The specific heat capacity of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 °C to 37 °C? a. 5.8 × 10-4 J b. 0.13 J c. 29 J d. 2.0 J e. -0.13 J

Answer:

A. 3.04×10^-19J

Explanation:

Hope this will help you.

Explanation:

here's the answer. I just plug the expression into my calculator and find the intercept to avoid the quadratic formula

Answer:

XCH₄ = 0.461

XCO₂ = 0.539

Explanation:

Step 1: Given data

Step 2: Calculate the total pressure in the container

We will sum both partial pressures.

P = pCH₄ + pCO₂

P = 431 mmHg + 504 mmHg = 935 mmHg

Step 3: Calculate the mole fraction of each gas

We will use the following expression.

Xi = pi / P

XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461

XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539

Answer:

A

Explanation:

it increase the rate of reaction when necessary

The chemical change that is  responsible for producing the heat that maintains the temperature of your body is metabolism.

Chemical changes are defined as changes which occur when a substance combines with another substance to form a new substance.Alternatively, when a substance breaks down or decomposes to give new substances it is also considered to be a chemical change.

There are several characteristics of chemical changes like change in color, change in state , change in odor and change in composition . During chemical change there is also formation of precipitate an insoluble mass of substance or even evolution of gases.

There are three types of chemical changes:

1) inorganic changes

2)organic changes

3) biochemical changes

During chemical changes atoms are rearranged and changes are accompanied by an energy change as new substances are formed.

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Answer:

56.5J

Explanation:

To find the heat energy required use the formula for the specific heat capacity which is

c=quantity of heat/mass×change in temperature

in this question c is 0.226j/g,the mass is 50g and the change in temperature is 30-25=5

therefore

0.226=Q/50×5

Q=0.226×250

=56.5J

I hope this helps

Answer:

D

Explanation:

We have to bear in mind that the acid is a weak acid. A weak acid does not dissociate completely in solution. We will have more concentration of undissociated acid than A^- and H3O^+ and OH^- in the system at equilibrium.

Being a weak acid, there is maximum concentration of water molecules followed by that of undissiociated acid.

Hence, for this solution, the concentration of ions in solution follows the order;

[H2O] > [HA] > [A-] ~ [H3O ] > [OH-]

Answer:

K = 3.3

Explanation:

Nitric acid, HNO3, reacts with nitrogen monoxide, NO, to produce nitrogen dioxide, NO2 and water H2O as follows:

2HNO3(g) + NO(g) → 3NO2(g) + H2O(g)

Where equilibrium constant, K, is:

K = [NO2]³[H2O] / [HNO3]²[NO]

[] is the molar concentration of each species at equilibrium.

To solve this question we need to find molarity of each gas and replace these in the equation as follows:

[NO2] -Molar mass NO2-46.0g/mol-

18.6g * (1mol/46.0g) = 0.404mol / 7.7L = 0.0525M

[H2O] -Molar mass:18.01g/mol-

236.7g * (1mol/18.01g) = 13.14 moles / 7.7L = 1.707M

[HNO3] -Molar mass:53.01g/mol-

16.2g * (1mol/53.01g) = 0.3056 moles / 7.7L = 0.0397M

[NO] -Molar mass: 30.0g/mol-

11.0g * (1mol/30.0g) = 0.367 moles / 7.7L = 0.0476M

Replacing:

K = [NO2]³[H2O] / [HNO3]²[NO]

K = [0.0525M]³[1.707M] / [0.0397M]²[0.0476M]

The nature of attack on sites in a molecule depends on the nature of such sites. The following are the nature of the sites mentioned in the question:

1) The indicated carbon in compound A is electrophilic.

2) The indicated bond in compound B is nucleophilic.

3) The indicated carbon in compound C is electrophilic.

4) The indicated carbon in compound D is neither electrophilic nor nucleophilic.

5) The indicated oxygen in compound E is nucleophilic.

The terms "electrophilic" and "nucleophilic" are very common in chemistry.

An electrophilic center is usually positively charged, has a positive dipole or is electron deficient hence it attacks negative centers. The term itself means "electron loving". That actually means that it has an affinity for negative charges.

The -I inductive effect of the bromine atom in the carbon in compound A makes that carbon atom to be electrophilic. Also, the carbonyl bond and the O C H 3 attached to the carbon in compound C also makes it electrophilic.

The term "nucleophilic" literately means "nucleus loving". That means a specie that has a high affinity for positive charges. This specie must be electron rich.

The carbon atom in compound B has a double bond which is electron rich and can attack any positive center hence it is nucleophilic. Also, the oxygen atom in E bears two lone pairs of electrons which can attack any positive center in a molecule hence the oxygen atom is also nucleophilic.

In compound D, the carbon atom is bonded to a methyl, two hydrogens and a carbon. There is a nitrogen atom two bonds away. There is no +I or -I inductive effect on this carbon atom because the nitrogen atom is far away. Therefore, the indicated carbon in compound D is neither electrophilic nor nucleophilic.

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Answer:

catenation

Explanation:

Carbon atoms have four electrons to share in bonding environments to get to the ideal octet. To do this, it bonds with other carbon molecules, called catenation. Catenation is the ability of an atom to bond and share electrons with other atoms of its kind.

Answer:

the % yield is 82%

Explanation:

Given the data in the question,

we know that;

Molar mass of benzil is 210.23 g·mol−1

Molar mass of dibenzyl ketone is 210.27 g·mol−1

Molar mass of tetraphenylcyclopentadienone is 384.5 g·mol−1

Now,

2.0 g benzil = 2 g / 210.23 g·mol−1 = 0.0095 mole

2.2 g dibenzyl ketone = 2.2 / 210.27 = 0.0105 mole

3.0 g of tetraphenylcyclopentadienone = 3 / 384.5  = 0.0078 mole

Now, the limiting reagent is benzil. 0.0095 mole can reacts wiyh 0.0095 mole of dibenzyl ketone

percentage yield = ( 0.0078 mole / 0.0095 mole ) × 100%

= 0.82 × 100%

= 82%

Therefore, the % yield is 82%

Explanation:

Amount of water required in each case:

(a)The mass% of the solution is:9.95

Mass of solute that is urea is 6.80 g

To determine the mass of solvent water use the formula:

[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\9.95=(6.80g/mass of solution )x100\\mass of solution =(6.80 /9.95)x100\\=68.3 g[/tex]

Hence the mass of solvent = mass of solution - the mass of solute

=68.3 g - 6.80g

=61.5 g

Hence, the answer is mass of solvent water required is 61.5 g.

(b) Given mass%=1.70

mass of solute MgBr2 = 29.3 g

The mass of solvent water required can be calculated as shown below:

[tex]mass percent=\frac{mass of solute}{mass of solution} x 100\\\\1.70=(29.3g/mass of solution )x100\\mass of solution =(29.3 g /1.70)x100\\=1720 g[/tex]

The mass of the solution is 1720 g.

Mass of solvent water = mass of solution - mass of solute

=1720 g - 29.3 g

=1690.7 g

Answer: The mass of water required is 1690.7 g.

Answer:

Ethyn gas ( acetylene gas )

Explanation:

All group II carbides react with water to form ethyn gas apart from beryllium which produces methane gas.

[tex]{ \sf{CaC_{2(s)} + 2H _{2} O_{(l)} → Ca(OH) _{2(s)} + C _{2} H _{2(g)} }}[/tex]

Answer:

See explanation

Explanation:

Mass of water lost = mass of hydrated salt - mass of anhydrous salt

Mass of water lost = 1.973 g - 1.196 g = 0.777g

Number of moles of water lost = 0.777g/18g/mol = 0.043 moles

Number of moles of anhydrous salt = 1.196 g /111g/mol = 0.011 moles

To obtain the number of moles of water of crystalization per hydrate molecule;

Number of moles of anhydrous salt = number of moles of hydrated salt

0.011 = 1.973 /111 + 18x

0.011(111 + 18x) = 1.973

1.221 + 0.198x = 1.973

0.198x = 1.973 - 1.221

x= 4

Hence, there are 4 moles of water per hydrate molecule. The formula of the hydrate is CaCl2.4H2O

Answer:

See explanation and image attached

Explanation:

The product of the Grignard reaction between methyl benzoate and excess phenylmagnesium bromide is triphenyl methanol.

The reaction proceeds by nucleophillic reaction as the carbonyl moiety is attacked. A tetrahedral intermediate is formed. Loss of the -OMe group is accompanied by the attack of the first molecule of PhMgBr.

Attack by a second PhMgBr molecule yields trimethyl phenoxide. Protonation of this specie yields the final product which is obtained by aqueous workup.

Answer:

Transition temperature is the temperature at which a substance changes from one state to another.

Allotropy is the existence of an element in many forms.

Solution :

We know that :

[tex]$\Delta T_f = k_f.m$[/tex]  and   [tex]$m=\frac{w_2}{m_2 \times w_1}$[/tex]

Then, [tex]$\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$[/tex]   ..................(1)

Where,

[tex]w_1[/tex] = amount of solvent (in kg)

[tex]w_2[/tex] = amount of solute (in kg)

[tex]m_2[/tex] = molar mass of solute (g/mole)

[tex]m[/tex] = molality of solution (mole/kg)

Given :

[tex]\Delta T_f[/tex] = [tex]3.14\ ^\circ C[/tex],   [tex]k_f= 5.12\ ^\circ C/m[/tex]

                              [tex]=5.12 \ ^\circ C/mole/kg[/tex]

                              [tex]=5.12 \ ^\circ C \ kg/mole[/tex]

[tex]w_1[/tex] = 0.250 kg,  [tex]w_2[/tex] = 24.3 g

Then putting this values in the equation is (1),

[tex]$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$[/tex]

[tex]$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$[/tex]

[tex]m_2= 158.49[/tex]  g/mole

So, the molar mass of the unknown compound is 158.49 g/mole.

Answer:

[tex]0.175\; \rm mol \cdot L^{-1}[/tex].

Explanation:

Magnesium chloride and silver nitrate reacts at a [tex]2:1[/tex] ratio:

[tex]\rm MgCl_2\, (aq) + 2\, AgNO_3\, (aq) \to Mg(NO_3)_2 \, (aq) + 2\, AgCl\, (s)[/tex].

In reality, the nitrate ion from silver nitrate did not take part in this reaction at all. Consider the ionic equation for this very reaction:

[tex]\begin{aligned}& \rm Mg^{2+} + 2\, Cl^{-} + 2\, Ag^{+} + 2\, {NO_3}^{-} \\&\to \rm Mg^{2+} + 2\, {NO_3}^{-} + 2\, AgCl\, (s)\end{aligned}[/tex].

The precipitate silver chloride [tex]\rm AgCl[/tex] is insoluble in water and barely ionizes. Hence, [tex]\rm AgCl\![/tex] isn't rewritten as ions.

Net ionic equation:

[tex]\begin{aligned}& \rm Ag^{+} + Cl^{-} \to AgCl\, (s)\end{aligned}[/tex].

Calculate the initial quantity of nitrate ions in the mixture.

[tex]\begin{aligned}n(\text{initial}) &= c(\text{initial}) \cdot V(\text{initial}) \\ &= 0.25\; \rm mol \cdot L^{-1} \times 0.150\; \rm L \\ &= 0.0375\; \rm mol \end{aligned}[/tex].

Since nitrate ions [tex]\rm {NO_3}^{-}[/tex] do not take part in any reaction in this mixture, the quantity of this ion would stay the same.

[tex]n(\text{final}) = n(\text{initial}) = 0.0375\; \rm mol[/tex].

However, the volume of the new solution is twice that of the original nitrate solution. Hence, the concentration of nitrate ions in the new solution would be [tex](1/2)[/tex] of the concentration in the original solution.

[tex]\begin{aligned} c(\text{final}) &= \frac{n(\text{final})}{V(\text{final})} \\ &= \frac{0.0375\; \rm mol}{0.300\; \rm L} = 0.175\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

Best practices for fume hoods: work 25 cm inside, organize items to one side, use raised ledge; avoid open sash and quick movements.

Laboratory fume hoods must be used safely. Workers should operate at least 25 cm within the hood to preserve ventilation and avoid dangerous chemicals. Place things on one side of the hood to preserve ventilation and prevent clogging.

A raised ledge on the rear of the hood prevents things from falling in and impeding airflow. Avoid fully opening the sash to maintain ventilation and containment. Fast, rapid motions can interrupt airflow, so prevent them. These practises guarantee the fume hood contains harmful compounds, making the lab safer. Therefore, option (B), (D) and (E) are correct.

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Answer: The mass of [tex]PH_3[/tex] produced is 45.22 g

Explanation:

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of [tex]P_4[/tex] = 62.0 g

Molar mass of [tex]P_4[/tex] = 124 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }P_4=\frac{62.0g}{124g/mol}=0.516mol[/tex]

Given mass of [tex]H_2[/tex] = 4.00 g

Molar mass of [tex]H_2[/tex] = 2 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2mol[/tex]

The chemical equation follows:

[tex]P_4(g)+6H_2(g)\rightarrow 4PH_3(g)[/tex]

By stoichiometry of the reaction:

If 6 moles of hydrogen gas reacts with 1 mole of [tex]P_4[/tex]

So, 2 moles of hydrogen gas will react with = [tex]\frac{1}{6}\times 2=0.333mol[/tex] of [tex]P_4[/tex]

As the given amount of [tex]P_4[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, hydrogen gas is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 6 moles of [tex]H_2[/tex] produces 4 mole of [tex]PH_3[/tex]

So, 2 moles of [tex]H_2[/tex] will produce = [tex]\frac{4}{6}\times 2=1.33mol[/tex] of [tex]PH_3[/tex]

We know, molar mass of [tex]PH_3[/tex] = 34 g/mol

Putting values in equation 1, we get:

[tex]\text{Mass of }PH_3=(1.33mol\times 34g/mol)=45.22g[/tex]

Hence, the mass of [tex]PH_3[/tex] produced is 45.22 g

Answer:

a. 181.5 kcal

Explanation:

Step 1: Calculate the enthalpy of the process (ΔH°).

Let's consider the following process.

CO₂(s) → CO₂(g)

We can calculate the enthalpy of the process using the following expression.

ΔH° = ∑ np × ΔH°f(p) - ∑ nr × ΔH°f(r)

ΔH° = 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CO₂(s))

ΔH° = 1 mol × (-393.5 kJ/mol) - 1 mol × (-427.4 kJ/mol) = 33.9 kJ

According to the balanced equation, 33.9 kJ are required to sublime 1 mole of CO₂.

Step 2: Convert 986 g of CO₂ to moles

The molar mass of CO₂ is 44.01 g/mol.

986 g × 1 mol/44.01 g = 22.4 mol

Step 3: Calculate the enthalpy needed to sublime 22.4 moles of CO₂

22.4 mol × 33.9 kJ/1 mol = 759 kJ

We can convert it to Kcal using the conversion factor 1 kcal = 4.184 kJ.

759 kJ × 1 kcal/4.184 kJ = 181.5 kcal

Explanation:

The given data is:

The mass of hydrogen is 6.54 kg.

The actual yield is 30.4 kg.

The balanced chemical equation of the reaction is:

[tex]N_2(g)+3H_2(g)<=>2NH_3(g)[/tex]

At first the theoretical yield should be calculated by using the balanced chemical equation:

3 mol. of hydrogen forms ---- 2 mol. of ammonia.

The molar mass of hydrogen is 2.0 g/mol.

The molar mass of ammonia is 17.0 g/mol.

Hence, the above statement can be rewritten as:

6g of hydrogen forms --- 34g of ammonia.

Then,

6.54g of hydrogen forms :

[tex]6.54 kg x 34 g / 6 g\\=37.1 kg[/tex]

% yield = (actual yield /theoretical yield )x 100

=(30.4 kg /37.1 kg )x100

=81.9

Hence, % yield is 81.9.

Answer:

A molecular formula consists of the chemical symbols for the constituent elements followed by numeric subscripts describing the number of atoms of each element present in the molecule.

Answer

Accordingly the order of electronegativity of the given elements would be: Fluorine > Chlorine > Bromine > Iodine. ( Fluorine has the highest electronegativity.)

Answer:

[tex]m_{isoborneol }=0.39g\\\\m_{Camphor}=0.38g\\[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to infer that the reaction whereby isoborneol goes to camphor occurs in a 1:1 mole ratio, that is why the theoretical yield of the latter is also 2.5 mmol (0.0025 mol) but the masses can be calculated as follows:

[tex]m_{isoborneol }=0.0025mol*\frac{154.25g}{1mol} =0.39g\\\\m_{Camphor}=0.0025mol*\frac{152.23 g}{1mol} =0.38g\\[/tex]

Because of the fact this is a rearrangement reaction whereas the number of atoms is not significantly modified.

Regards!

Answer: the correct option is that batteries (do NOT require a continuous source of fuel).

Explanation:

A battery can be classified as an electrochemical cell that has the ability to produce electric current. They do NOT require a continuous supply of fuel because it contains all the reactants needed to produce electricity. Below are some examples of batteries that are commonly used:

--> Primary battery: This is a single use battery because it can't be recharged. A typical example is the dry cell.

--> Secondary battery: This type of battery can be recharged. They are used as a power source for smartphones, electronic tablets, and automobiles.

A FUEL CELL is known as a device that converts chemical energy into electrical energy. Fuel cells are similar to batteries but require a continuous source of fuel, often hydrogen. They will continue to produce electricity as long as they are constantly resupplied with reactants. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines.

Answer:

accumulate reaction byproducts

Explanation:

Batteries accumulate reaction byproducts.  Fuel cells are similar to batteries but require a continuous source of fuel, often hydrogen. They will continue to produce electricity as long as fuel is available. Hydrogen fuel cells have been used to supply power for satellites, space capsules, automobiles, boats, and submarines.  A battery is an electrochemical cell or series of cells that produces an electric current.

Answer:

0

Explanation:

answer from gauth math

Answer:

2

Explanation:

i did this