11 grams of SO₂ react with how much mass of H₂S?

[tex]mass of H_{2} S=\frac{4.11 grams of SO_{2}x 68 grams of H_{2} S }{64 grams of SO_{2}}[/tex]

mass of H₂S= 4.37 grams

But 4.37 moles of H₂S are not available, 3.89 grams are available. Since you have less mass than you need to react with 4.11 grams of SO₂, H₂S will be the limiting reagent.

Using the limiting reagent, you can apply the following rule of three: if by stoichiometry 68 grams of H₂S produce 96 grams of S, 3.89 grams of H₂S will produce how much mass of S?

[tex]mass of S=\frac{96 grams of Sx 3.89 grams of H_{2} S }{68 grams of H_{2}S}[/tex]

mass of S= 5.49 grams

So, the maximum mass of sulfur that can be produced in the reaction is 5.49 grams.

The amount of product that is obtained when all the limiting reagent reacts.  This is called the theoretical yield of the reaction. That is, the theoretical yield is the maximum amount of product that can be produced in a reaction.

On the other hand, the actual yield is the amount of product actually obtained from a reaction.

The percent yield determines the efficiency of the  reaction, and describes the ratio of the actual yield to the theoretical yield:

[tex]percent yield=\frac{actual yield}{theoretical yield}x100[/tex]

In this case, you know:

So, the percent yield can be calculated as:

[tex]percent yield=\frac{4.89 grams}{5.49 grams}x100[/tex]

Solving:

percent yield= 89%

Finally, the percent yield of the reaction is 89%.

Learn more about reaction stoichiometry:

An ideal gas is one that follows the gas laws at all conditions of temperature and pressure. To do so, the gas would need to completely abide by the kinetic-molecular theory. On the other hand, a real gas is a gas that does not behave according to the assumptions of the kinetic-molecular theory.

Furthermore, the particles of an ideal gas are extremely small and have a mass equivalent to practically zero. Ideal gas particles also have no volume.

An example of a real gas is helium, oxygen, and nitrogen.

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Answer:

28.466256

Explanation:

Answer:

Explanation:

From the given information:

The energy of photons can be determined by using the formula:

[tex]E = \dfrac{hc}{\lambda}[/tex]

where;

planck's constant (h) = [tex]6.63 \times 10^ {-34}[/tex]

speed oflight (c) = [tex]3.0 \times 10^8 \ m/s[/tex]

wavelength λ = 58.4 nm

[tex]E = \dfrac{6.63 \times 10^{-34} \ J.s \times 3.0 \times 10^8 \ m/s}{58.4 \times 10^{-9 } \ m}[/tex]

[tex]E =0.34 \times 10^{-17} \ J[/tex]

[tex]E = 3.40 \times 10^{-18 } \ J[/tex]

To convert the energy of photon to (eV), we have:

[tex]1 eV = 1.602 \times 10^{-19} \ J[/tex]

Hence

[tex]3.40 \times 10^{-18 } \ J = \dfrac{1 eV}{1.602 \times 10^{-19 } \ J }\times 3.40 \times 10^{-18 } \ J[/tex]

[tex]E = 2.12 \times 10 \ eV[/tex]

E = 21.2 eV

b)

The equation that illustrates the process relating to the first ionization is:

[tex]Hg_{(g)} \to Hg^+ _{(g)} + e^-[/tex]

c)

The 1st ionization energy (I.E) of Hg can be calculated as follows:

Recall that:

I.E  = Initial energy - Kinetic Energy

I₁ (eV) = 21.2 eV - 10.75 eV

I₁ (eV) = 10.45 eV

Since ;

[tex]1 eV = 1.602 \times 10^{-19} \ J[/tex]

∴

[tex]10.45 \ eV = \dfrac{1.602 \times 10^{-19 } \ J }{ 1 \ eV}\times 10.45 \ eV[/tex]

Hence; the 1st ionization energy of Hg atom = [tex]1.67 \times 10^{-18} \ J[/tex]

[tex]1.67 \times 10^{-21} \ kJ[/tex]

Finally;

[tex]I_1 \ of \ the \ Hg (kJ/mol) = \dfrac{1.67 \times 10^{-21 \ kJ} \times 6.02 \times 10^{23} \ Hg \ atom }{1 \ Kg \ atom }[/tex]

[tex]\mathbf{= 1.005 \times 10^3 \ kJ/mol}[/tex]