Which of the following statements correctly describes the relationship
between frequency and pitch?

A higher frequency results in a higher pitch.

A higher frequency results in a lower pitch.

Frequency does not affect pitch.

None of the above.

The enthalpy change (ΔH) for the reaction CaC(s) + H2O(I) → Ca(OH)(ag) + CH4(g) is -3617.6 kJ.

To calculate ΔH for the reaction CaC(s) + H2O(l) → Ca(OH)2(ag) + CH4(g), we can use Hess's law, which states that the enthalpy change of a reaction is independent of the pathway taken and depends only on the initial and final states.

We can manipulate the given reactions to obtain the desired reaction:

(i) Ca(s) + 2C(graphite) → CaC2(s) ΔH = X (unknown value)

(ii) Ca(s) + 1/2O2(g) → CaO(s) ΔH = -635.5 kJ

(iii) CaO(s) + H2O(l) → Ca(OH)2(ag) ΔH = -653.1 kJ

(iv) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -1300.0 kJ

(v) C(graphite) + 1/2O2(g) → CO(g) ΔH = -393.5 kJ

Now, let's manipulate these equations to cancel out the common reactants and products and obtain the desired reaction:

(i) Ca(s) + 2C(graphite) → CaC2(s) ΔH = X

(ii) Ca(s) + 1/2O2(g) → CaO(s) ΔH = -635.5 kJ

(iii) CaO(s) + H2O(l) → Ca(OH)2(ag) ΔH = -653.1 kJ

(iv) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -1300.0 kJ

(v) C(graphite) + 1/2O2(g) → CO(g) ΔH = -393.5 kJ

Now, let's sum up the equations to obtain the desired reaction:(i) Ca(s) + 2C(graphite) → CaC2(s) ΔH = X

(ii) 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1271 kJ

(iii) CaO(s) + H2O(l) → Ca(OH)2(ag) ΔH = -653.1 kJ

(iv) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -1300.0 kJ

(v) C(graphite) + 1/2O2(g) → CO(g) ΔH = -393.5 kJ

By adding equations (ii), (iii), (iv), and (v), we can cancel out CaO(s), H2O(l), and O2(g):

2Ca(s) + 2C(graphite) + CH4(g) → 2Ca(OH)2(ag) + CO(g) ΔH = X -1271 -653.1 -1300.0 -393.5

2Ca(s) + 2C(graphite) + CH4(g) → 2Ca(OH)2(ag) + CO(g) ΔH = X -3617.6 kJ

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Answer: you have to talk to someone who wont mind wanting to wanting to like you a lot like that.

Explanation:  I wish I could be able to talk to someone who would want to get to like me like that, so its a very relatable situation.

Answer: I believe the answer is A, heat moves from warmer objects to cooler objects. I know for sure it isn’t C or D though so A

The heat is transferred from one object to another as heat moves from warmer objects to cooler objects. The correct option is A.

There are three ways to transfer heat. They are conduction, convection, and radiation. Heat travels from one body to another body. If the temperature of two objects is different, then the heat travels from higher temperature to lower temperature.

Conduction is the transfer of energy when two objects ate in contact with each other. Convection is a transfer between object and environment. Radiation is when transferred by emission of electromagnetic radiation.

Thus, the correct option is A. Heat moves from warmer objects to cooler objects.

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Answer:

a) equivalence point

b) direct titration

c) primary standard

d) titrand

e) Back titration

f) back titration

g) standard solution

h) standard solution

I) indirect titration

j) end point

Explanation:

A volumetric analysis is one in which a solution of unknown concentration is determined from its volume. This is commonly referred to as titration.

In titration, a standard solution is reacted with another solution of unknown concentration. The point at which the concentration of the standard solution is equal to that of the analyte is known as the equivalence point (usually indicated by a colour change). An indicator may be added to the analyte solution to help identify when the reaction is complete.

Answer:

298.9 million years ago - 251.902  million years ago

Explanation:

It might be in the 2011 Earth Science Review Packet so maybe go check that out

Answer:

The given equation is not balanced.

On the left-hand side, there are 3 atoms of potassium (3K), 3 atoms of bromine (3Br), and 1 atom of iron (Fe).

On the right-hand side, there is 1 atom of bromine (Br), 3 atoms of potassium (3K), and 1 atom of iron (Fe).

The number of atoms of each element should be equal on both sides of the equation. Therefore, to balance the equation, we need to adjust the coefficients of the molecules.

The balanced equation is:

3kBr + FeCl3 -> FeBr3 + 3KCl

The balanced equation has 3 atoms of potassium (3K), 3 atoms of bromine (3Br), 1 atom of iron (Fe), and 3 atoms of chlorine (3Cl) on both sides of the equation.

Rhenium oxide is a chemical compound composed of rhenium and oxygen atoms. Its chemical formula is ReO3, and it is a black or dark gray solid with a crystalline structure. Rhenium oxide is a highly refractory material with a very high melting point and excellent thermal stability, which makes it useful in high-temperature applications such as furnace linings and electrical contacts. It is also used as a catalyst in various chemical reactions, including the production of synthetic ammonia and in the dehydrogenation of alcohols. Rhenium oxide is a relatively rare material and is generally produced as a byproduct of other mining and refining processes. It has a number of unique properties that make it valuable in a variety of industrial and scientific applications.

Answer:

C3H7O + O2 → CO2 + H2O

Explanation:

a) The full symbol equation for the oxidation reaction of an isomer of C3H7O can be represented as:

C3H7O + O2 → CO2 + H2O

b) The proper reagent for this oxidation reaction is O2 (oxygen gas). The catalyst required for this reaction depends on the specific conditions. Common catalysts used for oxidation reactions include transition metals such as platinum (Pt), palladium (Pd), or copper (Cu).

c) There is no need to remove the product (CO2 and H2O) from the reaction vessel because they are typically in the gas or liquid phase and do not significantly interfere with the reaction. The product gases can be easily vented out of the vessel, while the liquid water can be left in the reaction mixture. Additionally, the product CO2 is a stable and inert gas, which does not pose any hazards in most cases. Therefore, it is often not necessary to remove the products after the reaction is complete.

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Answer: True

I hope this helped!

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- Zack Slocum

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Answer:

true

Explanation:

Answer:

310K

Explanation:

Rearrange PV=nRT to get T=PV/nR

T=(2.0L)(0.75atm)/(0.5mol)(0.08206)

=36.5 or 37

add 273 for K to get

310

Based on the given reactions, the compounds are as follows:

A: The specific product formed from the reaction between prop-1-yne and either 2HBr or H2O2.

B: The product formed when compound A reacts with 2H2O.

C: The product formed when compound B reacts with K2CO3(aq).

D: The product formed from the heat-induced reaction of compound C.

E: The product formed when compound D reacts with HBr.

Based on the given reactions, let's analyze the compounds involved:

Reaction 1: prop-1-yne + 2HBr/H2O2 = A

The reactant prop-1-yne reacts with either 2HBr or H2O2 to form compound A. The specific product formed will depend on the reaction conditions.

Reaction 2: A + 2H2O = B

Compound A reacts with 2H2O (water) to form compound B.

Reaction 3: B + K2CO3(aq) = C

Compound B reacts with K2CO3(aq) (potassium carbonate dissolved in water) to form compound C.

Reaction 4: C + heat = D

Compound C undergoes a heat-induced reaction to form compound D.

Reaction 5: D + HBr = E

Compound D reacts with HBr (hydrobromic acid) to form compound E.

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Answer:

The percentage is   k  \(= 20.8\)%

Explanation:

From the question we are told that

    The mass of the stibnite is  \(m_s = 5.86 \ g\)

   The volume of   KBrO3(aq) is  \(V = 26.6 mL = 26.6 *10^{-3} \ L\)

     The concentration  of   KBrO3(aq) is  \(C = 0.125 M\)

Now the balanced ionic  equation for this reaction is

        \(BrO_3 ^{-}+ 3Sb^{3+} + 6H^{+} \to Br^{1-} + 3Sb^{5+} + 3H_2O\)

The number of moles of   \(BrO_3 ^{-}\) is  

     \(n = C *V\)

substituting values

     \(n = 26.6*10^{-3} * 0.125\)

     \(n = 0.003325 \ mols\)

from the reaction we see that 1 mole of \(BrO_3 ^{-}\)  reacts with 3 moles of  \(Sb^{3+}\)

so 0.003325 moles will react with x moles of  \(Sb^{3+}\)

Therefore

               \(x = \frac{0.003325 * 3}{1}\)

              \(x = 0.009975 \ mols\)

Now the molar mass of \(Sb^{3+}\) is a constant with a values of  \(Z = 121.76 \ g/mol\)

Generally the mass of  \(Sb^{3+}\) is mathematically represented as

        \(m = x * Z\)

substituting values

        \(m = 1.215 \ g\)

The percentage of  Sb(antimony) in the overall mass of the stibnite is mathematically evaluated as

            k  \(= \frac{1.215}{5.85 } * 100\)

           k  \(= 20.8\)%

Answer:

Explanation:

Step 1: Data given

Mass of stibnite (Sb2S3) = 5.85 grams

The Sb3+(aq) is completely oxidized by 26.6 mL of a  0.125 M aqueous solution of KBrO3(aq).

Step 2: The balanced equation

BrO3-(aq)+ 3Sb^3+(aq) + 6 H+ → Br-(aq) + 3Sb^5+(aq) + 3H2O (l)

Step 3: Calculate moles KBrO3

Moles KBrO3 = molarity * volume

Moles KBrO3 = 0.125 M *0.0266 L

Moles KBrO3 = 0.003325 moles

Step 4: Calculate moles Bro3-

in 1 mol KBrO3 we have 1 mol K+ and 1 mol BrO3-

In 0.003325 moles KBrO3 we have 0.003325 moles BrO3-

Step 5: Calculate moles Sb

For 1 mol BrO3- we need 3 mol Sb^3+ to produce 1 mol Br- and 3 mol Sb^5+

For 0.029085 moles BrO3- we need 3*0.003325 = 0.009975 moles Sb

Step 6: Calculate mass Sb

Mass Sb = moles Sb * molar mass Sb

Mass Sb = 0.009975 moles * 121.76 g/mol

Mass Sb = 1.21 grams

Step 7: Calculate the percentage of Sb in the ore

% Sb = (mass Sb / total mass) * 100%

% Sb = (1.21 grams / 5.85 grams) * 100 %

% Sb = 20.76 %

Based on the Law of Conservation of Mass, the mass of products form when baking soda decomposes is 168 g/mole.

Law of conservation of mass is defined as chemical reactions and physical changes cannot build or remove mass in an isolated system. The mass of the reactants and products in a chemical reaction must equal each other according to the law of conservation of mass.

\(\rm 2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2\)

Mass of baking soda = 2 x molar mass

Molar mass of NaHCO₃ = 84.007 g/mole

Mass of baking soda = 2 x 84.007 g/mole

Mass of baking soda =  168.014 g/mole ≅ 168 g/mole

Thus, based on the Law of Conservation of Mass, the mass of products form when baking soda decomposes is 168 g/mole.

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447.8 liters of \(H_{2}\) can be produced at STP using 80.0 g of methane and 16.3 g of water, and water is the limiting reactant.

Limiting reactant is a chemical reactant used in a reaction that is completely used up before any other reactant is used. It limits the amount of product that can be produced in a given reaction.

Methane \((C*H_{4})\) reacts with water \((H_{2}O)\)to produce hydrogen \((H_{2})\) and carbon dioxide \((C*O_{2})\). The equation for this reaction is:
\(C*H_{4} + 2H_{2}O \rightarrow C*O_{2} + 4H_{2}\)
Using the given masses of reactants, we can calculate the number of moles of methane and water.
Methane: \(80.0 g C*H_{4} × 1 mol C*H_{4}/16.04 g C*H_{4} = 4.99 mol C*H_{4}\)

Water: \(16.3 g H_{2}O × 1 mol H_{2}O/18.02 g H_{2}O = 0.90 mol H_{2}O\)
Now we can use the mole ratio from the balanced equation to determine the limiting reactant.
Methane: \(4.99 mol C*H_{4}/1 mol C*H_{4} = 4.99 mol C*H_{4}\)
Water: \(0.90 mol H_{2}O/2 mol H_{2}O = 0.45 mol H_{2}O\)
Since the mole ratio of water is lower than that of methane, water is the limiting reactant.
Now we can use the molar ratio of the reaction to calculate the number of moles of hydrogen produced.
\(4.99 mol C*H_{4} × 4 mol H_{2}/1 mol C*H_{4} = 19.96 mol H_{2}\)
Finally, we can use the ideal gas law to calculate the number of liters of H_{2} at STP (standard temperature and pressure).
\(19.96 mol H_{2} × 22.4 L H_{2}/1 mol H_{2} = 447.8 L H_{2}\)
Therefore, 447.8 liters of \(H_{2}\) can be produced at STP using 80.0 g of methane and 16.3 g of water, and water is the limiting reactant.

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% of H in NH3= 17% and % of O in NO2 = 69%

             = 14 + (3 x 1)

             = 17

             % of H in NH3 = 3/17 x 100

                        = 17%

              = 14 + (16 x 2)

              = 14 +32 = 46

              % of O in NO2 = 32/46x 100

                        = 69%

The mass of a given molecule is its molecular mass (m), which is expressed in daltons (Da or u). Because they contain various isotopes of an element, multiple molecules of the same substance can have distinct molecular weights. According to IUPAC, the related quantity relative to molecular mass is a unitless comparison between the mass of a molecule and the unified atomic mass unit (also called the dalton). The molar mass is unrelated to, but distinct from, the molecular mass and relative molecular mass. The molar mass, which is stated in g/mol, is defined as the mass of a specific material divided by the amount of a substance

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The alkali metals belong to Group 1A of the periodic table and have one valence electron. They are among the most reactive metals because their outer electron configuration (ns') is one electron beyond a noble gas configuration.

They react to lose the ns electron, obtaining a noble gas configuration. This is why group 1A metals tend to form 1+ cations. Alkali metals are among the most reactive metals because their outer electron configuration (ns') is one electron beyond a noble gas configuration. Noble gases, such as helium, neon, and argon, have a filled valence shell, making them chemically inert or unreactive.

The alkali metals, with one valence electron, seek to obtain a noble gas configuration by losing that valence electron. When the alkali metals react, they tend to lose their valence electron to form a positively charged ion or cation. This is why group 1A metals tend to form 1+ cations.

By losing their valence electron, the alkali metals achieve a stable electron configuration, similar to that of the noble gas in the previous period, with a filled outer shell. This loss of valence electrons results in a decrease in atomic radius and an increase in electronegativity.

In summary, alkali metals' chemical behavior is heavily influenced by the number of valence electrons. Their tendency to lose one valence electron and form a 1+ cation is due to their one valence electron being one electron beyond a noble gas configuration. The resulting chemistry of alkali metals makes them highly reactive and important in a variety of industrial and biological applications.

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Answer:

Explanation:

ok if yes do follow

Answer:

salt water

Explanation:

beacuse it is made up of 2 elements,NaCl and H2O

Answer:

Yes cellular respiration is the only way to break down glucose.  Cellular respiration takes place by the cell using oxygen to break down glucose.  

Answer:

Metal is shapped in a blacksmith shop by heating the metal to the point it becomes plastic. The hot metal is then shaped by using different parts of the anvil and striking it with a hammer.

Answer: the answer is a

The final temperature if you mixed a 1,000 grams of 20 degree Celsius water with 2,000 grams of 40 degree Celsius water is 33.3 degree Celsius.

The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

It is a measure of how much energy it takes to raise the temperature of a substance. It is the amount of heat necessary to raise one mass unit of that substance by one temperature unit.

It is given by the formula -

                                                  Q = mcΔT

where, Q = amount of heat

m = mass

c = specific heat

ΔT = Change in temperature

Given,

On mixing the water at two different temperatures-

Q(gained) = Q (lost)

mcΔT = mcΔT

2000 × 4.184 × ( 40 - x ) = 1000 × 4.184 × ( x - 20)

x = 33.3⁰C

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The molar mass of the unknown gas in the 500 mL flask at stp is 89.7 g/mol

We'll begin by calculating the number of mole of the gas that occupied 500 mL at stp.

Recall at stp:

22400 mL = 1 mole of the gas

Therefore,

500 mL = 500 / 22400 = 0.0223 mole of gas.

Finally, we shall determine the molar mass of the gas. This can be obtained as follow:

Mass of gas = 2 g

Mole of gas = 0.0223 mole

Molar mass = mass / mole

Molar mass of gas = 2 / 0.0223

Therefore, the molar mass of the unknown gas is 89.7 g/mol

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Answer:

Differences in water density affect vertical ocean currents. Denser water tends to sink, while less dense water tends to rise. Other causes of currents include tides, rain, runoff, and ocean bottom topography. Topography is the surface features of a place. Ocean topography includes slopes, ridges, valleys, and mountains! All these things are found at the bottom of the ocean, and can influence currents.

The cause-and-effect relationship between density and ocean currents is the mixing and circulation are influenced by the density differences between the various layers of the water column.

The continuous, predictable, and directional movement of seawater known as ocean currents is caused by gravity and wind.

Ocean vertical currents are influenced by variations in water density. Less dense water tends to rise, while denser water sinks. Tides, rainfall, runoff, and the topography of the ocean bottom are additional causes of currents.

Thus, the mixing and circulation are influenced by the differences in densities between the various layers of the water column, which is the cause-and-effect relationship between density and ocean currents.

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The concentration of HBr used in the titration is found to be 0.0978M.

Volume of NaOH used is 27.50ml and it’s molarity used is 0.200M.

Volume of HBr used is 56.20ml.

On comparing molarity of both the solutions, we get the relation

HBr × VHbr  = MNaOH × VNaOH

MHBr = 0.200M × 27.50ml /56.20ml

MHBr = 0.0978M.

A colorless, stifling gas called hydrogen bromide is highly dissociated in water and exceedingly soluble in water. In the presence of damp air, it rapidly fumes. Hydrogen bromide gas is a very caustic chemical that, when contacted, can result in serious burns.

To determine the concentration of an identified analyte, titration is a typical laboratory technique for quantitative chemical analysis. A standard solution with a defined concentration and volume is prepared as a reagent, also known as the titrant.

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Answer:

: kinetic energy is the energy transformation that occurs in a hot balloon.

Explanation:

Hot air balloons use a propane burner that converts chemical energy to thermal energy. The hot air is less dense than than the colder air and it lifts the balloon

Answer:In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.

Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.

STP conditions are defined as a pressure of

100 kPa

and a temperature of

0

∘

C

. Under these conditions for pressure and temperature, one mole of any ideal gas occupies

22.7 L

- this is known as the molar volume of a gas at STP.

So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.

The balanced chemical equation for this decomposition reaction looks like this

2

KClO

3(s]

heat

×

−−−→

2

KCl

(s]

+

3

O

2(g]

↑

⏐

⏐

Notice that you have a

2

:

3

mole ratio between potassium chlorate and oxygen gas.

This tells you that the reaction will always produce

3

2

times more moles of oxygen gas than the number of moles of potassium chlorate that underwent decomposition.

Use potassium chlorate's molar mass to determine how many moles you have in that

231-g

sample

231

g

⋅

1 mole KClO

3

122.55

g

=

1.885 moles KClO

3

Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate

1.885

moles KClO

3

⋅

3

moles O

2

2

moles KClO

3

=

2.8275 moles O

2

So, what volume would this many moles occupy at STP?

2.8275

moles

⋅

22.7 L

1

mol

=

64.2 L

Explanation:

According to Charle's law, a balloon at 600 K has a volume of 3.83 L.

Charles' law states that, under constant pressure, the volume a given amount of gas fills is directly proportional to its absolute temperature. For instance, when the temperature drops during the winter, a basketball placed outside in the ground would contract.

Mathematically, Charles' law can be written as V1/T1 = V2/T2, where V1 denotes the volume at the beginning, T1 the temperature at that point, and V2 the volume at the end with T2 the temperature at that point.

The following examples of how Charles' law is applied in real life:

Helium balloons shrink when it's cold outside.

From Charles' law, V/T = constant

V₁/T₁ = V₂/T₂

or, 2/313 = V₂/600

or, V₂ = (2x600)/313

          = 3.83

Hence, form Charles' law we conclude the volume of balloon at 600 K is 3.83 L.

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Given the model from the question,

From the model given, we obtained the ffolowing

Thus, we can write the balanced equation as follow:

3H₂ + 2NO —> N₂ + 2H₂O + H₂

From the balanced equation above,

3H₂ + 2NO —> N₂ + 2H₂O + H₂

From the balanced equation above,

3 moles of H₂ reacted with 2 moles of NO.

Therefore,

5 moles of H₂ will react with = (5 × 2) / 3 = 3.33 moles of NO

From the calculation made above, we can see that only 3.33 moles of NO out of 4 moles given are required to react completely with 5 moles of H₂.

Thus, H₂ is the limiting reactant

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