If the elastic modulus of cobalt (co) is 200 gpa, and the elastic modulus of tungsten carbide (wc) is 700 gpa, calculate the upper and lower bound estimates of the elastic modulus of a 10 vol% co – 90vol% wc composite.

Answers

Answer 1
To determine the upper bond
Ec(u) = EmVm + EpVp
Em is the elastic modulus of cobalt.
E₁ is the elastic modulus of the particulate
Vm is the volume fraction of cobalt
Vp is the volume fraction of particulate
substitute
Ec(u) = 200 (Vm) + 700 (Vp)
To determine the lower bound
Ec (l) = EmEp/VmEp+ VpEm
Substitute
Ec (l) = 200(700)/Vm(700) + Vp (200)
Ec (l) = 1400/7Vm+2Vp
Answer 2

The upper bound estimate of the composite's elastic modulus is approximately 650 GPa, and the lower bound estimate is approximately 560 GPa, using the Reuss model.

To estimate the elastic modulus of the composite, we use the rule of mixtures. The Voight model and the Reuss model are the two estimates.

Upper Bound (Voigt Model): The Voigt model assumes that the strain in both materials is the same. The formula is:

[tex]E_{upper} = V_CE_C + V_{WC}E_{WC}[/tex][tex]V_C[/tex] = Volume fraction of cobalt = 0.10[tex]V_{WC}[/tex] = Volume fraction of tungsten carbide = 0.90[tex]E_C[/tex] = Modulus of cobalt = 200 GPa[tex]E_{WC[/tex] = Wrought iron carbide modulus = 700 GPa[tex]E_{upper}[/tex] = (0.10 x 200) + (0.90 x 700) = 20 + 630 = 650 GPa

Lower Bound (Reuss Model): The Reuss model assumes that the stress in both materials is the same. The formula is:

[tex]1/E_{lower} = V_C/E_C + V_{WC}/E_{WC}[/tex][tex]V_C[/tex] = Volume fraction of cobalt = 0.10[tex]V_{WC}[/tex] = Volume fraction of tungsten carbide = 0.90[tex]E_C[/tex] = Modulus of cobalt = 200 GPa[tex]E_{WC[/tex] = Wrought iron carbide modulus = 700 GPa[tex]1/E_{lower} = (0.10/200) + (0.90/700) = 0.0005 + 0.0012857 \approx 0.0017857[/tex][tex]E_{lower} \approx 1 / 0.0017857 \approx 560\ GPa[/tex]

Result: The upper bound estimate of the elastic modulus of the 10 vol% Co – 90 vol% WC composite is 650 GPa, and the lower bound estimate is 560 GPa.


Related Questions

.War nickels, produced from 1942-1945 are composed of 56% copper, 35% silver and 9% manganese. How many moles of each element are found in a 5.00 g nickel coin?

Answers

The moles  of each element  found in a 5.00 g nickel coin is calculated as below

moles =mass/molar mass

calculate the mass of each element  =% composition of element/100 x total mass of  nickel

Mn = 9/100 x5 = 0.45g
Cu=56/100 x5= 2.8 g
Ag= 35/100x5=  1.75 g
moles of each element is therefore=
Mn = 0.45g/54.94 =  8.19 x10^-3 moles

Ag=1.75g/107.87 g/mol = 0.0162 moles

Cu = 2.8 g/63.5 g/mol=0.0441  moles





ASAP HELP GIVING TO CORRECT PERSON A BRAINLIEST REWARD

Cellulose is a polymer that makes up plant cell walls. Cellulose is made from a chain of C6H10O5 molecules.

Which monomers are most likely used to produce cellulose?

a)nucleotides and amino acids
b)glucose molecules
c)nucleotides
d)amino acids

Answers

It's B: glucose molecules

Answer:

Answer is B

Explanation:

propose a reasonable synthesis to get benzylamine from toluene

Answers

Benzyl Amine is synthesized from Toluene in two steps;

Step 1: Bromination of Methyl Group in Toluene:
                                                                             Bromination is carried out on methyl group present on Benzene ring through free radical mechanism by treating toluene with N-Bromosuccinimide (NBS) and Benzyl Bromide is formed.

Step 2: Benzyl Bromide into Benzyl Amine;
                                                                    This step is carried out utilizing a neme reaction "Gabriel Reaction". Benzyl bromide is treated with Potassium Phthalamide and an intermediate is formed which on treatment with Hydrazine produces Benzyl amine.

Final answer:

The synthesis of benzylamine from toluene involves an oxidation step to convert toluene to benzaldehyde followed by reductive amination with an amine such as methylamine to yield the desired benzylamine.

Explanation:

To propose a reasonable synthesis to get benzylamine from toluene involves several steps. Initially, toluene can be oxidized to benzaldehyde through the oxidation reaction. This would typically involve a reaction with an oxidizing agent like KMnO4 in an aqueous base or CrO3 with acidic conditions. Once benzaldehyde is obtained, the next step is to convert it into benzylamine. This can be achieved by the reductive amination process, where benzaldehyde can react with an amine such as methylamine and then be reduced, often using a reducing agent like NaBH4 or H2 with a catalyst. To ensure that the amine added is solely the benzyl amine desired, the N-alkylation can be controled through the choice of appropriate reagents and conditions.

One possible method would utilize methylamine in the presence of a reductive agent to convert the aldehyde group to the desired primary amine. The overall reaction from toluene to benzylamine would likely proceed with good to high yields, assuming that the correct reaction parameters and purification techniques are employed.

A solution is made by dissolving 5.65 g of an unknown molecular compound in 110.0 g of benzene froze at 4.39 oc. what is the molar mass of the solute if pure benzene has a freezing point of 5.45 oc and the kf value of benzene is 5.07 oc/m

Answers

Final answer:

The correct answer is "245.76 g/mol". To find the molar mass of an unknown compound dissolved in benzene from the freezing point depression, the change in freezing point is calculated, then used with the freezing point depression formula to determine the molality. The molar mass is ultimately found by dividing the mass of the solute by the number of moles of solute calculated from the molality and mass of solvent.

Explanation:

To determine the molar mass of an unknown molecular compound from the freezing point depression in benzene, we first need to calculate the change in freezing point (ΔTf). Given that pure benzene freezes at 5.45 °C and the solution freezes at 4.39 °C, ΔTf is the difference between these two temperatures.

ΔTf = 5.45 °C - 4.39 °C = 1.06 °C

Using the formula for freezing point depression, ΔTf = i * Kf * m, where i is the van't Hoff factor (for a non-electrolyte, this is 1), Kf is the freezing point depression constant of benzene (5.07 °C/m), and m is the molality of the solution. Since we're looking for the molar mass of the solute, we rearrange the formula to find molality first: m = ΔTf / (i * Kf).

m = 1.06 °C / (1 * 5.07 °C/m) = 0.209 mol/kg

To find the molar mass, we need the number of moles of the solute, which is the mass of the solute divided by its molar mass. Given the mass of the solute is 5.65 g, and using the molality equation m = moles of solute / kg of solvent, we can find the number of moles of solute. Then, molar mass (M) = mass of solute / moles of solute.

Moles of solute = m * kg of solvent = 0.209 mol/kg * 0.110 kg = 0.02299 mol

Molar mass (M) = mass of solute / moles of solute = 5.65 g / 0.02299 mol ≈ 245.76 g/mol.

In processes that produce electricity, some of the energy used is wasted as _____ energy.
Select one:
a. heat
b. chemical
c. kinetic
d. light

Answers

In processes that produce electricity, some of the energy used is wasted as HEAT energy. :)
The answer is 
A) Heat

What is the total mass of products formed when 16 grams of ch4?

Answers

When CH₄ is burnt in excess O₂ following products are formed,

                           CH₄  +  2 O₂     →     CO₂  +  2 H₂O

According to equation 1 mole of CH₄ (16 g) reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O. Hence the products are,

                          1 mole of CO₂  and  2 moles of H₂O

Converting 1 mole CO₂ to grams;
As,
                           Mass  =  Moles × M.mass

                           Mass  =  1 mol ×  44 g.mol⁻¹

                           Mass  =  40 g of CO₂

Converting 2 moles of H₂O to grams,

                           Mass  =  2 mol ×  18 g.mol⁻¹
                         
                           Mass  =  36 g of H₂O

Total grams of products;

                           Mass of CO₂  =  44 g
                    +     Mass of H₂O  =  36 g
                                                  -------------
                           Total              =   80 g of Product

Result:
            80 grams of product
is formed when 16 grams of CH₄ is burnt in excess of Oxygen.
Final answer:

The total mass of the products formed when 16 grams of methane (CH4) is combusted is equal to the mass of the reactants, which is also 16 grams, when not considering the massless product, heat.

Explanation:

The student is asking about the total mass of products when 16 grams of methane (CH4) is combusted. Combustion of methane is represented by the balanced chemical equation CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) + energy. To find the total mass of the products, we would need to use stoichiometry to convert the mass of CH4 to moles, then use the balanced equation to find the moles of the products CO2 and H2O. Since mass is conserved, the total mass of reactants will equal the total mass of products. However, heat is also a product of the reaction, which is not measured by mass.

In this question, the student has mentioned specific molecular quantities, but since the aim is to determine the mass of products from 16 grams of methane, we don't need to consider these quantities. What we need is the concept that the mass of reactants equals the mass of products in a chemical reaction, where the mass of gaseous products is part of the overall calculation. Heat, despite being a part of the product side of the equation, does not contribute to the mass.

Therefore, the total mass of the CO2 and H2O produced from 16 grams of methane will also be 16 grams, neglecting energy. It is crucial to emphasize that this is a theoretical approach assuming complete combustion and no mass loss during the reaction.

___ + ___ = H2O + LiBrO3 Complete and balance the equation representing neutralization reaction.

Answers

when an acid reacts with a base its called a neutralisation reaction and products of the reaction are salt and water.
this is called neutralisation as hydrogen ions in acid react with hydroxide ions in base and form water. 
the cation in base reacts with anion in acid and forms a salt.
In the reaction given the salt formed is LiBrO₃
in the salt formed the cation is Li⁺ and anion is BrO₃⁻
therefore base is LiOH and acid is HBrO₃
neutralisation reaction is therefore 
LiOH + HBrO₃ ---> LiBrO₃ + H₂O
Final answer:

The complete and balanced neutralization reaction is HBrO₃(aq) + LiOH(aq) → H₂O(l) + LiBrO₃(aq), where hydrobromic acid reacts with lithium hydroxide to produce water and lithium bromate.

Explanation:

The student's question pertains to completing and balancing a chemical equation for a neutralization reaction. In a neutralization reaction, an acid and a base react to form water and a salt. To complete and balance the given equation ___ + ___ = H₂O + LiBrO₃, we need to identify the acid and the base that will produce lithium bromate (LiBrO₃) and water.

The balanced equation for this reaction is:

HBrO₃(aq) + LiOH(aq) → H₂O(l) + LiBrO3(aq)

Here, hydrobromic acid (HBrO₃) reacts with lithium hydroxide (LiOH), producing water (H₂O) and lithium bromate (LiBrO₃). This equation is balanced as written, with one mole of HBrO₃ reacting with one mole of LiOH to produce one mole of water and one mole of LiBrO₃.

Write a balanced half-reaction for the reduction of dichromate ion cr2o−27 to chromium ion cr 3 in basic aqueous solution. be sure to add physical state symbols where appropriate.

Answers

Explanation:

When electrons are added in a reaction then it means reduction is taking place. Whereas removal of electrons from a chemical reaction is known as oxidation.  

Oxidation state of chromium in [tex]Cr_{2}O^{2-}_{7}[/tex] is +6. So, its reduction half-reaction will be as follows.

         [tex]Cr_{2}O_{7}^{2-} + 3e^{-1} \rightarrow Cr^{3+}[/tex]

Since it is given that reaction is taking place in basic solution. So, we add [tex]H_{2}O[/tex] on reactant side and [tex]OH^{-}[/tex] on the product side.

        [tex]Cr_{2}O_{7}^{2-} + 3e^{-1} + H_{2}O \rightarrow Cr^{3+} + OH^{-}[/tex]

Now balancing Cr atom and charges on both sides, we get the following.

       [tex]Cr_{2}O_{7}^{2-} + 6e^{-1} + 7H_{2}O \rightarrow 2Cr^{3+} + 14OH^{-}[/tex]

The balanced half-reaction for the reduction of dichromate ion Cr2O2−7 to chromium ion Cr3+ in basic aqueous solution is:

Cr2O2−7(aq) + 14 OH−(aq) + 6 e− → 2 Cr3+(aq) + 7 H2O(l)

What's the reaction about?

Balance the chromium atoms. There are 2 chromium atoms on the left side and 2 on the right, so the equation is already balanced for chromium.

The final balanced equation is:

Cr2O2−7(aq) + 14 OH−(aq) + 6 e− → 2 Cr3+(aq) + 7 H2O(l)

The physical state symbols are also included in the balanced equation. The dichromate ion and the chromium ion are aqueous, the hydroxide ions are aqueous, and the water molecules are liquid.

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Write the balanced chemical equation for the reaction of the weak acid hcn with water. include the phase of each species.

Answers

Hydrogen Cyanide is considered as a weak acid. The pKa value of HCN is 9.21. When HCN is dissolved in water it slightly ionizes to produce H⁺ in the form of H₃O⁺ as water acts as a base and CN⁻ (Cyanide Ion).

The balanced chemical equation along with phases of species is as follow,

                     HCN ₍aq₎  +  H₂O ₍l₎    ⇄    H₃O⁺ ₍aq₎  +  CN⁻ ₍aq₎ 

In above equation the sign "⇄" indicates that HCN is not ionizing completely and the reaction is reversible in nature.
Final answer:

The weak acid HCN reacts with water through acid ionization, forming hydronium ions and cyanide ions. This reaction can be represented with the balanced chemical equation: HCN(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CN⁻(aq).

Explanation:

The weak acid HCN (hydrogen cyanide) reacts with water through a process called acid ionization. In this reaction, a hydrogen ion (H+) is transferred from the weak acid to a water molecule, leading to the formation of hydronium ions (H3O+), and the cyanide ion (CN-). The balanced chemical equation outlining this reaction is:

HCN(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CN⁻(aq)

In this equation, the (aq) annotation indicates that the species is in the aqueous - or water - phase, while (l) indicates the liquid phase for water.

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Write the structural formula for aspirin. label the ester group and the carboxylic acid group

Answers

The formula of the asprin is C₉H₈O₄. Asprin is also called as acetylsalicylic acid which is used as a drug in medical treatments. It is a synthetic derivative of salicylic acid which is extracted from willow bark.
As the functional groups, asprin has 3 groups as carboxylic acid group, ester group and aromatic ring. The structure is as the image.

The structural formula of aspirin is C9H8O4. The ester group in aspirin is -COO- and the carboxylic acid group is -COOH.

The structural formula of aspirin is C9H8O4. The ester group in aspirin is -COO- and the carboxylic acid group is -COOH. The structural formula can be represented as:

C6H4OH(CO)O using the circled atoms as reference points for the acid part of the molecule.

The (C4H1 0)3N represents a___ amine.
A. Primary
B. Secondary
C. Tertiary

Answers

C. tertiary :) hope this helps

Answer :The correct answer for (C₄H₁₀ )₃ N is C) TERTIARY AMINE .

The substitution level of nitrogen in amine or in simple words , description of carbon atoms attached to a given nitrogen in any amine is of 4 types :

1 ) Primary amine :

When the given nitrogen is attached to only one carbon then substitution level is known as PRIMARY amine.

Example : CH₃NH₂ : The nitrogen is attached to only one carbon (CH₃) . Hence it is primary amine

2) Secondary amine :

When the given nitrogen is attached to two carbons .

Example : (CH₃)₂NH : The nitrogen is attached to two carbon ( CH₃)₂ , hence it is secondary amine .

3) Tertiary amine :

When the given nitrogen is attached to three carbon atom .

Example : (CH₃)₃NH : The nitrogen is attached to 3 carbons atoms (CH₃)₃ . Hence it is a tertiary amine

4) Quaternary amine : when the nitrogen attached to 4 carbon atoms and amine posses a positive charge on it , its is known as quaternary amine .

Example : (CH₃)₄N⁺ : The nitrogen is attached to 4 carbons and nitrogen . hence it is quaternary amine .

(Image attached )

The given compound is (C₄H₁₀ )₃ N has 3 groups of C₄H₁₀ attached to N , means 3 carbons attached to N , ( image attached ) . Hence it can be said that (C₄H₁₀ )₃ N is TERTIARY AMINE .

Hence correct option is C) Tertiary amine .

List the following compounds in decreasing electronegativity difference. cl2 hcl nacl

Answers

Based on Pauling Scale, electro negativity of Cl = 3.2, Na = 0.9 and H = 2.1

Thus, Electronegativity difference  in [tex] Cl_{2} [/tex] = 3.2 -3.2 = 0
Electronegativity difference  in NaCl = 3.2-0.9 = 2.3
Similarly, Electronegativity difference  in HCl = 3.2 - 2.1 = 1.1

Thus, among the listed molecules following is the decreasing order of electronegativity difference: NaCl> HCl > [tex] Cl_{2} [/tex]
Final answer:

The decreasing electronegativity difference in the compounds is NaCl > HCl > Cl2. NaCl has the highest electronegativity difference, forming an ionic compound, then HCl forms a polar covalent bond, and Cl2, made of two identical atoms, has no electronegativity difference.

Explanation:

The compounds Cl2, HCl, and NaCl possess varying degrees of electronegativity difference depending on the atoms involved. Electronegativity is the ability of an atom to attract shared electrons in a bond. In NaCl (sodium chloride), the electronegativity difference is very high as it consists of a metal (sodium) and a non-metal (chlorine).

This forms an ionic compound, which is generally formed when there is a high electronegativity difference. On the other hand, Cl2 (chlorine gas) possesses no electronegativity difference as it is a molecule composed of two identical atoms. Finally, HCl (hydrogen chloride) has a considerable but not extreme electronegativity difference because it consists of two non-metals. This forms a polar covalent bond.

In summary, the decreasing electronegativity difference would be: NaCl > HCl > Cl2.

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At 7 degrees Celsius the volume of gas is 49 liters. At the same pressure its volume is 74 mL at what temperature

Answers

Charle's law gives the relationship between volume and temperature of gas.
It states that at constant pressure, volume of gas is directly proportional to temperature of gas.
V/T = k
where V - volume , T - temperature and k- constant 
[tex] \frac{V1}{T1} = \frac{V2}{T2} [/tex]
parameters for the first instance are on the left side of the equation and parameters for the second instance are on the right side of the equation 
T1 - 7 °C + 273 = 280 K
substituting these values in the equation 
[tex] \frac{49 mL}{280K} = \frac{74 mL}{T} [/tex]
T = 423 K
temperature in Celcius scale - 423 K - 273 = 150 °C

In the nuclear transmutation represented by 168o(p, \alpha) 137n, the emitted particle is ________.

Answers

Answer is: the emitted particle is an alpha particle.

Nuclear reaction: ¹⁶O + p⁺ → ¹³N + α (alpha particle).
Alpha decay is radioactive decay in which an atomic nucleus emits an alpha particle (helium nucleus) and transforms into an atom with an atomic number that is reduced by two and mass number that is reduced by four.
When oxygen-16 gain one proton, atomic mass is 17, but when lose alpha particle atomic mass reduces by four to 13.

A sample of gas occupies 10.0 l at 100.0 torr and 27.0
c. calculate the pressure if the temperature is changed to 127c while the volume remains constant. (133 torr)

Answers

The pressure  of a sample  of a gas if the temperature  is changed  to 127  c  while  the volume  remains  constant   is calculated  using gay lussac law formula

that is P1/T1 = P2/V2
P1 = 100 torr
T1 = 27+273 = 300 k
T2 =127 +273 =400 k
P2=?

by  making P2 the subject of the formula
P2=T2P1/T1

=100  x 400/300 = 133.3 torr

Final answer:

To calculate the new pressure of a gas when the temperature changes and volume remains constant, we apply Gay-Lussac's Law, which involves converting temperatures to Kelvin and using the relationship P1/T1 = P2/T2.

Explanation:

The student's question involves calculating the pressure of a sample of gas when the temperature is changed while keeping the volume constant. This is a typical problem you might encounter in a chemistry class when discussing gas laws, specifically relating to Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature, provided that the volume remains constant.

Since the volume is constant, we can use the Gay-Lussac's Law formula: P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature, respectively. Remember to convert temperatures to Kelvin. Initially, we have P1 = 100.0 torr and T1 = (27.0 + 273.15) K. The final temperature is T2 = (127 + 273.15) K. Solving for P2, we get the final pressure.

What volume of chlorine is required to produce 25.4?

Answers

from this reaction equation:

Cu + Cl2 → CuCl2

so 1 mol CuCl2 will produce from 1 mol Cl2

when the moles of CuCl2 = mass / molar mass 

                                            = 25.4 g / 134.45 g/mol

                                            = 0.189 moles 

∴0.189 moles of CuCl2 will produce from 0.189 moles Cl2

∴ Volume of 0.189 moles of Cl2 is :

 V = nRT/P

when n is the number of moles = 0.189 moles

R is ideal gas constant = 0.0821 

T temperature in Kelvin = 18 + 273 = 291 K

P is the pressure = 2.13 atm

by substitution:

∴ V = 0.189 * 0.0821 * 291 / 2.13

    = 2.12 L

The standard reduction potential for I2/I –is +0.54V, and the standard reduction potential for Br2/Br -is +1.07V. Which reaction occurs when bromine is added to an aqueous solution of iodide ions?
1. 2I –+ 2Br –→ I2+ Br2
2. I2+ 2Br –→ Br2+ 2I –
3. I2+ Br2→ 2I –+ 2Br –
4. 2I –+ Br2→ I2+ 2Br –

Answers

The reduction potential for Bromine is stronger than the one for Iodine, as seen by its larger charge. This means that Bromine can oxidize something else (by reducing itself) better than Iodine can, so Bromine gets reduced in the equation. This eliminates 1 and 2, because in these, Bromine is being oxidized (going from -1 to 0), rather than being reduced.

Number 3 doesn’t work because both Iodine and Bromine are being reduced in the equation, and that’s not how things work in the wonderful world of chemistry - one must be reduced and the other must be oxidized.

Therefore, 4 is the correct answer, as seen by the reduction of Bromine and the oxidation of Iodine.

Final answer:

When bromine is added to an aqueous solution of iodide ions, the reaction that occurs is the oxidation of iodide ions to iodine and the reduction of bromine to bromide ions, as bromine has a higher standard reduction potential and will thus be reduced. Hence, the reaction that occurs is 2I⁻ + Br₂ → I₂ + 2Br⁻.

Explanation:

To determine which reaction will occur when bromine is added to an aqueous solution of iodide ions, we can compare the standard reduction potentials of the Br₂/Br⁻ and I₂/I⁻ couples. The standard reduction potential for Br₂/Br⁻ is +1.07V, and for I₂/I⁻ it is +0.54V. Given that a higher potential indicates a greater tendency to gain electrons (undergo reduction), we can infer that Br₂ will be reduced (gain electrons) rather than I₂.

Therefore, iodide ions (I-) will be oxidized by bromine (Br₂) to form iodine (I₂) and bromide ions (Br⁻). The balanced half-reactions are:

Br₂ + 2e⁻ → 2Br− (Reduction)2I− → I₂ + 2e− (Oxidation)

Overall, the reaction that occurs is 2I⁻ + Br₂ → I₂ + 2Br⁻, which correlates to option 4 in the list provided by the student. This is because the reaction can only occur in the direction that allows bromine to be reduced since it has a higher reduction potential.

During an oxidation-reduction reaction, the number of electrons gained is

A) equal to the number of electrons lost

B) equal to the number of protons gained

C) less than the number of electrons lost

D) less than the number of protons gained

Answers

A balanced redox reaction maintains the number of electrons transferred throughout the reaction. So, the number of electrons gained is equal to the number of electrons lost, or A.
Final answer:

During a redox reaction, the number of electrons gained is equal to the number of electrons lost, demonstrating the law of conservation of charge.

Explanation:

In an oxidation-reduction reaction, also known as a redox reaction, the number of electrons gained is equal to the number of electrons lost. This principle is known as the law of conservation of charge which states that the total charge before a reaction must equal the total charge after the reaction.

So, when one substance loses electrons (oxidation), another must gain them (reduction). Therefore, the correct answer is A) equal to the number of electrons lost.

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If you start with 4.0 grams of sodium flouride (NaF), how many grams of magnesium flouride (MgF2) will be produced? The molar mass of NaF is 42g/mole and the molar mass of MgF2 is 62g/mole.

Answers

rgrudfogheruhguirgree

How many liters of 0.45 m hcl will be required to titrate completely 1.2 l solution of 0.22 m naoh? show your work?

Answers

The number of liters  of  0.45 m Hcl  that  will be  required to titrate  completely  is calculated as below

write the equation  for reaction

that is NaOH + HCl = NaCl + H2O
find the moles  of NaOH  used = molarity x volume
= 1.2 x0.22 = 0.264  moles

by  use mole ratio between  NaOH to HCl  which  is 1:1  the  moles of HCl =0.264  moles

volume =moles/molarity

=0.264/0.45= 0.587  liters  of HCL

What is the de broglie wavelength of a 0.17-kg hockey puck moving at 37 m/s ?

Answers

Im taking lambda as x (wavelength)

m = 0.17 kg

v=37m/s

h=6.62*10^-34m²kg/s

x= h/mv


[tex]x = 6.62 \times 10 ^{ - 34} \div 0.17 \times 37[/tex]
[tex]x =1.05 \times {10}^{ - 34} m[/tex]

The wavelength of a hockey of mass 0.17kg traveling at a speed of37 m/s can be calculated by using De Broglie's equation and it comes out to be  1.05m×10⁻³⁴m.

What is De Broglie's equation?

The De Broglie's equation is an equation which shows that a particle has a wave length and hence particle can show wave nature. Any matter can have two nature one is particle nature and other is wave nature.

The wave nature in a particle was given by De Broglie Particle nature of a matter can be described by the phenomenon of black body radiation. Photoelectric effect two shows particle nature nature of light.

Mathematically,

[tex]{\lambda}=\frac{h}{mv}[/tex]

where,

[tex]{\lambda}[/tex]=wavelength=?

m=mass of hockey=0.17kg

h=Plank's constant

v=velocity of hockey=37 m/s

Substituting all given values in the equation

[tex]\lambda[/tex]= (6.626×10⁻³⁴)÷(0.17kg ×37 m/s )=1.05m×10⁻³⁴m

Thus the wavelength of motorcycle is 1.05m×10⁻³⁴m.

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How many milliliters of 0.564 m hcl are required to react with 6.03 grams of caco3 ?

Answers

First write up a balanced equation
CaCO3 + 2 HCl = CO2 + CaCl2 + H2O

Find how many moles of CaCO3 you are dealing with
6.03 g x 1 mol / 100 g = .0603 moles

However, you need twice as much HCl
.0603 x 2 moles / .564 M = .21 L

When two molecules of methanol (ch3oh) react with oxygen, they combine with three o2 molecules to form two co2 molecules and four h2o molecules. how many h2o molecules are formed when 66 methanol molecules react?

Answers

[tex]2CH_{3}OH +3O_{2}--\ \textgreater \ 2CO_{2}+4H_{2}O From reaction 2 molecules CH_{3}OH give 4 molecules H_{2}O. [/tex]

[tex] \frac{2 molecules CH_{3}OH}{4 molecules H_{2}O} = \frac{66 molecules CH_{3}OH}{x molecules H_{2}O} x= 4*66/2=132 molecules of H_{2}O[/tex]

The number of water molecules formed when 66 molecules of ethanol react is; 132 molecules of water.

When two molecules of methanol react with oxygen.

They combine with three O2 molecules as implied in the question to form CO2 molecules and four H20 molecules.

The reaction between methanol and oxygen is as follows;

2CH3OH + 3O2 --> 2CO2 + 4H2O

According to the equation,

2 molecules of CH3OH = 4 molecules of H2O

Therefore,

66 molecules of CH3OH = x molecules of H2O

In essence, the number of x molecules of water formed when 66 molecules of ethanol react is;

x = (66 × 4)/2

x = 132 molecules of water.

The number of water molecules formed when 66 molecules of ethanol react is; 132 molecules of water.

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For a reaction to be spontaneous under standard conditions at all temperatures, the signs of δh° and δs° must be __________ and __________, respectively.

Answers

Final answer:

A reaction is spontaneous under standard conditions at all temperatures when ΔH° is negative and ΔS° is positive, leading to a negative Gibbs free energy change (ΔG°).

Explanation:

For a reaction to be spontaneous under standard conditions at all temperatures, the signs of ΔH° (ΔH degree) and ΔS° (ΔS degree) must be negative and positive, respectively. According to the Gibbs free energy equation, ΔG° = ΔH° - TΔS°, where T represents the absolute temperature in Kelvin. If ΔH° is negative (exothermic reaction) and ΔS° is positive (increase in entropy), the Gibbs free energy change (ΔG°) will always be negative, indicating that the reaction is spontaneous. Conversely, if ΔH° were positive and ΔS° negative, ΔG° would be positive and the reaction non-spontaneous at all temperatures.

which of the following naturally orccuing radioisotopes would be most useful in dating objects

Answers

Potassium -40half life = 1.28 ×10∧9.
Radioisotope can be used to determine the age of fossils, to treat skin disease, to sterilize foodstuffs, and also to sterilize surgical instruments.
Radioisotope dating is a way to estimate a fossil's age by a method of analyzing the elemental isotopes which are within the rocks which it is in.

Give the percent yield when 28.16 g of co2 are formed from the reaction of 4.000 moles of c8h18 with 4.000 moles of o2. 2 c8h18(l) + 25 o2(g) → 16 co2(g) + 18 h2o(g) molar mass co2 = 44.01 g/mol

Answers

Final answer:

The percent yield of CO2 is calculated by dividing the actual yield (28.16 g of CO2) by the theoretical yield determined from the stoichiometry of the balanced chemical equation and the limiting reactant, and then multiplying by 100 to get the percentage.

Explanation:Percent Yield Calculation

The question asks for the percent yield of CO2 produced from the reaction of C8H18 with O2. First, we need to consider the balanced chemical equation:

Write the balanced chemical equation: 2 C8H18(l) + 25 O2(g)
ightarrow 16 CO2(g) + 18 H2O(g).Identify the limiting reactant. Since 4.000 moles of C8H18 and 4.000 moles of O2 are provided, O2 is the limiting reactant because the reaction requires 25 moles of O2 for every 2 moles of C8H18.Calculate the theoretical yield of CO2. Based on the stoichiometry of the reaction, one mole of O2 would produce 16/25 moles of CO2, so 4.000 moles of O2 would produce 4.000 * (16/25) moles of CO2.Convert moles of CO2 to grams. The molar mass of CO2 is 44.01 g/mol, so the theoretical yield in grams is the number of moles times the molar mass.Calculate the percent yield using the formula: (actual yield / theoretical yield) * 100. The actual yield is given as 28.16 g of CO2.

Therefore, the percent yield can be calculated by dividing the actual yield of CO2 (28.16 g) by the theoretical yield (calculated in step 4), and then multiplying by 100 to get a percentage.

Predict which substance in each pair has greater molar entropy. (1) no2(g) n2o4(g) (2) ch3och3(l) ch3ch2oh(l) (3) hcl(g) hbr(g)

Answers

Entropy is measure of disorder in system. Higher the disorder, greater the entropy.

Pair 1: NO2 (g) and N2O4 (g)
In the above case, entropy of N2O4 is more as compared to NO2, because N2O4 has more number of bonds as compared to NO2. So large number of vibrational energy levels are available for energy distribution which results in increasing entropy.

Pair 2: CH3OCH3(l) and CH3CH2OH(l)
Above compounds are structure isomers. They have same number of bonds and vibrational energy levels. However, presence of -OH group in CH3CH2OH results in intermolecular interaction via H bonding. This results in an ordered structure in CH3CH2OH as compared to  CH3OCH3. Due to this entropy in CH3CH2OH is lower as compared to CH3OCH3.

Pair 3: HCl(g) and HBr(g)
In present case, HBr will have higher entropy as compared to HCl, because of larger number of sub-atomic particles in Br. Also, the higher molecular mass of HBr favors larger entropy. 
Final answer:

The substances with greater molar entropy in each case are: NO2(g), CH3OCH3(l), and HBr(g), due to factors such as number of atoms, molecule complexity and being heavier gases.

Explanation:

In predicting the substance with greater molar entropy in each pair, we have to consider the number of molecules and atoms, phase of matter and complexity of molecules. Here are the predicted substances:

NO2(g) - This molecule is monatomic when compared to N2O4. Molecules with more atoms have higher positional entropy, hence NO2 has higher molar entropy.CH3OCH3(l) - This is a more complex molecule as compared to CH3CH2OH. More complex molecules have higher entropy as there are more possible configurations of their atoms.HBr(g) - This substance is a gas and contains bromine which is heavier than chlorine in HCl. Heavier gases have larger molar entropy.

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Bombardment of uranium-235 with a neutron generates tellurium-135, 3 neutrons, and ________.

Answers

Answer is: zirconium-98.
Nuclear reaction: ²³⁵U + n°→ ¹³⁵Te + ⁹⁸Zr + 3n°.
Atomic number of uranium is 92, atomic number of zirconium (Zr) is 40 and atomic number of tellurium (Te) is 52, 92 = 40 + 52.
Uranium has atomic mass 235, zirconium has 98 and tellurium has 135:
235 + 1 = 135 + 98 +3, 236 = 236.
n° is neutron.
Symbol for atomic number is Z and for atomic mass is A.

Bombardment of uranium-235 with a neutron produces tellurium-135, three neutrons, and molybdenum-98.

When uranium-235 (U-235) is bombarded with a neutron, it first forms U-236, which is unstable.

This unstable isotope undergoes nuclear fission, resulting in the production of tellurium-135 (Te-135), three neutrons, and another element, which we need to determine.

In fission reactions, the sum of the atomic masses and the number of protons must be equal on both sides of the equation.

Here’s the reaction:

U-235 + neutron ⟶ Te-135 + 3 neutrons + X

Starting with 235 (U-235) + 1 (neutron) = 236 mass units total.

The mass of Te-135 is 135 units.

Since three neutrons (3 x 1 = 3 units) are produced, the remaining mass is:

236 - 135 - 3 = 98 units.

Since tellurium (Te) has an atomic number of 52, and assuming the fission product balances the protons and neutrons in the equation, the remaining fission product is molybdenum-98 (Mo-98) with an atomic number of 42.

Completing the equation:

U-235 + neutron ⟶ Te-135 + 3 neutrons + Mo-98.

what is the name of this molecule? h3c-c=c-ch3

Answers

Answer : The name of the given molecule is 2-butene or but-2-ene

Explanation :

Step 1 : Find the longest carbon chain.

The longest carbon chain for the given structure contains 4 carbons.

This will help us to find out the prefix for our parent compound.

The alkane with 4 carbons is "Butane"

Therefore our parent compound will have the prefix is "but"

Step 2 : Find the functional group

The functional group is the group that gives specific characteristics to the compound. In this case, we have a double bond in the structure which behaves as a functional group.

Therefore the functional group here is "Alkene"

Functional groups helps us to find the suffix of the name. In this case it is "ene"

Therefore our parent compound is but+ene = butene

Step 3 : Find the position of the double bond

The double bond is present between second and third carbon atom.

We always select the lower number to denote the position of the bond.

So we have 2-butene OR but-2-ene

There are no substituents present in this compound.

Therefore the name of the compound is 2-butene or But-2-ene

Final answer:

The molecule represented by H3C-C=C-CH3 is named But-2-ene, according to the IUPAC nomenclature system.

Explanation:

H3C-C=C-CH3 is named But-2-ene, according to the IUPAC nomenclature system. The molecule represented by the formula H3C-C=C-CH3 is named But-2-ene. In this name, 'But' refers to the four carbon atoms that form the base of the molecule, '-2-' indicates that the double bond is between the second and third carbon atom, and 'ene' signifies the presence of a carbon-carbon double bond in the molecule. This nomenclature is part of the International Union of Pure and Applied Chemistry (IUPAC) system, which is a standardized method for naming chemical compounds.

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A 10.0-ml sample of 0.200 m hydrocyanic acid (hcn) is titrated with 0.0998 m naoh. what is the ph at the equivalence point? for hydrocyanic acid, pka = 9.31

Answers

when the titration of HCN with NaOH is:

HCN (aq) + OH- (aq) → CN-(aq) + H2O(l)

So we can see that the molar ratio between HCN: OH-: CN- is 1:1 :1

we need to get number of mmol of HCN = molarity * volume 

                      = 0.2 mmol / mL* 10 mL = 2 mmol

so the number of mmol of NaOH = 2 mmol according to the molar ratio

so, the volume of NaOH = moles/molarity

                                          = 2 mmol / 0.0998mL

                                          = 20 mL

and according to the molar ratio so, moles of CN- = 2 mmol

∴the molarity of CN- =  moles / total volume 

                                   = 2 mmol / (10mL + 20mL ) = 0.0662 M

when we have the value of PKa = 9.31 and we need to get Pkb

so, Pkb= 14 - Pka

            = 14 - 9.31 = 4.69 

when Pkb = -㏒Kb

         4.69 = -㏒ Kb 

∴ Kb = 2 x 10^-5

and when the dissociation reaction of CN- is:

CN-(aq) + H2O(l) ↔ HCN(aq) + OH- (aq) 

by using the ICE table:

∴ the initials concentration are:

[CN-] = 0.0662 M

and [HCN] = [OH]- = 0 M

and the equilibrium concentrations are:

[CN-] = (0.0662- X)

[HCN] = [OH-]= X

when Kb expression = [HCN][OH-] /[CN-]

by substitution:

2 x 10^-5 = X^2 / (0.0662 - X)

X = 0.00114 

∴[OH-] = X = 0.00114

when POH = -㏒[OH]

                    = -㏒ 0.00114

POH = 2.94

∴PH = 14 - 2.94 = 11.06



 

Final answer:

The pH at the equivalence point of the titration can be calculated using the Henderson-Hasselbalch equation. The pH at the equivalence point is 9.31.

Explanation:

The pH at the equivalence point of the titration can be calculated using the Henderson-Hasselbalch equation. The equivalence point occurs when the moles of hydrocyanic acid (HCN) is equal to the moles of sodium hydroxide (NaOH).

In this case, the hydrocyanic acid (HCN) is a weak acid, and the sodium hydroxide (NaOH) is a strong base. At the equivalence point, the weak acid is completely neutralized by the strong base, forming the conjugate base of the acid.

Using the Henderson-Hasselbalch equation, we can calculate the pH at the equivalence point:

pH = pKa + log([A-]/[HA])

The concentration of the hydrocyanic acid (HCN) is 0.200 M, and since hydrocyanic acid is a weak acid, its concentration can be assumed to remain nearly constant during the titration. Therefore, the concentration of the conjugate base at the equivalence point is also 0.200 M.

Plugging in the values, we have:

pH = 9.31 + log(0.200/0.200) = 9.31

So, the pH at the equivalence point is 9.31.

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