I need help with my chemistry homework. SHOW WORK

1) The density of tin (Sn) is the 5.75 g/cm^3. How many cubic decimeters (dm^3) does 17.2 kg of Sn occupy?

2) If half a liter (500 mL) of olive oil weights 425g, what is its density?

3) The density of silver is 10.5 g/cm^3. What is the mass of 0.987 dm^3 of silver ?

Answer: 35.72 % of Barium ions will be present in the original unknown compound.

Explanation: The reaction of Barium ions and sodium sulfate is:

[tex]Na_2SO_4(aq.)+Ba^{2+}(aq.)\rightarrow BaSO_4(s)+2Na^+(aq.)[/tex]

Here, Sodium sulfate is present in excess, Barium ions are the limiting reagent because it limits the formation of product.

Now, 1 mole of barium sulfate is produced by 1 mole of Barium ions.

Molar mass of Barium sulfate = 233.38 g/mol

Molar mass of Barium ions = 137.327 g/mol

233.38 g/mol of barium sulfate will be produced by 137.323 g/mol of Barium ions, so

0.4105 grams of barium sulfate will be produced by = [tex]\frac{137.327g/mol}{233.38g/mol}\times 0.4105g[/tex] of Barium ions

Mass of barium ions = 0.2415 grams

To calculate percentage by mass, we use the formula:

[tex]\% mass=\frac{\text{Mass of solute (in grams)}}{\text{Total mass of the solution(in grams)}}\times 100[/tex]

Mass of the solution = 0.6760 grams

Putting the value in above equation, we get

[tex]\% \text{ mass of }Ba^{2+}\text{ ions}=\frac{0.2415g}{0.6760g}\times 100[/tex]

% mass of Barium ions = 35.72%.

The percentage by mass of barium ion (Ba²⁺) in the unknown compound is 35.7%

We'll begin by writing the balanced equation for the reaction. This is given below:

Ba²⁺(aq) + Na₂SO₄(aq) —> BaSO₄(s) + 2Na⁺(aq)

Molar mass of Ba²⁺ = 137 g/mol

Mass of Ba²⁺ from the balanced equation = 1 × 137 = 137 g

Molar mass of BaSO₄ = 137 + 32 + (16×4)

= 233 g/mol

Mass of BaSO₄ from the balanced equation = 1 × 233 = 233 g

Thus,

From the balanced equation above,

137 g of Ba²⁺ reacted to produce 233 g of BaSO₄.

Next, we shall determine the mass of Ba²⁺ that reacted to produce 0.4105 g of BaSO₄. This can be obtained as follow:

From the balanced equation above,

137 g of Ba²⁺ reacted to produce 233 g of BaSO₄.

Therefore,

X g of Ba²⁺ will react to produce 0.4105 g of BaSO₄ i.e

X g of Ba²⁺ = [tex]\frac{137 * 0.4105}{233} \\\\[/tex]

X g of Ba²⁺ = 0.2414 g

Finally, we shall determine the percentage of Ba²⁺ in the unknown compound. This can be obtained as follow:

Mass of Ba²⁺ = 0.2414 g

Mass of compound = 0.6760 g

[tex]Percentage = \frac{mass of ion}{mass of compound} * 100\\\\= \frac{0.2414}{0.6760} * 100\\\\[/tex]

Therefore, the percentage by mass of barium ion (Ba²⁺) in the unknown compound is 35.7%

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Hey there!:

C6H12O6 + C6H12O6 ----> H2O + C12H22O11

it is a dehydration synthesis reaction, because H2O is released and C12H22O11  is formed .

Answer A

Hope  that helps!

The process of dehydration synthesis combines two smaller molecules into a larger one, eliminating a molecule of water. In this case, two glucose molecules combine to create one sucrose molecule and water, thus the correct answer is A) H2O + C12H22O11.

In the process of dehydration synthesis, two smaller molecules combine to create a larger molecule by eliminating one molecule of water. In the reaction given, two molecules of glucose (C6H12O6) undergo dehydration synthesis to produce a molecule of sucrose (C12H22O11) and water (H2O). Therefore, the correct answer for the missing molecules is A) H2O + C12H22O11.

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If the density of an object is affected by its mass, then osmium will have the highest density and hydrogen will have the lowest density

Density = mass/volume or D = m/V

If we hold V constant, we can write D = km. where k = 1/V.

Thus, D ∝ m. As the mass of a fixed volume of a substance increases, the density increases.

The mass of H₂ in 1 cm³ 0.080 g. The mass of Os in 1 cm³ is 22.6 g.

However, there are substances with even higher densities. For example, the material at the centre of a neutron star is so dense that 1 cm³ of the material has a mass of 10¹² kg!

C) [tex]\text{HCl}[/tex] hydrochloric acid and, in case that the question allows for more than one choices, D) [tex]\text{CO}_2[/tex] carbon dioxide as well.

Methane molecules [tex]\text{CH}_4[/tex] are nonpolar and barely dissolve in water.

Sodium hydrocarbonate [tex]\text{Na}\text{HCO}_3[/tex] undergoes hydrolysis to release hydroxide ions [tex]\text{OH}^{-}[/tex] that can end up consuming [tex]\text{H}^{+} \; (aq)[/tex]:

[tex]\text{NaHCO}_3 \; (aq) \to \text{Na}^{+} \; (aq) + \text{HCO}_3^{-} \; (aq)\\\text{HCO}_3^{-} \; (aq) + \text{H}_2\text{O} \; (l) \leftrightharpoons \text{H}_2\text{CO}_3\; (aq) + \text{OH}^{-} \; (aq)[/tex]

Hydrochloric acid, a strong acid, ionizes to produce protons [tex]\text{H}^{+} \; (aq)[/tex] and chloride ions when dissolved in water:

[tex]\text{HCl}\; (aq) \to \text{H}^{+} \; (aq) + \text{Cl}^{-} \; (aq)[/tex]

Carbon dioxide reacts with water to produce hydrocarbonic acid, a weak acid that slightly ionizes, also producing protons [tex]\text{H}^{+} \; (aq)[/tex] but to a significantly lesser extent than hydrochloric acid does.

[tex]\text{CO}_2\; (g) + \text{H}_2\text{O} \; (l) \leftrightharpoons \text{H}_2\text{CO}_3\; (aq)\\\text{H}_2\text{CO}_3\; (aq) \leftrightharpoons \text{H}^{+}\; (aq) + \text{HCO}_3^{-} \; (aq)[/tex]

D) Co2!

it reacts with water to form carbonic acid...

that has H+ ions!

Hey There!

Ideal gas equation: PV = nRT where P is pressure, V is volume, n is moles of gas, R is molar gas constant and T is temperature.

Since P, R and T are constant, V/n = constant

V1 = 2.00 L, V2 = 3.30 L

n1 = (2.00 g)/(4.00 g/mol) = 0.5 mol, n2 = ?

V1/n1 = V2/n2

2.00/0.5 = 3.30/n2

n2 = 0.825 mol

Moles of He added = n2 - n1

= 0.825 - 0.5 = 0.325 mol

Mass of He added = moles x atomic mass

= 0.325 * 4.00 = 1.30 g of He

The amount of helium added to the cylinder, while keeping the pressure and temperature constant and increasing the volume from 2.00L to 3.70L, is approximately 1.70 grams.

The problem you've presented involves the concept of an ideal gas law which states that the number of moles of gas is directly proportional to the volume it occupies, provided that temperature and pressure remain constant. Considering that you are keeping the pressure and temperature constant, and you have increased the volume from 2.00 L to 3.70 L, the mass of helium will also increase proportionally.

To find out how much helium was added, we first must determine the initial number of moles of helium using its given mass (2.00 g) and the molar mass of helium (around 4 g/mol). That is, initial moles of helium = 2.00 g / 4 g/mol = 0.5 mol. Since number of moles is directly proportional to volume, we can set up a proportion: initial volume/initial moles = final volume/final moles (2.00 L / 0.5 mol) = (3.70 L / x mol), which yields a solution of roughly 0.925 mol for 'x', the final number of moles.

 Subtracting the initial moles (0.5 mol) from the final moles (0.925 mol), we find that around 0.425 mol of helium has been added. The mass of the added helium equals the moles of added helium multiplied by the molar mass, i.e., 0.425 mol * 4 g/mol = 1.70 g.

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Answer:

C. transition metals

Explanation:

Answer:

c

Explanation:

i did it on usatp

answer D)

explanation:

Bases turn the red litmus to Blue

while aciDs turn blue litmus to reD.

Answer: The correct answer is Option d.

Explanation:

Litmus paper is defined as the indicator which is used to determine the nature of the solution, whether it is acidic or basic.

There are 2 types of litmus paper:

Hence, the correct answer is Option d.

The answer is D. 3.15 × 10−4

Now the dissolution of methylamine can be represented as follows:

CH3NH2 + H2O → CH3NH3OH  

The product formed dissociates and it is represented as given below

CH3NH3OH ↔ CH3CH3+ + OH-  

Now pH = 11.4 ,

Then pOH = 2.6  

[OH-] = 10^-2.6  

[OH-] = 2.51*10^-3  

Considering the below equation again

CH3NH3OH ↔ CH3CH3+ + OH-  

We can calculate Kb = [CH3NH3] [ OH-] / [CH3NH3OH]  

where Kb is the equilibrium constant.

Now [CH3NH3] = [OH-] = 2.51 × 10^-3

and [CH3NH3OH] = 0.020M (given)

Substituting these values we get  

Kb = ( 2.51 × 10^-3)² / 0.02  

Kb = 6.31 × 10^-6 / 0.02  

Kb = 3.15 × 10^-4

Mixtures of gases such as argon and helium are more likely to form an ideal solution because they experience insignificant intermolecular interactions, making their mixture thermally neutral with an increase in entropy. Option C is correct.

The intermolecular forces in ideal gases are considered to be nonexistent, and gases behave more ideally than solids or liquids. In an ideal solution, gases mix without any energy change due to similar intermolecular forces of attraction between the mixed gases as those present in the separated components.

This process is thermally neutral, and the formation of the gas solution results in increased disorder, or entropy, as gases like helium and argon will spread out and occupy a larger volume when mixed.

Hence, C. is the correct option.

The order of the electronegativity for the given elements is:

Fluorine > Phosphorous > Copper > Aluminium > Barium > Francium

Electronegativity measures the tendency of an element to attract electrons of bonded pairs towards itself. The value of electronegativity of an atom is with respect to another bonded atom.

When two atoms are bonded with each other through a bond and one of them is more electronegative than another bonded atom. Then the electron density of the bond will lie closer to the more electronegative atom.

Firstly, Fluorine is the most electronegative element in the modern periodic table. Aluminium and phosphorous are both present in the 2nd period. As we know, the electronegativity increases as we move from left to right in the periodic table.

Therefore phosphorous is more electronegative than aluminium. Copper is more electronegative than aluminium as the value of the electronegativity for copper is 1.92 and for aluminium is 1.6.

Barium and Francium lie at the bottom of the periodic table, Therefore they are the least electronegative. But the electronegativity of the Braium is more than that of Francium.

Learn more about electronegativity, here:

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Given:

Density of iron:7.874g/mL

Mass of iron : 7.75 gms

Now we know that

Volume = Mass/density

Substituting the given values in the above formula we get

Volume of iron= 7.874/7.75= 1.016 mL

Volume of iron = 0.01016 dL

The volume of a 7.75 gram iron sample, with a density of 7.874 g/mL, is approximately 0.9843 mL, which converts to approximately 0.009843 dL.

To calculate the volume of an iron sample with a mass of 7.75 grams, given the density of iron is 7.874 g/mL, use the density formula:

Density = Mass/Volume

By rearranging the formula to solve for volume, we get:

Volume = Mass/Density

Substitute in the known values:

Volume = 7.75 g / 7.874 g/mL

Volume ≈ 0.9843 mL

To convert milliliters to deciliters (dL), use the conversion factor that 1 dL = 100 mL:

Volume ≈ 0.9843 mL × (1 dL / 100 mL)

Volume ≈ 0.009843 dL

Therefore, the volume of the sample of iron in deciliters is approximately 0.009843 dL.

1 mole has 6.02*10^23 molecules in it.

1 nickel (II) chloride molecule, NiCl2, has 1 Ni atom in it.

so 1 mole of nickel (II) chloride molecule has 1 mole of Ni atom in it.

so 100 moles of nickel (II) chloride molecule has 100*6.02*10^23

= 6.02*10^25 Ni atom in it.

Nickel (II) chloride has the chemical formula of NiCl2. So 1 mole of NiCl2 contains 1 mole of Ni atoms.

1 mole contains 6.02*10^23 atoms. So 100 moles of Nickel (II) chloride contains 6.02*10^25 Ni atoms.

1. The position of equilibrium would shift to the left, making more reactants.

Cl₂(g) + I₂(g) + heat ⇌ 2ICl(g)

Le Châtelier’s Principle states that, if you add a stress to a system at equilibrium, the system will respond in a direction that will remove the stress.

Here, the stress is the removal of Cl₂. The system will respond by generating more Cl₂ to replace what was removed.

The position of equilibriumwill move to the left.

Removing Cl2(g) from the system would cause the equilibrium to shift to the left according to Le Chatelier's principle, resulting in the production of more reactants.

Removing Cl2(g) from the reaction Cl2(g) + I2(g) <=> 2 ICl(g), which is endothermic, would cause the equilibrium position to shift to the left. This is because according to Le Chatelier's principle, when a reactant is removed, the equilibrium shifts in the direction that replaces the removed substance, in this case, to the left to replenish the lost Cl2(g).

When considering the endothermic reaction Cl2(g) + I2(g) <=> 2 ICl(g), and analyzing the effect of removing Cl2(g) from the system, we apply Le Chatelier's principle. This principle states that if a change is imposed on a system at equilibrium, the system will adjust to counteract that change. In this case, removing Cl2(g) will result in the system shifting to the left to produce more reactants and re-establish equilibrium. Therefore, the correct answer is (1) the reaction would go to the left, making more "reactants".

Metals conduct heat from one to another end.....this property will help you

Heat is used to measure a mineral's thermal conductivity and specific heat to determine its metallic nature. By comparing these values with known metallic minerals and observing properties like color and luster, one can identify if a mineral is metallic.

Heat can be utilized to determine if a mineral has a metallic nature by examining its thermal conductivity and specific heat. Metallic minerals generally conduct heat more efficiently than non-metallic minerals due to the movement of free electrons. To quantify this, the specific heat of a mineral could be measured by transferring a known quantity of heat to the mineral and observing the change in its temperature.

For example, consider a scenario where a piece of metal at a known temperature is placed in water at a different temperature. By measuring the resulting temperature change in the water, and knowing the mass and specific heat of the water, the specific heat of the metal can be calculated. This value can then be compared against known values for metallic minerals to help identify the mineral in question.

If the measured specific heat is close to known values for metals such as gold or lead, additional observations such as color and luster can help refine the identification. For instance, a silver/gray color, when paired with an appropriate specific heat value, suggests a lead mineral over gold, which has a yellowish color.

A) deposition is the processes where particles of rock or laid down in sections with heavier sediments building up first

Answer:

The correct answer is option A, that is, deposition.

Explanation:

The erosion, weathering, and deposition refer to the procedures, which works spontaneously on or close to the surface of the Earth. With time, these procedures lead to the generation of sedimentary rocks.  

Deposition refers to the procedure where the rocks particles are laid down in segments with the heavier sediments forming up initially. It is the geological phenomenon in which the soil, sediments, and rocks are supplemented to a land mass or landform. Water, ice, wind, and gravity carry the previously weathered surface material that at the expense of adequate kinetic energy in the fluid gets deposited, forming up the sediment layers.  

Hey there!:

absorbance = log 100 - log Transmitance

absorbace = 0.85

log 100 = 2

- log transmitance = absorbace / log 100

0.85 / 2= 0.425

transmitance =  10 ^ ( - 0.425 )

transmitance = 0.376

D the belguim people atacked using it

correct answer is B) equal to the number of protons

Yes- as long as the roller coaster car is not powered once it leaves the station and energy loss due to resistance can be ignored.

Energies of a system:

A roller coaster car moving along a level track with an initial velocity has mechanical energy in the form of kinetic energy.

The car would slow down while gaining height as its kinetic energy converts to gravitational potential energy when moving upwards along the track. It would have no kinetic energy since all its mechanical energy are now in the form of gravitational potential energy in case it comes to a stop before reaching the top of the track.

Similarly, the car gains speed while losing height as some of its gravitational energy is converted to kinetic energy when it travels downwards.

The mechanical energy of this vehicle conserves since any of the movements, moving upslope, downslope, or level, would ensure that

Loss in Potential Energy = Gain in Kinetic Energy

and vice versa.

Final answer:

Energy is conserved in a roller coaster system when ignoring friction and air resistance. The total mechanical energy remains constant as potential energy is converted to kinetic energy and vice versa throughout the ride.

Explanation:

Yes, energy is conserved in the system of a roller coaster car moving along a track, provided that we neglect any losses due to friction or air resistance. According to the law of conservation of energy, the total mechanical energy of a closed system remains constant. This is represented by the equation E = T + U, where E stands for the total mechanical energy, T is the kinetic energy, and U is the potential energy.

At the top of the first hill, the roller coaster car has its maximum potential energy, which is then converted to kinetic energy as it descends. The conversions between potential and kinetic energy continue throughout the ride, with energy transitioning as the car goes up and down the track. At the bottom of each descent, the car reaches its maximum kinetic energy, which becomes potential energy again as the car rises.

The total mechanical energy at any point in the ride should be the same as it was at the start of the coaster's trip, assuming no energy is lost. Remember that both kinetic and potential energies are forms of mechanical energy and are measured in joules, indicating the capacity for doing work.

1) The gold will occupy a volume of 2.99 dm³.

[tex]\text{Volume = 17 200 g} \times \frac{\text{1 cm}^{3}}{\text{5.75 g}} = \text{2990 cm}^{3} = \textbf{2.99 dm}^{3}\\[/tex]

2) The density of the olive oil is 0.850 g/cm³.

[tex]\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{\text{425 g}}{\text{500 cm}^{3}}= \textbf{0.850 g/cm}^{3}[/tex]

3) The mass of the silver is 10.4 kg.

[tex]\text{Mass} = \text{987 cm}^{3} \times \frac{\text{10.5 g} }{\text{1 cm}^{3 }} = \textbf{10 400 g = 10.4 kg}\\[/tex]

The metals are elements which have low ionization potential and can lose electrons easily

the other physical characteristics associated with them and hence with Aluminium are

a) they can conduct electricity in molten state, hence It would only conduct electricity if it were melted.

b) It could be stretched into a thin wire. It means it is ductile.

In case of given aluminium metal the correct answer is

it could be stretched to thin wire

(E) It could be stretched into a thin wire.

- Aluminum is a ductile metal, meaning it can be stretched into thin wires without breaking.

- Let's analyze each option to see why (E) is the most reasonable assumption:

 - (A) Aluminum conducts electricity well even in its solid state, not just when melted, so this assumption is not accurate.

 - (B) Aluminum has a high melting point (660.32°C or 1220.58°F), much higher than typical campfire temperatures, so it would not melt or boil if thrown into a campfire.

 - (C) Aluminum is malleable and ductile, meaning it can be hammered into different shapes without breaking easily.

 - (D) Aluminum can maintain its shape under normal conditions, but it can be reshaped.

 - (E) Aluminum is known for its ability to be drawn into thin wires, which is a characteristic of ductile metals.

Therefore, option (E) is the most reasonable assumption about aluminum metal among the given choices.

[tex]\Delta E = 808 \; \text{kJ} \cdot \text{mol}^{-1}[/tex]

Forming chemical bonds releases energy whereas breaking them requires an input of energy.

Typical chemical reaction involves breaking preexisting bonds and forming new ones. The amount of energy absorbed in the first process is not necessarily equal to the energy released in the second. Depending on the relationship between the two energies, their difference is either released or absorbed in the chemical reaction.

Start by identifying bonds broken and formed in this chemical process:

[tex]\phantom{\text{H}-}\;\;\text{H}\\\phantom{\text{H}-}\;\;\;|\\\text{H}-\text{C}-\text{H} + 2 \; \text{O}-\text{O} \to \text{O}=\text{C}=\text{O} + 2\; \text{H}-\text{O}-\text{H}\\\phantom{\text{H}-}\;\;\;|\\\phantom{\text{H}-}\;\;\text{H}[/tex]

Bonds Broken per reaction:

Bonds Formed per reaction:

[tex]\text{Energy Change} = \text{Energy Released} - \text{Energy Absorbed}\\\phantom{\text{Energy Change}} = E[\text{Bonds Formed}] - E[\text{Bonds Broken}][/tex]

Referring to a thermodynamic data table of bond energies,

Note that due to the conservation of energy, the energy required to break a chemical bond shall be the same as the energy released on its formation.

[tex]\text{Energy Change} =(2 \times 804 + 4 \times 463) - (4 \times 414 + 2 \times 498)\\\phantom{\text{Energy Change}} =3460 - 2652\\\phantom{\text{Energy Change}} = 808 \; \text{kJ} \cdot \text{mol}^{-1}[/tex]

3460 kJ, the energy released in the formation of bonds is greater than 2652 kJ, the energy absorbed to break bonds in the reactant. The energy change is thus positive and 808 kJ of energy is released per mole reaction.

Hey there!

Stock solution:

Concentracion bromide = 0.128 M

initial solution in volumetric flask =  450 µl

So , moles of bromide present:

450  µl in liters :

1 µl  =   = 1*10⁻⁶ liters

450 * ( 1*10⁻⁶ )  = 0.00045

0.128 * 0.00045 =>  57.6 * 10⁻⁶ moles

Now volume final is 25 mL , in liters : 0.025 L ou 25*10⁻³

so  new  bromide concentration:

57.6*10⁻⁶ / 25*10⁻³=> 2.304*10⁻³ M

To find the bromide concentration in the diluted solution, use the formula for dilution C1V1 = C2V2. Your values are C1= 0.128 M, V1= 0.45 ml, and V2= 25 ml. Solve the equation for C2 to obtain the required bromide concentration.

The bromide concentration in the diluted solution can be calculated using the formula for dilution: C1V1 = C2V2. Where C is concentration, V is volume, 1 refers to initial conditions, and 2 refers to conditions after dilution.

Here, C1 is 0.128 M (the initial concentration of bromide in the stock solution), V1 is 450 μl (the volume of the aliquot), V2 is 25 ml (the volume of the diluted solution in the volumetric flask). Inserting these values into the formula, we need to find C2 (unknown concentration of diluted solution).

Therefore, C2 = C1V1 / V2.

Before doing calculations, make sure that all volumes are in the same unit. In this case, convert μl to ml: 1μl = 0.001 ml, so 450 μl = 0.45 ml.

The calculation becomes: C2 = (0.128 M * 0.45 ml) / 25 ml, which, when calculated, provides the concentration of bromide in the diluted solution.

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Hey There!

ideal gas law  :

PV = nRT

P = nRT/V

P = (1*0.082)(18+273)/(0.5) = 47.724 atm

For VDW :

(P + a(n/V)²)(V - nB) = nRT

P = nRT/(V - nB) - a(n/V)²

P = 1*0.082*(18+273)  / (0.5-1*0.03219) - 1.345*( 1/0.5 )² = 45.62

P = 45.62 atm

Pdif = P2-P1 = 47.724 - 45.62 = 2.104 atm

Final answer:

The difference between the ideal pressure and the real pressure of argon in the given conditions is found by using both the ideal gas law and the van der Waals equation, then subtracting the ideal pressure from the real pressure.

Explanation:

When comparing the ideal gas law with the van der Waals equation, we are looking at two different models for describing the behavior of gases. The ideal gas law assumes that gas particles do not interact and have no volume, while the van der Waals equation accounts for the finite volume of gas particles and the attractive forces between them, which can cause deviations from ideal behavior at high pressures and low volumes or high temperatures.

To find the ideal pressure, use the ideal gas law PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(K·mol)), and T is the temperature in Kelvin. Substituting the values into this equation, we can calculate the ideal pressure for argon.

The real pressure of argon can be estimated using the van der Waals equation, which is given by (P + n²a/V²)(V - nb) = nRT, with a and b being the van der Waals constants for argon. Subtracting the ideal pressure from the real pressure will give us the difference between the two.

BaSO₄ is relatively harmless, but BaS is highly toxic.

BaSO₄ is quite insoluble (240 µg/100 mL). It is a mild irritant in cases of skin contact and inhalation. However, it is safe enough that health professionals ask patients to drink a suspension of BaSO₄. The Ba is opaque to X-rays, so it makes the stomach and intestines more visible to radiographers.

BaS is soluble (7.7 g/100 mL). It reacts slowly with water and more rapidly in the acid conditions of the stomach to release H₂S.

BaS + 2HCl ⟶ BaCl₂ + H₂S

An H₂S concentration of 60 mg/100 mL can be fatal within 30 min.

Don’t eat barium sulfide!

Hey There!

Assuming the part of the round flask containing liquid to be perfect sphere, we can calculate it's volume.

Given diameter of sphere = 24 cm  

Therefore radius of the sphere, r= 24/2= 12 cm

Volume of the sphere of diameter 24 cm = (4/3)*pi*r³ = 7,238.23 cm3

Conversion of cm³ to L

1 L= 1 dm³

1 dm = 10 cm

Therefore 1L= 1 dm³ = 1000 cm³

So the volume of liquid in the round flask = 7.24 L( round of to nearest 0.01 L)

The volume of a round flask can be calculated using the formula: Volume = 4/3 * π * radius³. The calculated volume should then be converted from cubic meters to liters, and rounded to the nearest 0.01 liters.

The volume of a round flask or sphere can be calculated using the mathematical formula: Volume = 4/3 * π * radius³. Assuming you have the radius of the flask, you plug it into this formula, calculate the volume in cubic meters, and then convert to liters (since 1 cubic meter equals 1000 liters). If you are given the diameter, remember to divide it by 2 to find the radius. Once you have your answer, round to the nearest  0.01 liters as per the question's instructions.

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The earth's surface will split due to the rising currents of magma. When two oceanic plates drift apart a ridge is formed. When two continental plates drift apart a ridge push is formed.

The Magma rises, spreads outwards,cools and forms new rock and the older rocks are pushed away.

Hey there!

Molar mass C2H6O = 46.0684 g/mol

Number of moles:

n = mass of solute / molar mass

n = 70.6 / 46.0684

n = 1.532 moles

Therefore:

M = number of moles / volume ( L )

M = 1.532 / 2.25

= 0.680 M

Hope that helps!

The molarity of a solution containing 70.6 g of C2H6O in 2.25 L of solution is calculated by determining the number of moles of C2H6O and dividing by the volume in liters, resulting in a molarity of 0.6818 M.

The molarity of a solution is calculated by dividing the number of moles of solute by the volume of the solution in liters. First, you need to find the molar mass of C2H6O, which is (2 × 12.01) + (6 × 1.008) + (1 × 16.00) = 46.07 g/mol. Next, calculate the number of moles of C2H6O in 70.6 g by using the molar mass:

Then, divide the number of moles by the volume of the solution in liters to find the molarity:

Therefore, the molarity of the solution is 0.6818 M.

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I agree with what you've chosen for question two and three.

I would choose C for the fourth question and argue that you shall also go with C for question five.

One of the main ideas about repetition is to have multiple Trials- or independent experiments- for a single independent variable using identical experimental setup. The table presented in option C is the only one among the four choices that allots spaces for measurements from multiple independent experiments.

Hypotheses are predictions of the outcome of a particular experiment. A hypothesis is correct only if it accurately describes the outcome of an experiment. The hypothesis, not the data, would have to be revised if the two doesn't fit. The researcher could keep developing and testing new hypotheses until arriving at a correct one.

Additionally, I would prefer option B over option D in the first question- responses to that question can depends on what your teacher says about the validity of different sources, given that neither B nor D are reliable as sources that one would like to cite in a paper.