Explain how you would determine the number of grams of Cu(NO3)2 that would be needed to make 1052 mL of a 2.50 M solution of Cu(NO3)2

Answer:

793 kJ

Explanation:

Given that :

[tex]T_1 = 1760^0C\\\\T_2 = 2060^0C\\\\ c_s = 44.8 \ J/g \ ^0C[/tex]

Then; the heat needed to convert solid from [tex]1760^0C \ to \ 1860^0C[/tex] is calculated as:

[tex]Q_1 = mc_s \delta T[/tex]

[tex]Q_1 = mc_s (T_f - T_i)[/tex]

[tex]Q_1 = 159.6*44.8(1860-1760)^0C \\ \\ Q_1 = 715008 \ J[/tex]

Also;

[tex]\delta H_f = 20.5 \ kJ/mol = 20500 \ J/mol \\ \\ Molar \ mass \ of \ Cr = 52.0 g/mol \\ \\ number \ of \ moles\ of \ Cr = \frac{mass \ of \ Cr}{molar \ mass} \\ \\ = \frac{159.6}{52.0}\\\\= 3.069 \ mol[/tex]

Now; the heat required to convert solid to liquid at [tex]1860^0C[/tex] is;

[tex]Q_2 = n* \delta H_f[/tex]

= 3.069 × 20500

= 62914.5 J

Also ; given that :

[tex]c__l}} = 0.94 \ J/g \ ^0C \\ \\[/tex]

Then the heat needed to convert liquid from [tex]1860^0C[/tex]to [tex]2060^0C[/tex] is;

[tex]Q_3 = m*c__l} (T_f-T_l)[/tex]

[tex]Q_3 = 159.6*0.94*(2060-1760)[/tex]

[tex]Q_3 = 15002.4 \ J[/tex]

∴ The total heat required = [tex]Q_1 + Q_2 + Q_3[/tex]

= (715008 + 62914.5 + 15002.4) J

= 792924.9 J

= 793 kJ

Teking into accoun the definition of calorimetry, sensible heat and latent heat, the amount of heat required is 808 kJ.

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

In firts place, you know that the melting point is 1860°C. So, first of all you must increase the temperature from 1760 ° C (in solid state) to 1860 ° C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

Replacing:

Q1= 44.8 [tex]\frac{J}{gC}[/tex]× 159.6 g× 100 °C

Solving:

Q1=715,008 J= 715.008 kJ

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

Replacing:

Q2= 3.069 moles×20.5 [tex]\frac{kJ}{mol}[/tex]

Solving:

Q2=62.9145 J

Similar to sensible heat previously calculated, you know:

Replacing:

Q3= 0.94 [tex]\frac{J}{gC}[/tex]× 159.6 g× 200 °C

Solving:

Q3=30,004.8 J= 30.0048 kJ

The total heat required is calculated as:  

Total heat required= 715.008 kJ + 62.9145 kJ + 30.0048 kJ

Total heat required= 807.9273 kJ ≅ 808 kJ

In summary, the amount of heat required is 808 kJ.

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Answer:

C. NaOH acts as a reactant in the reaction

Explanation:

Because during the saponification process, Na+ replaces the H+ in the fatty acid been used for the saponification process

Answer: Option:C

Explanation:

Answer:

The question is incomplete, no worries I got you.

Here is the complete question;

__________ reactions are used to detect antibodies for relatively large pathogens, such as bacteria. For these tests, the antigen is mixed with the test sample at various dilutions. Reaction mixes are then monitored for the formation of visible aggregates.

Explanation:

AGGLUTINATION is the reaction used.

Agglutination is the reaction in which there is the clumping of particles. A agglutination reaction is the visible clumping of the bacterial cells as an antigen reacts with its corresponding antibody. This type is often used as an initial confirmation of specific pathogens. Agglutination tests is is used to detect antibody or antigen and it involves the  agglutination of bacteria, red cells, or antigen- or antibody-coated latex particles. It is therefor used for large pathogens like bacteria. In this reaction, antigens are introduced into various dilutions of antibodies in test tubes or surfaces of glass slides, visible clumping is observed which depends on the size of the antibodies, amount and acidity of the antibody molecule, time of incubation and as well as the environment of the reaction which includes optimum pH, protein concentration among others.  

Answer: [tex]2CH_4+4H_2O\rightarrow 2CO_2+4H_2O[/tex] is not a combination reaction.

Explanation:

Synthesis reaction or combination reaction is defined as the reaction where two or more substances combine to form a single product.

[tex]2N_2+3H_2\rightarrow 2NH_3[/tex]

[tex]2Mg+O_2\rightarrow 2MgO[/tex]

[tex]CaO+H_2O\rightarrow Ca(OH)_2[/tex]

Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.

[tex]2CH_4+4H_2O\rightarrow 2CO_2+4H_2O[/tex]

Among the provided reactions, the reaction 2CH4 + 4O2 → 2CO2 + 4H2O is not a combination reaction, but rather a combustion reaction.

In the given reactions, the reaction 2CH4 + 4O2 → 2CO2 + 4H2O is not a combination reaction. In a combination reaction, two or more substances (either elements or compounds) combine to form a single compound. However, the mentioned reaction is actually an example of a combustion reaction where methane (CH4) undergoes combustion in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O).

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Final answer:

There are 4 dozens in 48 doughnuts, as a dozen is a group of 12 objects. This quantity would require four boxes with a volume of 324 cubic inches each to hold all the doughnuts.

Explanation:

To find out how many dozens of doughnuts are in 48 doughnuts, we can simply divide the total number of doughnuts by the number in a dozen. Since a dozen refers to 12 objects, we divide 48 by 12.

48 ÷ 12 = 4

Therefore, there are 4 dozens in 48 doughnuts. When buying items like doughnuts, it is common to acquire them in groups, like a dozen, because it is more convenient and efficient.

Let's consider the dimensional aspect for a moment. If you have a 9x9x4 inch box for a dozen doughnuts, this box would have a volume of 324 cubic inches. For 48 doughnuts, which we have established are 4 dozens, we would need four of these boxes, equating to 4 × 324 in3 or 1296 in3 in total to hold all of them.

0.091 moles are contained in 2.0 L of N2 at standard temperature and pressure.

Explanation:

Data given:

volume of the nitrogen gas = 2 litres

Standard temperature = 273 K

Standard pressure = 1 atm

number of moles =?

R (gas constant) = 0.08201 L atm/mole K

Assuming nitrogen to be an ideal gas at STP, we will use Ideal Gas law

PV = nRT

rearranging the  equation to calculate number of moles:

PV = nRT

n = [tex]\frac{PV}{RT}[/tex]

putting the values in the equation:

n = [tex]\frac{1X2}{0.08201 X 273}[/tex]

n = 0.091 moles

0.091 moles of nitrogen gas is contained in a container at STP.

Final answer:

To find how many moles are in 2.0 L of N2 at standard temperature and pressure, we use the fact that 1 mole of any gas occupies 22.4 L at STP. By dividing 2.0 L by 22.4 L, we find that there are approximately 0.089 moles of N2.

Explanation:

The question asks, How many moles are contained in 2.0 L of N2 at standard temperature and pressure? To answer this, it's essential to know that at standard temperature and pressure (STP), which is defined as 0°C (273K) and 1 atmosphere (atm) of pressure, 1 mole of any ideal gas occupies 22.4 liters of volume. This is a key concept in understanding gas laws and mol calculations in chemistry.

Given that we have 2.0 liters of N2 gas at STP, we can use the proportionality given by the molar volume at STP to find the number of moles. Since 22.4 liters is equivalent to 1 mole of a gas, we can set up a simple calculation to find the moles of N2 in 2.0 liters:

Moles of N2 = Volume of N2 (L) / Volume of 1 mole of gas at STP (L)

= 2.0 L / 22.4 L

= 0.089 moles of N2

This calculation demonstrates the direct application of molar volume at STP to determine the number of moles in a given volume of gas.

Answer:

it is called pollutant

Explanation:

this means a subtance that pollutes something especially the environment.

Answer:

the first and last

Explanation:

the enthalpy Change is positive.

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Answer:

One (2H2O → 2H2 + O2, ΔH = 484 kJ )

Four (2NH3 → N2 + 3H2, ΔH = 92 kJ)

Explanation:

An endothermic process is any process with an increase in the enthalpy H of the system. So if it has a positive result then the heat is going up which means it is endothermic .

Answer:

A.

Once it rains the pesticide goes down the runoff and pollute the water

Answer:

The percent yield of this reaction is 92.7 %

Explanation:

Step 1: Data given

Mass of nitrogen gas (N2) = 34.0 grams

Mass of ammonia (NH3 produced = 41.0 grams

Molar mass of N2 = 28.0 g/mol

Molar mass of NH3 = 17.02 g/mol

Actual yield of ammonia = 38 grams

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles N2 = 34.0 grams / 28.0 g/mol

Moles N2 = 1.214 moles

Step 4: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 1.214 moles N2 we'll have 2* 1.214 = 2.428 moles NH3

Step 5: Calculate mass NH3

Mass NH3 = moles * molar mass

Mass NH3 = 2.428 moles * 17.02 g/mol

Mass NH3 = 41 grams

Step 6: Calculate percent yield for the reaction

Percent yield = (actuald yield / theoretical yield) * 100 %

Percent yield = (38 grams / 41 grams ) * 100 %

Percent yield = 92.7 %

The percent yield of this reaction is 92.7 %

Answer:

[tex]Y=92\%[/tex]

Explanation:

Hello,

In this case, by considering the given chemical reaction, with given mass of nitrogen, one could compute the theoretical yield of ammonia as shown below and considering their 1 to 2 molar relationship in the chemical reaction:

[tex]m_{NH_3}^{theoretical}=34gN_2*\frac{1molN_2}{28gN_2}*\frac{2molNH_3}{1molN_2}*\frac{17gNH_3}{1molNH_3} \\m_{NH_3}^{theoretical}=41.3gNH_3[/tex]

In such a way, the percent yield is obtained as shown below:

[tex]Y=\frac{m_{NH_3}^{actual}}{m_{NH_3}^{theoretical}} *100\%=\frac{38g}{41.3g} *100\%\\\\Y=92.0\%[/tex]

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Answer:

arterial hypertension

Explanation:

When increasing the arterial pressure, in the renal glomerulus the filtration of liquids increases since the liquids diffuse from zones of greater pressure to less pressure.

In this way, the liquid content of the urine increases and it is more diuluid because this does not apply to proteins or solutes, since these are affected by discrepancies in concentrations and not in pressures.

Excessive excretion of dilute urine, or polyuria, can be caused by conditions such as diabetes insipidus or diabetes mellitus, as well as the excessive use of diuretics, kidney disease, and excessive water intake. Diabetes insipidus reduces the number of water channels leading to the loss of water, while diabetes mellitus results in glucose in the urine which also causes the loss of water.

If a patient is excreting a large volume of very dilute urine on a continuing basis, this can be due to a condition known as polyuria. Polyuria is characterized by urine production in excess of 2.5 L per day, and it can be caused by a number of potential factors.

One such factor could be diabetes insipidus, a condition caused by an insufficient release of the pituitary hormone known as antidiuretic hormone (ADH), or insufficient numbers of ADH receptors in the kidneys. This results in an insufficient number of water channels, known as aquaporins, which reduces water absorption and leads to high volumes of very dilute urine.

Another potential cause is diabetes mellitus, a condition where blood glucose levels exceed the number of available sodium-glucose transporters in the kidneys. This causes glucose to appear in the urine, and its osmotic properties attract water, leading to its loss in urine. Other potential causes of polyuria can also include the excessive use of diuretics, kidney disease, and excessive water intake.

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Explanation:

Given

mass (m) = 44 kg

velocity (v) = 31 m/s

Now

Kinetic energy

= 1/2 mv²

= 1/2 * 44 * 31²

= 0.5 * 44 * 961

= 21142 joule

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Question:

The question is incomplete. What is required to calculate was not added.The equilibrium data was not also added. Below is the additional questions and the answers.

1. Calculate the minimum solvent that can be used.

2.Using a solvent rate of 1.5 times the minimum, calculate the number of

theoretical stages.

Answer:

1. Minimum solvent = 411.047

2. N = 5

Explanation:

See the attached files for explanations.

An aqueous feed solution.

The aqua solution is a solvent on water that has a chemical equation of aq. The example includes the solution of the salt table and NaCl. Having positive and negative ions. The solution feed id of 1000 KG/hand consists of 23.5 wt% acetone and has a 76.5 wt% of water by volume which is being used.

Thus the answer is minimum solvent = 411.047 and the N equals 5.

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Answer: [tex]481^0C[/tex]

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]

where,

[tex]P_1[/tex] = initial pressure of gas = 64.7 kPa

[tex]P_2[/tex] = final pressure of gas = 50.8 kPa

[tex]V_1[/tex] = initial volume of gas = 5.45 L

[tex]V_2[/tex] = final volume of gas = 18.50 L

[tex]T_1[/tex] = initial temperature of gas = [tex]10^oC=273+10=283K[/tex]

[tex]T_2[/tex] = final temperature of gas = ?

Now put all the given values in the above equation, we get:

[tex]\frac{64.7kPa\times 5.45}{283K}=\frac{50.8kPa\times 18.50}{T_2}[/tex]

[tex]T_2=754K=(754-273)^0C=481^0C[/tex]

Thus final temperature of the gas in degrees Celsius is 481

Answer:

8.77g

Explanation:

Step 1:

Data obtained from the question.

This includes the following:

Concentration of A (C1) =.?

Volume of A (V1) = 16 mL

Volume of B (V2) = 300 mL

Concentration of B (C2) = 0.50 M

Molar Mass of NaCl = 58.443 g/mol

Mass of NaCl =.?

Step 2:

Determination of the concentration of A.

Applying the dilution formula:

C1V1 = C2V2

The concentration of A i.e C1 can be obtained as follow:

C1V1 = C2V2

C1 x 16 = 0.5 x 300

Divide both side by 16

C1 = (0.5 x 300) / 16

C1 = 9.375 M

Therefore, the concentration of A is 9.375 M

Step 3:

Determination of the number of mole of NaCl in 9.375 M NaCl solution. This is illustrated below:

Molarity = 9.375 M

Volume = 16 mL = 16/1000 = 0.016 L

Mole of NaCl =?

Molarity = mole /Volume

Mole =Molarity x Volume

Mole of NaCl = 9.375 x 0.016

Mole of NaCl = 0.15 mole

Step 4:

Determination of the mass of NaCl. This is illustrated below:

Mole of NaCl = 0.15 mole

Molar Mass of NaCl = 58.443 g/mol

Mass of NaCl =?

Mass = number of mole x molar Mass

Mass of NaCl = 0.15 x 58.443

Mass of NaCl = 8.77g

Therefore, 8.77g of NaCl is needed to make 1 L of the original solution A.

Answer:

We need 8.77 grams of NaCl

Explanation:

Step 1: Data given

Volume of solution A = 16 mL = 0.016 L

Volume of solution B = 300 mL = 0.300 L

Concentration of solution B = 0.50 M

Step 2: Calculate molarity of solution A

C1*V1 = C2*V2

⇒with C1 = the concentration of solution A = TO BE DETERMINED

⇒with V1 = the volume of solution A = 0.016 L

⇒with C2 = the concentration of solution B = 0.50 M

⇒with V2 = the volume of solution B = 0.300 L

C1 * 0.016 L = 0.50 M * 0.300 L

C1 = 9.375 M

Step 3: Calculate moles of solution A

Moles = molarity * volume

Moles = 9.375 M * 0.016 L

Moles = 0.15 moles

Step 4: Calculate mass of NaCl needed

Mass NaCl = moles / molar mass

Mass NaCl = 0.15 moles * 58.443 g/mol

Mass NaCl = 8.77 grams

We need 8.77 grams of NaCl

Answer:

P1V1 = P2V2

(1.5 atm)(5.6 L) = (x)(4.8 L) x = 1.8 atm

If we have 5.6 liters of gas in a piston at a pressure of 150 kPa and compress the gas until its volume is 4.8 L, the new pressure inside the piston will be 175 kPa.

The combined gas law is the law of of gaseous state which is made by combination of Boyle's law, Charle's law, Avogadro's law and Gay Lussac's law.

It is a mathematical expression that relates Pressure, Volume and Temperature.

(P1 × V1)÷T1 = (P2 × V2)÷T2

Here temperature doesn't change, hence-

(P1 × V1) = (P2 × V2)

P1 = 150 kPa

V1 = 5.6 litres

P2 = ?

V2 = 4.8 litres

P2 = 175 kPa

Therefore, If we have 5.6 liters of gas in a piston at a pressure of 150 kPa and compress the gas until its volume is 4.8 L, the new pressure inside the piston will be 175 kPa.

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Answer: 1 dissociation. 2. Hydration. 3. Dissolving.

Explanation:

Answer:

44 mL of Na2SO4

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

Step 2:

Determination of the number of mole of Ba(NO3)2 in 20.00 mL of 0.500 M barium nitrate (Ba(NO3)2). This is illustrated below:

Molarity of Ba(NO3)2 = 0.5 M

Volume of solution = 20 mL = 20/1000 = 0.02 L

Mole of solute (Ba(NO3)2) =?

Molarity = mole /Volume

0.5 = Mole of Ba(NO3)2 / 0.02

Cross multiply to express in linear form

Mole of Ba(NO3)2 = 0.5 x 0.02

Mole of Ba(NO3)2 = 0.01 mole

Step 3:

Determination of the number of mole of Na2SO4 that reacted.

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

From the balanced equation above,

1 mole of Ba(NO3)2 reacted with 1 mole of Na2SO4.

Therefore, 0.01 mole of Ba(NO3)2 will also react with 0.01 mole of Na2SO4.

Step 4:

Determination of the volume of Na2SO4 needed for the reaction. This is illustrated below:

Mole of Na2SO4 = 0.01 mole

Molarity of Na2SO4 = 0.225M

Volume =?

Molarity = mole /Volume

0.225 = 0.01 / volume

Cross multiply to express in linear form

0.225 x Volume = 0.01

Divide both side by 0.225

Volume = 0.01/0.225

Volume of Na2SO4 = 0.044 L

Converting 0.044 L to mL, we have

Volume of Na2SO4 = 0.044 x 1000

Volume of Na2SO4 = 44 mL

Therefore, 44 mL of Na2SO4 is needed for the reaction

Answer:

She has to add 44.44 mL of Na2SO4

Explanation:

Step 1: Data given

Volume of barium nitrate = 20.00 mL = 0.020 L

Molarity of barium nitrate = 0.500 M

Molarity of sodium sulfate = 0.225 M

Step 2: The balanced equation

Ba(NO3)2 (aq) + Na2SO4 (aq) → BaSO4 (s) + 2 NaNO3 (aq)

Step 3: Calculate volume of the sodium sulfate solution needed

For 1 mol Ba(NO3)2 we need 1 mol Na2SO4 to produce 1 mol basO4 and 2 moles NaNO3

C1*V1 = C2*V2

⇒with C1 = the molarity of Ba(NO3)2 = 0.500 M

⇒with V1 = the volume of Ba(NO3)2 = 0.020 L

⇒with C2 = the molarity of Na2SO4 =0.225 M

⇒with V2 = the volume of Na2SO4 = TO BE DETERMINED

0.500 M * 0.020 L = 0.225 M * V2

V2 = (0.500 M * 0.020 L) / 0.225 M

V2 = 0.04444 L = 44.44 mL

She has to add 44.44 mL of Na2SO4

The letter that represents the activation energy from the diagram is letter A.

A chemical reaction occurs only when there is collision between the particles of reactants.

These colliding particles become activated with increased kinetic energy.

Activation energy is the energy barrier that must be overcome before a reaction takes place.

From the diagram,

C is the reactants

D is the products

B is the heat of reaction

While A is the activation energy.

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Answer:

The answer to your question is butanal

Explanation:

To name this compound we must consider:

1.- Identify the functional group. The functional group of this molecule is the first carbon to the right and its oxygen.

When carbon is attached to oxygen is a border, this functional group is called Aldehyde.

2.- Count the total number of carbons starting from the right. This molecule has 4 carbons.

3.- Name the compound

    An organic molecule with 4 carbons is called butane but change the ending for al, then the name will be butanal

Answer:

A noncompetitive inhibitor can only bind to an enzyme with or without a substrate at several places at a particular point in time

Explanation:

this is because It changes the conformation of an enzyme as well as its active site, which makes the substrate unable to bind to the enzyme effectively so that the efficiency of the enzyme decreases. A noncompetitive inhibitor binds to the enzyme away from the active site, altering/distorting the shape of the enzyme so that even if the substrate can bind, the active site functions less effectively and most of the time also the inhibitor is reversible

Answer:

Theoretical yield = 4.09 g water.

75% yield = 3.07g water.

Explanation:

C3H8 + 5O2 --->  3CO2 + 4H2O

Using atomic masses:

(12 *3 + 8 * 1.008) g C3H8  yield (theoretically) (4* (2*1.008 + 16) g water

44.06g C3H8 yields 72.06g water

So 2.5 g yields  (72.06/44.06) * 2.5

= 4.09 g  water.

If the yield is 75% then it is 4.09 * 0.75

= 3.07 g water.

If I correct, answer is C, 75%.

Answer:

0.0812 grams of nitrate ions are there in the final solution.

Explanation:

Mass of cobalt (II) nitrate = 3.00 g

Moles of cobalt(II) nitrate = [tex]\frac{3.00 g}{183 g/mol}=0.0164 mol[/tex]

Volume of the solution = 100 mL = 0.100 L

1 mL = 0.001 L

[tex]Molarity=\frac{Moles}{Volume(L)}[/tex]

Molarity of the solution = [tex]\frac{0.0164 mol}{0.100 L}=0.164 M[/tex]

Cobalt (II) nitrate in its aqueous solution gives 1 mole of cobalt(II) ion and 2 moles of nitrate ions.

[tex][NO_3^{-}]=2\times [Co(NO_3)_2]=2\times 0.164 M=0.328 M[/tex]

Molarity of the nitrate ion before solution = [tex]M_1=0.328 M[/tex]

Volume of the  nitrate ion  before solution = [tex]V_1=4.00 mL[/tex]

Molarity of the  nitrate ion after solution = [tex]M_2=?[/tex]

Volume of the  nitrate ion after solution = [tex]V_2=275 mL[/tex]

[tex]M_1V_1=M_2V_2[/tex] ( Dilution)

[tex]M_2=\frac{0.328 M\times 4.00 mL}{275 mL}=0.00477 M[/tex]

Moles of nitrate ions in 275 ml = n

Molarity of the  nitrate ion after solution =0.00477 M

volume of the final solution = 275 mL = 0.275 L

[tex]n=0.00477 M\times 0.275 L=0.00131 mol[/tex]

Mass of 0.00131 moles of nitrate ions:

0.00131 mol × 62 g/mol = 0.0812 g

0.0812 grams of nitrate ions are there in the final solution.

Answer:

entropy increases when a glass of water is set in the sun until the water evaporates.

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Explanation:

In the water cycle, evaporation occurs when sunlight warms the surface of the water. The heat from the sun makes the water molecules move faster and faster, until they move so fast they escape as a gas. ... When it is cool enough, the water vapor condenses and returns to liquid water.

Answer:

(B.) What is the molar solubility of barium chloride, BaCl2 in water?

Explanation:

Molar solubility is the number of moles of a substance that can dissolve in a liter of solution to the point of the solution's saturation. It can be calculated stoichiometrically from a substance's solubility product constant in mol/L.

Since all the [tex]BaCl_{2}[/tex] reacted all the [tex]Na_{2} SO_{4}[/tex] from the information, we can easily assume all the substances were consumed in the reaction, and hence account for their purity. Furthermore, [tex]BaSO_{4}[/tex] is insoluble in water, the most probable scientific query would be the molar solubility of the [tex]BaCl_{2}[/tex] used in the experiment.

Answer:

The best question is A Is the BaCl2(s) used in the experiment pure?Explanation:

Step 1: Data given

Mass of the BaCl2 sample = 10.0 grams

Volume of water = 50.0 mL

We add excess Na2SO4

A precipitate BaSO4 will be formed

Step 2: The balanced equation

BaCl2(aq) → Ba^2+(aq) + 2Cl-(aq)

Ba^2+(aq) + SO4^2-(aq) →BaSO4(s)

Step 3: Calculate moles BaCl2

Moles BaCl2 = 10.0 grams / 208.23 g/mol

Moles BaCl2 = 0.048 moles

Step 4: Calculate moles Ba^2+

For 1 mol BaCl2 we have 1 mol Ba^2+

For 0.048 moles BaCl2 we have 0.048 moles Ba^2+

Step 5: Calculate mass Ba^2+

Mass Ba^2+ = moles Ba^2+ * molar mass Ba^2+

Mass Ba^2+ = 0.048 moles * 137.33 g/mol

Mass Ba^2+ = 6.59 grams

After measuring the mass of barium in BaSO4 we can determine if the BaCl2 was pure or not.

If the mass = 6.59 grams the BaCl2 was pure

If the mass <6.59 grams the BaCl2 wasn't pure

The best question is A Is the BaCl2(s) used in the experiment pure?

Answer:

A)  acetic acid ten times greater than acetate pH = 3.74

B)  acetate ten times greater than acetic acid pH = 5.74

C) For solution 3: acetate=acetic acid pH = 4.74

Mass NH4Cl = 40.30 grams

Explanation:

Step 1: Data given

Ka is: 1.8 * 10^-5

Solution 1: acetic acid ten times greater than acetate

Solution 2: acetate ten times greater than acetic acid

Solution 3: acetate=acetic acid

Step 2: The pH formula

pH = pKa + log[CH3COO-]/[CH3COOH]

For solution 1:  acetic acid ten times greater than acetate this means ) [[CH3COO-]/[CH3COOH]) has a value of 1/10

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(1/10)

pH = 4.74 -1

pH = 3.74

For solution 2: acetate ten times greater than acetic acid

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(10)

pH = 4.74 + 1

pH = 5.74

For solution 3: acetate=acetic acid

pH = pKa + log[CH3COO-]/[CH3COOH]

pH = -log(1.8*10^-5) + log(1)

pH = 4.74 + 0

pH = 4.74

Part B: Calculate molarity

pOH = pKb + log [H

5.29= -log (1.8*10^-5)  + log [BH+]/[0.600 M]

5.29 = 4.744 + log [BH+]/[0.600 M]

0.546 = [BH+]/0.600

[BH+] = 0.3276 M

Moles NH4+ = 0.3276  M * 2.30 L

Moles NH4+ = 0.75348 moles

Moles NH4Cl = 0.75348 moles

Mass NH4Cl = 0.75348 moles * 53.49 g/mol

Mass NH4Cl = 40.30 grams

Final answer:

The group that contains elements that are gaseous at room temperature are noble gases and halogens. Noble gases include helium, neon, argon, krypton, xenon, and radon, and halogens like fluorine and chlorine are gases under standard conditions. So the correct option is D.

Explanation:

The question at hand involves identifying which groups of elements are gaseous at room temperature. Of the options provided, d) noble gases and halogens contain elements that are gaseous at room temperature. Noble gases, which include elements like helium, neon, argon, krypton, xenon, and radon, are all gases under standard conditions. Halogens such as fluorine and chlorine are also gaseous at room temperature. Alkali metals and alkaline earth metals are solid at room temperature. Furthermore, most transition metals are also solid at room temperature, with mercury being an exception as it is liquid.

To define the subject for the questions based on the examples given, we can use the keywords: alkali metal, halogen, noble gas, and alkaline earth metal as they relate to the classifications on the periodic table. The given examples classify elements like lithium as an alkali metal (Group 1), argon as a noble gas (Group 18), and chlorine as a halogen (Group 17), which are key aspects of elementary Chemistry covered in high school.

Answer: C. Acts as energy reserve in muscle tissue.

Explanation:

Creatine Phosphate is a high energy compound that can be found in muscle tissue.

It serves as an extra energy storage, since the normal metabolism cannot supply the high energy requirement of a muscle at work.

Creatine Phosphate is quickly transfered to ADP, for the production of ATP needed for muscle contraction.

Answer: 2.07 x 1020 tons

Explanation:

Answer:

6.022X10^23

Explanation:

ehh i had this and this is what i answered your welcome?