Calculate the energy required to heat 566.0mg of graphite from 5.2°C to 23.2°C. Assume the specific heat capacity of graphite under these conditions is ·0.710J·g−1K−1 . Be sure your answer has the correct number of significant digits.

Answer:

1.543 pounds = 700 grams

Answer:

The correct answer is 1.60 Liters.

Explanation:

The given reaction:

CO (g) + H₂(g) ⇔ CH₃OH (g)

Based on the given reaction, two moles of H₂ reacts with one mole of CO and produce one mole of CH₃OH.

It is mentioned that 3.20 L of H₂ is reacted, therefore, there is a need to convert it into moles.

As 22.4 L at standard temperature and pressure is equivalent to 1 mole.

Therefore, 1 L at STP will be, 1/22.4 mole

Now 3.20 L at STP will be,

= 1/22.4 × 3.20

= 0.1428 mole

And as mentioned in the reaction that 2 moles of H₂ gives 1 mole of CH₃OH, therefore, 1 mole of H₂ will give 1/2 mole of CH₃OH

Now, 0.1428 mole of H₂ will give,

= 0.1428/2 = 0.071 mole of CH₃OH

= 0.071 × 22.4 = 1.60 L

The volume, in liters, of CH₃OH gas formed is 1.60 L

From the question,

We are to determine the volume of CH₃OH formed

The given chemical equation for the reaction is

CO(g)+ H₂(g) → CH₃OH

The balanced chemical equation for the reaction is

CO(g)+ 2H₂(g) → CH₃OH

This means

1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH

Now, we will determine the number of moles of H₂ present in the 3.20 L H₂ at STP

1 mol of an ideal gas has a volume of 22.4 L at STP

Then,

x mole of the H₂ gas will have a volume of 3.20 L at STP

x = [tex]\frac{3.20 \times 1}{22.4}[/tex]

x = 0.142857 mole

∴ The number of mole of H₂ present is 0.142857 mole

Since

2 moles of H₂ reacts to produce 1 mole of CH₃OH

Then,

0.142857 mole of H₂ will react to produce [tex]\frac{0.142857}{2}[/tex] mole of CH₃OH

[tex]\frac{0.142857}{2} = 0.0714285[/tex]

∴ The number of moles of CH₃OH produced = 0.0714285 mole

Now, for the volume of CH₃OH formed

Since

1 mol of an ideal gas has a volume of 22.4 L at STP

Then,

0.0714285 mol of CH₃OH will have a volume of 22.4 × 0.0714285  at STP

22.4 × 0.0714285 = 1.5999984 L ≅ 1.60 L

Hence, the volume of CH₃OH gas formed is 1.60 L

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Answer:

See explanation

Explanation:

1)

Mg3N2(s) + 6H2O(l) ------------> 3Mg(OH)2 + 2NH3(g)

Number of moles of magnesium nitride= mass/molar mass= 4.86g/100.9494 g/mol = 0.048 moles

1 mole of magnesium nitride yields 3 moles of magnesium hydroxide

0.048 moles of magnesium nitride yields 0.048 moles × 3= 0.144 moles of magnesium hydroxide

Theoretical yield of magnesium hydroxide = 0.144 moles × 58.3197 g/mol = 8.398 g

Percent yield= actual yield/ theoretical yield × 100

Percent yield= 7.18/8.398 × 100/1 = 85.5%

2)

N2(g) + 3H2(g) -------> 2NH3(g)

Number of moles of hydrogen gas = mass/ molar mass = 6.41g/ 2gmol-1 = 3.205 moles of hydrogen gas.

From the balanced reaction equation;

3 moles of hydrogen gas yields 2 moles of ammonia

3.205 moles of hydrogen gas yields 3.205 × 2/3 = 2.1367 moles of ammonia

Theoretical yield of ammonia = 2.1367 moles × 17 gmol-1 = 36.3 g

Percent yield = actual yield/ theoretical yield ×100

Percent yield = 26.2/36.3 ×100

Percent yield = 72.2%

3)

3Cl2(g) + 3H2O(l) ------> HOCl3(aq) + 5HCl(aq)

Number of moles of water= mass/ molar mass = 3.79g/18 gmol-1 = 0.21 moles

Since

3 mole of water yields 5 mole of HCl

0.21 moles of water yields 0.21 × 5/3 = 0.35 moles of HCl

Theoretical yield of HCl = 0.35 moles × 36.5 gmol-1 = 12.775 g

Percent yield = actual yield/ theoretical yield × 100/1

Percent yield = 8.70/12.775 ×100

Percent yield = 68.1%

Answer:

[tex]1.25~mol~H_2O[/tex] and [tex]0.627~mol~N_2[/tex]

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ([tex]N_2H_4[/tex]) and oxygen ([tex]O_2[/tex]). So, we can start with the reaction between these compounds:

[tex]N_2H_4~+~O_2~->~N_2~+~H_2O[/tex]

Now we can balance the reaction:

[tex]N_2H_4~+~O_2~->~N_2~+~2H_2O[/tex]

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for [tex]N_2H_4[/tex] and 31.99 g/mol for [tex]O_2[/tex]):

[tex]20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4[/tex]

[tex]20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2[/tex]

In the balanced reaction we have 1 mol for each reagent (the numbers in front of [tex]O_2[/tex] and [tex]N_2H_4[/tex] are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is [tex]N_2H_4[/tex].

With this in mind, we can calculate the number of moles for each product. In the case of [tex]N_2[/tex] we have a 1:1 molar ratio (1 mol of [tex]N_2[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2[/tex]

We can follow the same logic for the other compound. In the case of [tex]H_2O[/tex] we have a 1:2 molar ratio (2 mol of [tex]H_2O[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:

[tex]0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O[/tex]

I hope it helps!

Answer:

Gay lussacs law

Explanation:

Answer:

[tex]x_{et}=0.6068[/tex]

Explanation:

Hello,

In this case, since the mole fraction of a compound, in this case ethanol, in a binary mixture, in this constituted by both water and ethanol, is mathematically defined as follows:

[tex]x_{et}=\frac{n_{et}}{n_{et}+n_{w}}[/tex]

Whereas [tex]n[/tex] accounts for the moles in the solution for each species, we must first compute the moles of both ethanol (density: 0.789 g/mL and molar mass: 46.07 g/mol) and water (density: 1g/mL and molar mass: 18.02 g/mol)

[tex]n_{et}=10.00mL\ et*\frac{0.789g\ et}{mL\ et} *\frac{1mol\ et}{46.07g\ et}=0.1713mol\ et\\ \\n_w=2.00mL\ w*\frac{1g\ w}{mL\ w} *\frac{1mol\ w}{18.02g\ w}=0.1110mol\ w[/tex]

Therefore, the mole fraction turns out:

[tex]x_{et}=\frac{0.1713mol}{0.1713mol+0.1110mol}\\\\x_{et}=0.6068[/tex]

Best regards.

Answer:

The empirical formulae for the compound is CH3.

Answer:

Al(s)<NaF(s)<HF(g)<SiH4(g)<SiF4(g)

Explanation:

Hello,

In this case, we can arrange the increasing order of entropy at 25 \°C by taking into account, at first, that since solids are more molecularly organized than gases, the first we have solid sodium fluoride and solid aluminium, but in this case, as the higher the molar mass, the higher the entropy, the molar mass of aluminium is 27 g/mol and 42 g/mol for sodium fluoride, therefore, we first have:

Al(s)<NaF(s)

Afterwards, since the molar mass of hydrogen fluoride (HF), silicon fluoride (SiF4) and silane (SiH4) are 20, 104 and 32 g/mol respctively, since silicon fluoride has the greater molar mass, it also has the higher entropy. In such a way, the overall order turns out:

Al(s)<NaF(s)<HF(g)<SiH4(g)<SiF4(g)

Best regards.

Answer: Reaction (1) , (3) and (4) are accompanied by an increase in entropy.

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from  an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

(1) [tex]2SO_2(g)+O_2(g)\rightarrow SO_3(g)[/tex]

3 moles of reactant are changing to 1 mole of product , thus the randomness is increasing. Thus the entropy also increases.

2) [tex]H_2O(l)\rightarrow H_2O(s)[/tex]

1 mole of Liquid reactant is changing to 1 mole of solid product , thus the randomness is decreasing. Thus the entropy also decreases.

3) [tex]Br_2(l)\rightarrow Br_2(g)[/tex]

1 mole of Liquid reactant is changing to 1 mole of gaseous product , thus the randomness is increasing. Thus the entropy also increases.

4)  [tex]H_2O_2(l)\rightarrow H_2O(l)+\frac{1}{2}O_2(g)[/tex]

1 mole of Liquid reactant is changing to half mole of gaseous product and 1 mole of liquid product, thus the randomness is increasing. Thus the entropy also increases.

Answer:

Difference of the enthalpy (of a system) minus the product of the entropy and absolute temperature

Explanation:

The basis of spontaneity in a chemical reaction is that ∆G must be negative. ¡∆G is known as the change in free energy of a system. If ∆G is negative, then the reaction will occur without any external help (the reaction is spontaneous at room temperature).

∆G is given by;

∆G= ∆H -T∆S

Where;

∆H= change in enthalpy of the system

T= absolute temperature of the system

∆S= change in entropy

Hence; when ∆H -T∆S gives a negative result, the reaction proceeds without any external help.

Answer:

b.) Put on protective gloves

Answer:

2020 Put on protective gloves.

Explanation:

The element which can not loose electron easily and having electronagtive character is called non-metal it has following property-

1. it can not conduct heat and electricity

2. it is netiher ductile not malleable

3. it is not lsuturous and also not sonorous

Explanation:

a nonmetal (or non-metal) is a chemical element that mostly lacks the characteristics of a metal. Physically, a nonmetal tends to have a relatively low melting point, boiling point, and density. A nonmetal is typically brittle when solid and usually has poor thermal conductivity and electrical conductivity. Chemically, nonmetals tend to have relatively high ionization energy, electron affinity, and electronegativity. They gain or share electrons when they react with other elements and chemical compounds. Seventeen elements are generally classified as nonmetals: most are gases (hydrogen, helium, nitrogen, oxygen, fluorine, neon, chlorine, argon, krypton, xenon and radon); one is a liquid (bromine); and a few are solids (carbon, phosphorus, sulfur, selenium, and iodine). Metalloids such as boron, silicon, and germanium are sometimes counted as nonmetals.

Answer:

Molar solubility is 1.12x10⁻⁴M

Explanation:

The dissolution of magnesium hydroxide is:

Mg(OH)₂(s) ⇄ Mg²⁺ + 2OH⁻

The molar solubility represents the moles of the solid that the solution can dissolve, that could be written as:

Mg(OH)₂(s) ⇄ X + 2X

Where X is solubility.

If you obtained a [OH⁻] = 2.24x10⁻⁴M and you know [OH⁻] = 2X:

2X = 2.24x10⁻⁴M

X = 2.24x10⁻⁴M/2

X =1.12x10⁻⁴M

Answer:

Ca²⁺ and Cl⁻

Explanation:

In a chemical reaction, spectator ions are ions that are not involved in the reaction, that means are the same before and after the reaction.

In water, calcium hydroxide, Ca(OH)₂ is dissociated in Ca²⁺ and OH⁻. Also, hydrochloric acid, HCl, dissociates in H⁺ and Cl⁻. The reaction is:

Ca²⁺ + 2OH⁻ + 2H⁺ + 2Cl⁻ → 2H₂O + Ca²⁺ + 2Cl⁻

The ions that react are H⁺ and OH⁻ (Acid and base producing water)

And the ions that are not reacting, spectator ions, are:

Answer:

The rate would have doubled

Explanation:

Answer:

NaN3 = 47.2 g

Explanation:

Given:

2 NaN3 ⇒ 2 Na  + 3 N2

Find:

Amount of NaN3

Computation:

N2 moles = Product of N2 / molar mass of N2

N2 moles =30.5/28

N2 moles = 1.0893

2NaN3 makes 3(N2 )

So,

NaN3 moles = (2/3) moles of N2  

NaN3 moles = ( 2/3) × 1.0893

NaN3 moles = = 0.7262

NaN3 mass = 0.7262 x 65

NaN3 = 47.2 g

Answer:

NaN3 = 47.2 g

Explanation:

Given:

2 NaN3 ⇒ 2 Na  + 3 N2

Find:

Amount of NaN3

Computation:

N2 moles = Product of N2 / molar mass of N2

N2 moles =30.5/28

N2 moles = 1.0893

2NaN3 makes 3(N2 )

So,

NaN3 moles = (2/3) moles of N2  

NaN3 moles = ( 2/3) × 1.0893

NaN3 moles = = 0.7262

NaN3 mass = 0.7262 x 65

NaN3 = 47.2 g

Explanation:

Answer:

7.4 × 10⁻⁹ M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for Li₃PO₄: 5.9 × 10⁻¹⁷

Concentration of lithium ion: 0.0020 M

Step 2: Write the reaction for the solution of Li₃PO₄

Li₃PO₄(s) ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

Step 3: Calculate the phosphate concentration required for a precipitate to occur

The solubility product constant is:

Ksp = 5.9 × 10⁻¹⁷ = [Li⁺]³ × [PO₄³⁻]

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / [Li⁺]³

[PO₄³⁻] = 5.9 × 10⁻¹⁷ / 0.0020³

[PO₄³⁻] = 7.4 × 10⁻⁹ M

Answer:

226Ra88→222Rn86+4He2

Explanation:

An α-particle usually consists of a helium nucleus which indicates the type of decay that was undergone in this radioactive process.

During α-decay(alpha decay), an atomic nucleus emits an alpha particle.

Answer:

b) the Ka of NH₄⁺ to find the hydronium concentration

Explanation:

The equilbrium of NH₄⁺ (The conjugate acid of NH₃, a weak base), is:

NH₄⁺ ⇄ NH₃ + H⁺

Where Ka of the conjugate acid is:

Ka = [NH₃] [H⁺] / [NH₄⁺]

Thus, if you know Ka of NH₄⁺ and its molar concentration you can calculate  [H⁺], the hydronium concentration, to find pH (Because pH =  -log [H⁺])

Thus, right option is:

Answer:

The correct answer is - option B -  lowest b.p.... isobutane < acetone < isopropanol ...highest b.p.

Explanation:

Isobutane has lowest boiling point due to no hydrogen bonding and no diole to dipole interaction found in them. Isobutane only shows weak dispersion force.

Acetone has dipole dipole interaction but due to lack of Hydrogen bonding they have low boiling point than isopropanol but higher than isobutanol.

Isopropanol is the compound that has ability to form hydrogen bonding with other molecule its boiling point is maximum among all three.

Thus, the correct answer is - option B -  lowest b.p.... isobutane < acetone < isopropanol ...highest b.p.

Answer:

I hope it works

Explanation:

As we know that

w=m*g

given m=0.145 , g=9.8

hence we get

w= (9.8)*(0.145)

w=1.421 m/sec 2

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3 2 1 is the set of quantum numbers.

The set of numbers used to describe the position and energy of the electron in an atom is called quantum numbers. There are four quantum numbers, namely, principal, azimuthal, magnetic, and spin quantum numbers.

The rules for quantum numbers are: (n) can be any positive, nonzero integral value. (l) can be zero or any positive integer but not larger than (n-1). l = 0, 1, 2, 3, 4, …. (n-1) (ml) values follow the equation.

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Answer:

[tex]r_A=-1\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, given the rate of production of C, we can compute the rate of consumption of A by using the rate relationships which include the stoichiometric coefficients at the denominators (-1 for A and 2 for C) as follows:

[tex]\frac{1}{-1} r_A=\frac{1}{2}r_C[/tex]

In such a way, solving the rate of consumption of A, we obtain:

[tex]r_A=-\frac{1}{2} r_C=-\frac{1}{2}*2\frac{M}{s}\\ \\r_A=-1\frac{M}{s}[/tex]

Clearly, such rate is negative which account for consumption process.

Regards.

Answer:

4.46 mol of NH3

Explanation:

The equation of he reaction is given as;

2N + 3H2 --> 2NH3

From the stochiometry of the reaction, 1 mol of Nitrogen produces 2 mol of Ammonia.

Mass of Nitrogen = 31.2g

Molar mass of Nitrogen = 14g/mol

Number of moles = Mass / Molar mass = 31.2 / 14 = 2.23 mol

Since 1 mol of N = 2 mol of NH3;

2.23 mol of N2 would produce x

x = 2.23 * 2 = 4.46 mol of NH3

Answer:

False

Explanation:

The reaction rate increases if the temperature also increases, as does the concentration of ractives or the presence of catalysts.

The reaction rate talks about how reagents are converted into products as a function of time, this process can take less or more depending on the factors to which the reaction is exposed.

The increasing temperature generates an increase in the kinetic energy of the particles, promoting their proximity and their reaction between them to be able to give the final product, so a faster reaction occurs, which is why it has promoted when the particles collide.

Answer:

b. the amount of heat given off or absorbed

Explanation:

Hello,

In this case, we should take into account a formal definition of enthalpy change such as an energetic change that occurs in a system when matter is transformed by a given chemical reaction from reactants to products. Thus, such energetic change is macroscopically exhibited and it is related with either a temperature increase or decrease; it means that if a reaction exhibits a temperature increase, we say that heat was given off and if the temperature exhibits a decrease, we say that heat is absorbed. For that reason, answer is b. the amount of heat given off or absorbed.

Regards.

Answer:

Major product does not undergo oxidation since it is a tertiary alcohol whereas minor product undergoes oxidation to ketone as it is  secondary alcohol.

Explanation:

Hello,

In this case, given the attached picture, the hydration of the 1 methylcyclohexene yields to alcohols; 1-methylcyclohexan-1-ol and 1-methylcyclohexan-2-ol. Thus, since the OH in the 1-methylcyclohexan-1-ol (major product) is bonded to a tertiary carbon (bonded with other three carbon atoms) it is not able to increase the number of oxygen bonds (oxidation) as it already attained the octet whereas the 1-methylcyclohexan-2-ol (minor product) is able to undergo oxidation to ketone as the carbon bonded to it is secondary (bonded with other two carbon atoms), so one extra bond the oxygen is allowed to be formed to carbonyl.

Best regards.

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

[tex]MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-[/tex]

And the undergoing chemical reaction:

[tex]MgCl_2+2NaF\rightarrow MgF_2+2NaCl[/tex]

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

[tex]n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2[/tex]

Next, the moles of magnesium chloride consumed by the sodium fluoride:

[tex]n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2[/tex]

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

[tex]n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2[/tex]

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

[tex][Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M[/tex]

[tex][F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M[/tex]

Thereby, the reaction quotient is:

[tex]Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}[/tex]

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

Answer:

12.5g

Explanation:

Half life = 2.4 Minutes.

The half life of a compound is the time it takes to decay to half of it's original concentration or mass.

Time lapsed= 7.2 minutes. This is equivalent to 3 half lives ( 3 * 2.4)

Initial mass = 100g

First half life;

100g --> 50g

Second half life;

50g --> 25g

Third half life;

25g --> 12.5g

The amount of Zn-71 that remains after 7.2 mins has elapsed is 12.5 g

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 2.4 mins

Time (t) = 7.2 mins

[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{7.2}{2.4} \\\\[/tex]

Thus, 3 half-lives has elapsed.

Original amount (N₀) = 100 g

Number of half-lives (n) = 3

[tex]N = \frac{N_{0}}{2^{n}} \\\\N = \frac{100}{2^{3 }}\\\\N = \frac{100}{8}\\\\[/tex]

Thus, the amount of Zn-71 that remains after 7.2 mins is 12.5 g

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Answer: The number of moles of a solute per kilogram of solvent

Explanation: