Given that S is the central atom, draw a Lewis structure of OSF4 in which the formal charges of all atoms are zero. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons.

Given That S Is The Central Atom, Draw A Lewis Structure Of OSF4 In Which The Formal Charges Of All Atoms

Answers

Answer 1

Answer:

Here's what I get  

Explanation:

A Lewis structure shows the valence electrons surrounding the atoms.

Your structure has two problems:

It shows too many valence electrons It violates the octet rule for O — there are 10 electrons around the O atom.

Here's one way to draw a Lewis structure.

1. Draw a trial structure

Make F and O terminal atoms and give each one an octet (Fig. 1).

2. Count the valence electrons in the trial structure

5 BP + 15 LP  = 10 + 30 = 40 electrons

3. Check the number of valence electrons available  

1 S =   1 × 6 =  6 electrons

1 O =   1 × 6 =   6

4 F  = 4 × 7 = 28                    

     TOTAL = 40 electrons

The trial structure has the correct number of electrons.

4. Determine the formal charge on each atom.

To get the formal charges, we cut the covalent bonds in half.

Each atom gets the electrons on its side of the cut.

Formal charge = valence electrons in isolated atom - electrons on bonded atom

FC = VE - BE  

(a) On S

VE = 6

BE = 5 bonding electrons = 5

FC = 6 - 5 = +1

(b) On O:

VE = 6

BE = 3 LP(six electrons) + 1 bonding electron  = 7

FC = 6 - 7 = -1

(c) On F:

VE = 6

BE = 3 lone pairs(6 electrons) + 1 bonding electron = 6 + 1 =7

FC = 7 - 7 = 0

5. Minimize the formal charges

We must rearrange the valence electrons so that S gets one more and O gets one fewer.

Move a lone pair from the O to make an S=O double bond (Fig. 2).

6. Recalculate the formal charges

(a) On S

VE = 6

BE =  (3 bonding electrons) = 6

FC = 6 - 6 = 0

(b) On O:

VE = 6

BE = 2 LP(four electrons) + 2 bonding electrons = 6  

FC = 6 - 6 = 0

Fig. 2 shows the Lewis structure in which all atoms have a formal charge of  zero.

Given That S Is The Central Atom, Draw A Lewis Structure Of OSF4 In Which The Formal Charges Of All Atoms
Answer 2

The formal charge of the atoms can be concluded zero with the bond formation between the sulfur and oxygen atom.

The lewis structure can be defined as the dot structure of the valence bond with the bonded atoms. The formal charge can be calculated with the difference in the valence electrons and the bonding electrons.

The formal charge of an atom can be zero when the valence electrons and the bonding electrons are equal. In the structure of [tex]\rm OSF_4[/tex], the formal charge has been assigned zero with the bond formation resulting in the valence electrons and bonding electrons being equal.

The lewis structure with the central S atom has been attached.

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Given That S Is The Central Atom, Draw A Lewis Structure Of OSF4 In Which The Formal Charges Of All Atoms

Related Questions

How many moles of h2 can be formed if a 3.25g sample of Mg reacts with excess HCl

Answers

Answer:

0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

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

Mg: 1 moleHCl: 2 molesMgCl₂: 1 moleH₂: 1 mole

Being:

Mg: 24. 31 g/moleH: 1 g/moleCl: 35.45 g/mole

the molar mass of the compounds participating in the reaction is:

Mg: 24.31 g/moleHCl: 1 g/mole + 35.45 g/mole= 36.45 g/moleMgCl₂: 24.31 g/mole + 2*35.45 g/mole= 95.21 g/moleH₂: 2*1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:

Mg: 1 mole* 24.31 g/mole= 24.31 gHCl: 2 moles* 36.45 g/mole= 72.9 gMgCl₂: 1 mole* 95.21 g/mole= 95.21 gH₂: 1 mole* 2 g/mole= 2 g

Then you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?

[tex]moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}[/tex]

moles of H₂= 0.134

0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl

0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.

Let's consider the balanced equation between Mg and HCl.

Mg + 2 HCl ⇒ MgCl₂ + H₂

The molar mass of Mg is 24.3 g/mol. The moles corresponding to 3.25 g of Mg are:

[tex]3.25 g \times \frac{1mol}{24.3g} = 0.134 mol[/tex]

The molar ratio of Mg to H₂ is 1:1. The moles of H₂ formed by 0.134 moles of Mg are:

[tex]0.134 mol Mg \times \frac{1molH_2}{1molMg} = 0.134molH_2[/tex]

0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.

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A boy with pneumonia has lungs with a volume of 1.7 L that fill with 0.070 mol of air when he inhales. When he exhales, his lung volume decreases to 1.3 L. Enter the number of moles of gas that remain in his lungs after he exhales. Assume constant temperature and pressure.

Answers

Answer:

0.053moles

Explanation:

Hello,

To calculate the number of moles of gas remaining in his after he exhale, we'll have to use Avogadro's law which states that the volume of a given mass of gas is directly proportional to its number of moles provided that temperature and pressure are kept constant. Mathematically,

V = kN, k = V / N

V1 / N1 = V2 / N2= V3 / N3 = Vx / Nx

V1 = 1.7L

N1 = 0.070mol

V2 = 1.3L

N2 = ?

From the above equation,

V1 / N1 = V2 / N2

Make N2 the subject of formula

N2 = (N1 × V2) / V1

N2 = (0.07 × 1.3) / 1.7

N2 = 0.053mol

The number of moles of gas in his lungs when he exhale is 0.053 moles

A sample of carbon dioxide gas at a pressure of 879 mm Hg and a temperature of 65°C, occupies a volume of 14.2 liters. Of the gas is cooled at constant pressure to a temperature of 23°C, the volume of the gas sample will be

Answers

Answer:

The correct answer is 12.43 Liters.

Explanation:

Based on the given question, the volume V₁ occupied by the sample of carbon dioxide gas is 14.2 liters at temperature (T₁) 65 degree C or 65+273 K = 338 K.  

The gas is cooled at a temperature (T₂) 23 degree C or 273+23 K = 296 K

The volume of the gas (V₂) after cooling can be determined by using the formula,  

V₁/T₁ = V₂/T₂

14.2/338 = V₂/296

0.0420 = V₂/296

V₂ = 0.0420 * 296  

V₂ = 12.43 Liters.  

¿What are the units that make up the 3 quantities (mass, volume of a substance and density)?

Answers

Answer:

Grams , centimeters cubed, and grams per centimeter

Explanation:

ions always have the same electronic structure as elements in which group of the periodic table?​

Answers

Answer:

In 0 group of the periodic table

Explanation:

So they will not react with other atoms because they have a full outer shell of electrons and an overall charge of 0.

Hope it helps.

Commercially available hot packs are simple in design: a pouch with water on one side, isolated by a barrier from a specific salt on the other side. When the barrier is broken, the salt and water mix and the solution becomes hot. Select the statement that is true about the reaction occurring in hot packs.A. The hot pack absorbs heat from the environment.
B. The reaction between the salt and water is exothermic.
C. The reaction between the salt and water is endothermic.
D. The reaction between the salt and water is neither endothermic nor exothermic.

Answers

Answer:

b :)

Explanation:

an exothermic reaction is when heat/light is produced. heart is produced from this reaction so it is exothermic

The correct answer is option c. The reaction between salt and water is endothermic.

How are salt and water endothermic?

Chemical reaction

Salt is a hygroscopic material that attracts water and heat from its surroundings. This is why leather shoes are "dry" and wrinkled when covered with salt. This type of reaction is known as an endothermic reaction, which is a chemical reaction that absorbs heat.

Endothermic reaction: A chemical reaction that absorbs heat is called an endothermic reaction. The temperature will drop. For example, (i) heating nitrogen and oxygen together to a temperature of about 3000 ° C produces nitrogen oxide gas. N2 + O2 + heat → 2NO (g)

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Phosphofructokinase is a four‑subunit protein with four active sites. Phosphofructokinase catalyzes step 3 of glycolysis, converting fructose‑6‑phosphate to fructose‑1,6‑bisphosphate. Phosphoenolpyruvate (PEP) is the product of step 9 of glycolysis. The PEP concentration in the cell affects phosphofructokinase activity.Select the true statements about PEP regulation of phosphofructokinase.
1. PEP is a feedback inhibitor of phosphofructokinase.
2. The apparent affinity of phosphofructokinase for its substrate increases when PEP binds.
3. PEP is a positive effector of phosphofructokinase.
4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.
5. PEP competes with fructose-6-phosphate for the active site of phosphofructokinase.
6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.

Answers

Answer:

1. PEP is a feedback inhibitor of phosphofructokinase.

4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.

6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.

Explanation:

Phosphofructokinase-1, PFK-1, is an allosteric enzymes composed of four protein subunits.

Allosteric enzymes are enzymes that function through non-covalent binding of allosteric modulators which may be activators or inhibitors. They produce a characteristic velocity versus substrate sigmoidal curve. PFK-1 has a separate binding site for its substrate, fructose-6-phosphate and it's allosteric modulators: ATP, ADP or phosphoenolpyruvate, PEP.

The enzyme can exist in two conformations, the T-state (tense) or the R-state (resting). Binding of substrate causes a conformational change from T-state to R-state, whereas binding of allosteric inhibitors returns it to the T-state.

PEP, the product of step 9 in glycolysis, is an allosteric inhibitor of PFK-1. When it binds to the the allosteric site, it leads to conformational changes in PFK-1 from the R-state to the T-state which reduces the enzymes ability to bind the substrate. These changes are responsible for the sigmoidal velocity/substrate curve in allosteric enzymes.

Therefore, the true statements from the options above are 1, 4, 6.

Options 2,3 and 5 are wrong because PEP is a negative effector of PFK-1, thus its binding reduces the affinity of PFK-1 for its substrate. Also, PFK-1 being an allosteric enzyme has separate binding sites for its substrate and its modulators. Thus, there is no competition for active site binding by substrate and modulators.

A tank at is filled with of dinitrogen monoxide gas and of boron trifluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Round each of your answers to significant digits.

Answers

Answer:

(1). Mole fraction = 0.152 for sulfur tetrafluoride gas.

Mole fraction = 0.848 For dinitrogen monoxide gas.

(2). Partial Pressure for dinitrogen monoxide gas = 187 kPa

Partial Pressure for sulfur tetrafluoride gas = 33.4 kpa.

(3). Total Partial Pressure = 220.4 kpa.

Explanation:

So, we are given the following data or parameters or information in the question above;

• Volume of the tank = 5.00L per tank;

• Temperature of the tank = 7.03°C;

• The mass of the content in the tank =

17.7g of dinitrogen monoxide gas and

7.77g of sulfur tetrafluoride gas.

So, we will be making use of the formulae below to calculate the MOLE FRACTION:

Moles, n= mass/molar mass and mole fraction = n(1)/ n(1) + n(2) per each constituents.

Moles, n1 = 17.7g of dinitrogen monoxide gas/ 44 grams per mole. =0.4023 moles.

Moles, n2 = 7.77g of sulfur tetrafluoride gas/ 108.1 grams per mole. = 0.07188 moles.

Total numbers of moles = n1 + n2 = 0.47415 moles

Mole fraction =0.4023 / 0.47415 = 0.848 of dinitrogen monoxide gas.

Mole fraction = 0.07188/0.47415 = 0.152 of sulfur tetrafluoride gas.

PART TWO: CALCULATE THE PARTIAL PRESSURE AND TOTAL PRESSURE BY USING THE FORMULA BELOW;

pressure × volume = number of moles × gas constant, R × temperature.

Pressure = n × R × T/ V.

For dinitrogen monoxide gas. ;

Partial Pressure = 0.4023 × 8.314 × 280.03 / 5 × 10^-3 = 187 kPa.

For sulfur tetrafluoride gas

Partial Pressure = 0.07188 × 8.314 ( × 280.03 / 5 × 10^-3. = 33.4 kpa.

(3). Total pressure = (187 + 33.4)kpa = 220.4 kpa

The acetate ion is the conjugate base of the weak acid acetic acid. The value of Kb for CH3COO-, is 5.56×10-10. Write the equation for the reaction that goes with this equilibrium constant.

Answers

Answer: The equation for the reaction that goes with this equilibrium constant is [tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

Explanation:

[tex]CH_3COOH\rightarrow CH_3COO^-+H^+[/tex]

Here [tex]CH_3COOH[/tex] donates a proton and thus behaves as an acid and forms [tex]CH_3COO^-[/tex] which is called as the conjugate base of [tex]CH_3COOH[/tex]

The dissociation constant of acids is given by the term [tex]K_a[/tex] and the dissociation constant of bases is given by the term [tex]K_b[/tex] and is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

[tex]K_a[/tex] for  [tex]CH_3COOH[/tex] :

[tex]K_a=\frac{[CH_3COO^-]\times [H^+]}{[CH_3COOH]}[/tex]

[tex]CH_3COO^-+H^+\rightarrow CH_3COOH[/tex]

[tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

[tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

The equation for the reaction that goes with this equilibrium constant is [tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

Draw the Lewis structure of H2O. Include any nonbonding electron pairs. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. - CHONSPFBrClIXMore Request Answer Part B What is the electron geometry of H2O

Answers

Answer:

Concepts and reason

Lewis structure is a structure that explains the bonding between atoms of a molecule and lone pair of electrons that is present in the molecule is called a Lewis structure.

With the help  of Lewis structure the electronic geometry of a molecule can be determine.

Fundamentals

According to Lewis structure, every atom and their position in the structure of a molecule by using its chemical symbol.

Lines connecting the atoms that are bonded to them are drawn. Lone pairs are expressed by pairs of dots and are located beside the atoms.

Lewis structure  of [tex]H_{2}O[/tex] is, the total number of valence electrons is eight in [tex]H_{2}O[/tex].

A gas company in Massachusetts charges $2.80 for 15.0 ft3 of natural gas (CH4) measured at 20.0°C and 1.00 atm. Calculate the cost of heating 2.00 × 102 mL of water (enough to make a cup of coffee or tea) from 20.0°C to 100.0°C. Assume that only 50.0% of the heat generated by the combustion is used to heat the water; the rest of the heat is lost to the surroundings. Assume that the products of the combustion of methane are CO2(g) and H2O(l).

Answers

Answer:

$0.0238

Explanation:

The energy you need to increase the temperature of water from 20°C to 100°C is obtained from:

Q = C×m×ΔT

Where Q is the energy, C is specific heat of water (4.184J/g°C), m is mass of water (2.00x10²g - Density of water 1g/mL), ΔT is change in temperature (100.0°C - 20.0°C)

Replacing:

Q = 4.184J/g°C × 2.00x10²g × 80.0°C

Q = 66944J = 66.944kJ

As you are assuming the energy of combustion will be just 50.0% to heat the water the energy you need is 66.944kJ × 2 = 133.888kJ

The combustion of methane is:

CH4(g) + 2O2(g) ⟶ CO2(g) + 2H2O(l) ΔH = −890.8kJ

That means 1 mole of methane produce 890.8kJ. As you need 133.888kJ, moles of methane are:

133.888kJ × (1 mol CH₄ / 890.8kJ) = 0.150 moles of CH₄.

Using PV = nRT, moles of 15.0ft³ (424.8L) at 20.0°C (293.15K) and 1.00atm:

1.00atmₓ424.8L = moles CH₄ₓ0.082atmL/molKₓ293.15K

17.67 = moles CH₄

As 17.67 moles of CH₄ cost $2.80, the cost of 0.150 moles of CH₄ is:

0.150 moles CH₄ ₓ ($2.80 / 17.67 moles) =

$0.0238

Question 8
1 pts
A closed flask contains a 0.25 moles of O2 which exerts a pressure of
0.50 atm. If 0.75 moles of CO, is added to the container what is the
total pressure in the flask?​

Answers

Answer:

\large \boxed{\text{2.0 atm}}  

Explanation:

We can use Dalton's Law of Partial Pressures:

Each gas in a mixture of gases exerts its pressure separately from the other gases.

0.25 mol of O₂ exerts 0.50 atm.

If you add 0.75 mol of CO, the total amount of gas is  

0.25 mol + 0.75 mol = 1.00 mol

[tex]p_{\text{total}} = \text{1.00 mol} \times \dfrac{\text{0.50 atm}}{\text{0.25 mol}}= \textbf{2.0 atm}\\\\\text{The total pressure in the flask is $\large \boxed{\textbf{2.0 atm}}$}[/tex]

 

The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.

Partial pressure can be defined as the pressure exerted by each gas in a given solution.

The total moles of gas in the container by the addition of CO has been:

Total moles = moles of oxygen + moles of CO

Total moles = 0.25 + 0.75

Total moles = 1 mol.

By using Dalton's law of partial pressure:

Total pressure = total moles [tex]\rm \times\;\dfrac{pressure\;of\;oxygen}{moles\;of\;oxygen}[/tex]

Total pressure = 1 [tex]\rm \times\;\dfrac{0.50}{0.25}[/tex]

Total pressure = 2 atm.

The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.

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The transfer of surface water into the ground to become groundwater is known as
and it can replenish an aquifer.

Answers

Answer: Recharge

Explanation:

To solve this we must be knowing each and every concept related to groundwater recharge. Therefore, the transfer of surface water into the ground to become groundwater is known as groundwater recharge.

What is groundwater recharge?

The water that is added to the aquifer and through unsaturated zone after percolation (or infiltration) following any storm rainfall event is known as groundwater recharge.

In the natural world, rivers, lakes, streams, rain, and snowmelt all contribute to groundwater recharge. Other surface water trickles and through soil, eventually connecting with a source of water underneath the surface, while other surface water has evaporated or enters another watershed.

Therefore, the transfer of surface water into the ground to become groundwater is known as groundwater recharge.

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Balance the following chemical equation:
NH4NO3
N20+
H2O

Answers

Answer:

NH4NO3 = N2O + 2(H2O)

Explanation:

there are 2 N, 4 H, 3 O

Answer:

NH4NO3=N2O+2H2O

Explanation:

N-2,O-3,H-4



Pick the odd one out?


Ethanol

Hexane

Oil

Carbon tetrachloride

Answers

Answer: Ethanol is the odd one out.

Explanation:

A polar compound is defined as the compound which is formed when there is a difference of electronegativities between the atoms. It is also defined as the bond which is formed due to the unequal sharing of electrons between the atoms.

Non-polar compound is defined as the compound which is formed when there is no difference of electronegativities between the atoms or the polarities cancel out.

Hexane [tex](C_6H_{14}), Oil (mixture of hydrocarbons) and carbon tetrachloride [tex](CCl_4)[/tex] all are non polar whereas ethanol is polar due to electronegative difference between hydrogen and oxygen.

Identify the person who made the correct statement.

Mike said petrified fossils are hard and heavy like rock.

Bobby said that petrified fossils have the same appearance as when they were alive.

Neither Mike nor Bobby is correct.
Mike is correct.
Bobby is correct.
Both Mike and Bobby are correct.

Answers

Answer: Both Mike and Bobby are correct.

Explanation:

Petrifcation can be defined as the process in which the organic material of the dead living being becomes fossil by the replacement of mineral deposition in the bony, hard material.

Thus although the body components gets decomposed wiped out due to this process. The body shape of the dead organism remains the same as that was in living.

Thus the statements made by Mike and Bobby both are correct. The fossils are hard and have the same appearance as when they were alive.

Calculate the Kc for the following reaction if an initial reaction mixture of 0.500 mole of CO and 1.500 mole of H2 in a 5.00 liter container forms an equilibrium mixture containing 0.198 mole of H2O and corresponding amounts of CO, H2, and CH4.

Answers

Answer:

4.41

Explanation:

Step 1: Write the balanced equation

CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

Step 2: Calculate the respective concentrations

[tex][CO]_i = \frac{0.500mol}{5.00L} = 0.100M[/tex]

[tex][H_2]_i = \frac{1.500mol}{5.00L} = 0.300M[/tex]

[tex][H_2O]_{eq} = \frac{0.198mol}{5.00L} = 0.0396M[/tex]

Step 3: Make an ICE chart

        CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

I       0.100      0.300        0            0

C         -x           -3x          +x          +x

E    0.100-x    0.300-3x     x            x

Step 4: Find the value of x

Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396

Step 5: Find the concentrations at equilibrium

[CO] = 0.100-x = 0.100-0.0396 = 0.060 M

[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M

[CH₄] = x = 0.0396 M

[H₂O] = x = 0.0396 M

Step 6: Calculate the equilibrium constant (Kc)

[tex]Kc = \frac{[CH_4] \times [H_2O] }{[CO] \times [H_2]^{3} } = \frac{0.0396 \times 0.0396 }{0.060 \times 0.181^{3} } = 4.41[/tex]

Which of the following is named using the unmodified element name and adding the word "ion"? Select the correct answer below:

a. simple cations (monatomic cations of elements of only one possible charge)
b. simple anions (monatomic anions of elements of only one possible charge)
c. simple protons
d. simple neutrons

Answers

Answer:

simple cations (monatomic cations of elements of only one possible charge)

Explanation:

Simple cations (monatomic cations of elements of only one possible charge)  are named using the unmodified element name and adding the word "ion"

For example, the Na+ is named the sodium ion.

An atom or molecule with a net electric charge as a result of the loss or gain of one or more electrons is known as an ion.

Consider the following system at equilibrium: P(aq)+Q(aq)⇌3R(aq) Classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the net reaction. Drag the appropriate items to their respective bins.
Items:1) Increase [P]2) Increase [Q]3) Increase [R]4) Decrease [P]5) Decrease [Q]6) Decrease [R]7) Triple [P] and reduce [Q] to one third8) Triple both [Q] and [R]

Answers

Explanation:

P(aq)+Q(aq)⇌3R(aq)

This problem involves applying LeChatelier's principle.

LeChatelier's principle states that whenever a system in equilibrium is disturbed, the equilibrium position would change in order to annul that change.

1) Increase [P]

This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.

2) Increase [Q]

This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.

3) Increase [R]

This would cause the equlibrium position to shift to the left. This is because more products have been formed, to annul that change more reactants have to be formed.

4) Decrease [P]

This would cause the equlibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.

5) Decrease [Q]

This would cause the equilibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.

6) Decrease [R]

This would cause the equilibrium position to shift to the right. This is because there are now less products, to annul that change more products have to be formed.

7) Triple [P] and reduce [Q] to one third

No shift in the direction of the net reaction because both changes cancels each other.

8) Triple both [Q] and [R]

No shift in the direction of the net reaction because both changes cancels each other.

A laser is used in eye surgery to weld a detached retina back into place. The wavelength of the laser beam is 503 nm, while the power is 1.4 W. During surgery, the laser beam is turned on for 0.070 s. During this time, how many photons are emitted by the laser?

Answers

Answer:

Number of proton emmitted by laser=[tex]2.48*10^17proton[/tex]

Explanation:

Energy is the ability to cause change; power is directly proportional to energy and its the rate energy is utilized.

Power=energy/time.

First we need to calculate the total energy used which is equal to the total power utilized.

E(total)= P( total) = 1.4W × 0.070 s =[tex]0.098J[/tex]

CHECK THE ATTACHMENT FOR THE REMAINING DETAILED CALCULATION

Which of the following is an example of a mechanical wave?
O A. A light ray
B. A seismic wave
C. A radio wave
D. An X-ray

Answers

Answer:

A seismic wave

Explanation:

It requires a medium for its propagation.

Consider each pair of compounds listed below and determine whether a fractional distillation would be necessary to separate them or if a simple distillation would be sufficient.

a. Ethyl acetate and hexane
b. Diethyl Ether and 1-butanol
c. Bromobenzene and 1,2-dibromobenzene

Answers

It’s b the answer Distillationis a process of separation of liquids having significantly different boiling points.SimpleDistillationis used if the components have widely different

Consider the following reaction where Kc = 2.90×10-2 at 1150 K: 2 SO3 (g) 2 SO2 (g) + O2 (g) A reaction mixture was found to contain 4.71×10-2 moles of SO3 (g), 5.00×10-2 moles of SO2 (g), and 4.53×10-2 moles of O2 (g), in a 1.00 liter container.

Answers

Answer:

The reaction is not in equilibrium and  must move in a backward manner i.e towards the reactant so that it will attain equilibrium

Explanation:

The complete question is as follows;

Consider the following reaction where Kc = 2.90×10-2 at 1150 K: 2 SO3 (g) 2 SO2 (g) + O2 (g) A reaction mixture was found to contain 4.71×10-2 moles of SO3 (g), 5.00×10-2 moles of SO2 (g), and 4.53×10-2 moles of O2 (g), in a 1.00 liter container.

Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc, equals . The reaction A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.

Solution

The first thing to do here is to calculate the pressure of each of the gases. This would be useful in the equilibrium calculations. We calculate this by dividing the respective number of moles by the volume of the container.

Now, since the volume of the container is 1L, then the number of moles will be equal to the pressure of the gaseous substances, although units will be different.

So, [SO3] = 4.71 * 10^-2 mol/L

[SO2] = 5.00 * 10^-2 mol/L

[O2] = 4.53 * 10^-2 mol/L

The equation of the reaction is as follows;

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

We proceed to calculate the reaction quotient Qc

Mathematically Qc for this reaction = [[tex]SO_{2}[/tex]]^2 × [[tex]O_{2}[/tex]]/ [[tex]SO_{3}[/tex]]^2

Qc = {(5 * 10^-2)^2 * (4.53 * 10^-2)}/ (4.71 * 10^-2)^2 = 5.11 × 10^-2 mol/L

Now, we are given that the value of Kc = 2.9 * 10^-2 which is less than Qc

Since Kc < Qc, the backward reaction is favored.

Now to the question;

The reaction is not in equilibrium and  must move in a backward manner i.e towards the reactant  so that it will attain equilibrium

Three different students determined the density of a metal object. Here are their results: 15.12 g/mL, 15.09 g/mL, and 15.12 g/mL. The actual density of the object was 14.41 g/mL. Calculate the percent error. Make sure to include units with your answer, units are %.

Answers

Answer:

The correct answers are 4.93 %, 4.72 % and 4.93 %.

Explanation:

Based on the given question, 14.41 g per ml is the actual density of the object. However, the density determined by three different students of the object is 15.12 g per ml, 15.09 g per ml, and 15.12 g per ml. The percent error can be calculated by using the formula,  

% error = (actual value - calculated value) / actual value * 100

By 1st student, the calculated value is 15.12 g per ml, the percent error will be,  

% error = (14.41 - 15.12) / 14.41 * 100  

= 0.71/14.41 * 100

= 4.93 %

By 2nd student, the calculated value is 15.09 g per ml, the percent error will be,  

% error = (14.41-15.09)/14.41 * 100

= 0.68/14.41 * 100

= 4.72 %

By 3rd student, the calculated value is 15.12 g per ml, the percent error will be,  

% error = (14.41-15.12)/14.41 * 100

= 0.71/14.41 * 100

= 4.93 %

A patient arrives in the emergency room with a burn caused by steam. Calculate the heat that is released when 14.5 g of steam at 100. ∘C hits the skin, condenses, and cools to body temperature of 37.0 ∘C. (For water, 540 cal (2260 J) is needed to convert 1 g of water to vapor at 100 ∘C.)

Answers

Answer:36,592.1J or 36.5921KJ

Explanation:

first convert to steam

14.5 g of steam  at 100∘C

To covert to water vaporor steam, becomes

14.5g x 2260 J of energy per gram of steam

=32,770J

Also, Quantity of heat released when the temperature is reduced from 100 ∘C to 37 ∘C, we will use the formulae,

q= m C ΔT

Where specific heat capacity of water C = 4.184 J/g.C

mass= 14.5g

Change in temperature= 100∘C-37∘C= 63∘C

we will now have  

= 14.5 g x 4.184 J/g.Cx ( 100 - 37) C = 3,822.084 J= 3822.1J

Therefore total energy released = 32,770 J + 3822.1 J =   36,592.1J

OR converting to KJ becomes=36,592.1/1000=36.592KJ

A balanced equation has

Answers

Answer:

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge is the same for both the reactants and the products.In other words, the mass and the charge are balanced on both sides of the reaction.

Explanation:

molar mass of A1C1 3

Answers

Answer:

Gold(III) chloride

Experiment predicted observation A student has two unopened cans containing carbonated water. Can A has been stored in the garage () and can B has been stored in the fridge (). The student opens one can at the time, both cans make a fizz.
A) The fizz will be the same for both cans
B) There is not enough information to predict which can will make the louder fizz
C) Can A will make a louder and stronger fizz than can B.
D) Can B will make a louder and stronger fizz than can A.

Answers

Answer:

Can A will make a louder and stronger fizz than can B.

Explanation:

Temperature has a direct effect on gas solubility. We know that carbonated water contains carbon dioxide dissolved in water. The extent of dissolution or solubility of this gas is dependent on the temperature of the system.

As the temperature of the system rises, the solubility of gas in solution decreases. It follows that can A, having been stored in a garage is definitely at a higher temperature than can B stored in the refrigerator.

Since solubility of gases decreases with increasing temperature, the carbon dioxide in can A will be less soluble than in can B. This will cause can A to make a louder and stronger fizz when opened than can B.

what type of bonds do compounds formed from non metal consist of?​

Answers

Compounds formed from non-metals consist of molecules. The atoms in a molecule are joined together by covalent bonds. These bonds form when atoms share pairs of electrons.

How many moles of solute are contained in the following solution: 15.25 mL of a 2.10 M CaCl₂

Answers

Answer:

0.032moles

Explanation:

2.10moles in 1000ml what about 15.25ml

(15.25×2.10)÷1000

0.032moles

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