Consider the following precipitation reaction: 2Na3PO4(aq)+3CuCl2(aq)→Cu3(PO4)2(s)+6NaCl(aq)2Na3PO4(aq)+3CuCl2(aq)→Cu3(PO4)2(s)+6NaCl(aq) Part A What volume of 0.174 MNa3PO4MNa3PO4 solution is necessary to completely react with 90.8 mLmL of 0.102 MCuCl2MCuCl2? Express your answer in milliliters.

Answer:

Dear user,

Answer to your query is provided below

When small amount of acid was added to buffered solution, pH will change very less.

Explanation:

Buffer solution resists change in ph on adding small amount of acid or base but when we calculate the value of buffer capacity we take the change in ph when we add acid or base to 1 lit solution of buffer.This contradicts the definition of buffer solution.

The strongest acid is b, the acid of intermediate strength is c and the weakest acid is a Then this leads to the H2S (Hydrogen sulfide). Therefore, option C is correct.

A chemical that, when combined with certain metals, emits hydrogen ions in water and forms salts. Acids have a sour taste and cause certain dyes to turn red.

An acid is a substance that releases hydrogen ions into solution, whereas a base or alkali absorbs hydrogen ions.

In an aqueous solution or water, a weak acid is an acid that has been partially dissociated into its ions. A strong acid, on the other hand, completely dissociates into its ions in water. The pH of weak acids is higher than that of strong acids.

Thus, The strongest acid is b, the acid of intermediate strength is c and the weakest acid is a Then this leads to the H2S, option C is correct.

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Answer:

{ Carbon, Oxygen, Nitrogen }

Explanation:

Elements can only have more than an octet of electrons if they demonstrate an expanded octet. This is if they belong to groups in or beyond the third group. Why? Well these elements have d - orbitals that they can rely on to expand the number of electrons that could otherwise be limited. * Here we are focusing on main group elements, P - block elements more specifically. *

Carbon belongs to the 2 group, and thus doesn't have an empty d - orbital. Thus, it can't have more than an octet of electrons. Sulfur belongs to group 3, hence has an empty d - orbital, and can have more than an octet of electrons. Oxygen belongs to the 2 group, and thus doesn't have an empty d - orbital, so it can't have more than an octet of electrons. Same goes for Nitrogen. Bromine belongs to group 4, thus has empty d - orbitals, and can expand further than Sulfur can - it can have more than an octet of electrons.

Solution = { Carbon, Oxygen, Nitrogen }

Answer:

[tex]\% O=27.6\%[/tex]

Explanation:

Hello,

In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:

- Moles of carbon are contained in the 9.582 grams of carbon dioxide:

[tex]n_C=9.582gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2} =0.218molC[/tex]

- Moles of hydrogen are contained in the 3.922 grams of water:

[tex]n_H=3.922gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =0.436molH[/tex]

- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:

[tex]m_O=4.215g-0.218molC*\frac{12gC}{1molC} -0.436molH*\frac{1gH}{1molH} =1.163gO[/tex]

Finally, we compute the percent by mass of oxygen:

[tex]\% O=\frac{1.163g}{4.215g}*100\% \\\\\% O=27.6\%[/tex]

Regards.

Explanation:

1g = 1000mg

154000mg = 154g

No of days she can drink = 154÷ 11 = 14days

Answer:

Alpha- ketoglutarate --> NADP+ --> Oxaloacetate --> O2.

(O2 has the HIGHEST tendency to donate electrons).

Explanation:

Okay, we are to Arrange Alpha- ketoglutarate, Oxaloacetate ,O2, NADP+ and Malate in order of increasing tendency to donate electrons, therefore, the correct order is;

Alpha- ketoglutarate --> NADP+ --> Oxaloacetate --> O2.

And this result can be gotten if we consider these compounds in their oxidative mechanism and check their potential values.

(1). When Alpha- ketoglutarate react with CO2, it gives isocitrate and standard redox potential value of -0.38.

(2). NADP+ with react with two moles of electrons and one mole of hydrogen ion to produce NADPH and standard redox potential value of -0.32.

(3). Oxaloacetate will react with two moles of hydrogen ion and electrons respectively to produce standard redox potential value of -0.18 and Malate.

(4). O2 will react with 4 moles of hydrogen ion and electrons respectively to produce 2 moles of water and standard redox potential value of +0.82.

CONCLUSION: the higher the negative values of the potential, the lesser the electron affinity and vice versa.

Answer:

c

Explanation:

when the weight of the object is greater than the buoyant force sinking is occurred.

Answer:Crushed, decreased

Explanation:

Just got it right

Answer:

The 2nd Picture represents the particles in liquids.

Explanation:

Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

Explanation:

mass of nitrogen = 37.8 g

mass of oxygen = (100-37.8) g = 62.2 g

Using the equation given by Raoult's law, we get:

[tex]p_A=\chi_A\times P_T[/tex]

[tex]p_{O_2}[/tex] = partial pressure of [tex]O_2[/tex] = ?

[tex]\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}[/tex]

[tex]P_{T}[/tex] = total pressure of mixture  = 525 mmHg

[tex]{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles[/tex]

[tex]{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles[/tex]

Total moles = 1.94 + 1.35 = 3.29 moles

[tex]\chi_{O_2}=\frac{1.94}{3.29}=0.59[/tex]

[tex]p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg[/tex]

Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg

A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.

A gaseous mixture of O₂ and N₂ contains 37.8% nitrogen by mass, that is, in 100 g of the mixture, there are 37.8 g of N₂. The mass of O₂ in 100 g of the mixture is:

[tex]mO_2 = 100 g - 37.8 g = 62.2 g[/tex]

We will convert both masses to moles using their molar masses.

[tex]N_2: 37.8 g \times 1 mol/28.00 g = 1.35 mol\\\\O_2: 62.2 g \times 1 mol/32.00 g = 1.94 mol[/tex]

The mole fraction of O₂ is:

[tex]\chi(O_2) = \frac{nO_2}{nN_2+nO_2} = \frac{1.94mol}{1.35mol+1.94mol} = 0.590[/tex]

Given the total pressure (P) is 525 mmHg, we can calculate the partial pressure of oxygen using the following expression.

[tex]pO_2 = P \times \chi(O_2) = 525 mmHg \times 0.590 = 310 mmHg[/tex]

A gaseous mixture of O₂ and N₂ that contains 37.8% nitrogen by mass, and whose total pressure is 525 mmHg, has a partial pressure of oxygen of 310 mmHg.

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Answer:

[tex]\large \boxed{2 \, \%}[/tex]

Explanation:

1. Data

Mass of graduated cylinder              =  47.229 g

Mass of graduated cylinder + water =  71.821  g

Actual volume of water                     = 25.0     mL

2. Calculations

(a) Mass of water

Mass = 71.821 g -47.229 g  = 24.592 g

(b) Volume of water

[tex]\text{Volume} = \dfrac{\text{mass}}{\text{volume }} = \dfrac{\text{24.592 g}}{\text{ 1 g/mL}} = \text{24.592 mL}[/tex]

(c) Percent Difference

[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{Measured - Calculated}\lvert}{ \text{Calculated}} \times 100 \,\%\\\\& = & \dfrac{\lvert 25.0 - 24.492\lvert}{24.492} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.5\lvert}{24.492} \times 100 \, \%\\ \\& = & 0.02 \times 100 \, \%\\& = & \mathbf{2 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{2 \, \%} }$}[/tex]

Answer:

- 61.2°C = Initial T°

Explanation:

This is a calorimetry problem, where the heat from the water was gained by the marble.

Q = m . C . ΔT

where ΔT is final T° - initial T°, C is the specific heat and m, mass.

By the volume of water, we realize the mass (we apply density):

1 g/mL = mass / 38.7 mL

38.7 g = mass of water

Now, we need to find out the specific heat for the marble and we have Heat Capacity data

Heat Capacity = C . m

3.52 J/°C / 4 g = C → specific heat → 0.88 J/g °C

We make the equations for both heats:

m . C . ΔT from water = m . C . ΔT from the marble

38.7 g . 4.18 J/g°C ( 26.1°C - 28°C) = 4 g .  0.88 J/g °C . (26.1°C - Initial T°)

307.35 J = 4 g .  0.88 J/g °C . (26.1°C - Initial T°)

- 307.35 is a negative value, because the was has decreased the temperature, is a loss of heat, but we have to work with the positive number.

307.35 J / (4 . 0.88 °C/J) = 26.1°C - Initial T°

87.32°C = 26.1°C - Initial T°

- 61.2°C = Initial T°

Answer:

PF3 : trigonal bipyramidal

Explanation:

PF3 has 4 domains around the central phosphorus (3 shared pairs and one lone pair of electrons), thus the electron geometry that has 4 domains is tetrahedral not  trigonal bipyramidal

From the options the specie that is incorrectly matched is ( 5 ) ; PF₃ : trigonal bipyramidal

The specie PF₃ is composed of 3 shared pairs and one unshared pair of electrons ( i.e. It has 4 domains ) as seen in the  Lewis structure of  PF₃. therefore  when writing its electronic geometry, it should expressed/written as tetrahedral and not trigonal bipyramidal.

Hence we can conclude that The specie that is incorrectly matched is PF3 : trigonal bipyramidal

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Answer:

[tex]m=107.8g[/tex]

Explanation:

Hello,

In this case, for the given information, we can compute the gained grams by firstly compute the energy consumed by 14.3 h of being sit relaxed and 9.70 h  of sleeping:

[tex]E_{sit}=120\frac{J}{s}*\frac{3600s}{1h} *14.3h*\frac{1kJ}{1000J} =6177.6kJ\\\\E_{sleep}=83\frac{J}{s}*\frac{3600s}{1h} *9.70h*\frac{1kJ}{1000J} =2898.36kJ[/tex]

Then, we compute the energy used that day:

[tex]E_T=10000kJ-2898.36kJ-6177.6kJ=4203.28kJ[/tex]

Finally, the mass by considering the consumed fat:

[tex]m=\frac{4203.28kJ}{39kJ/g} \\\\m=107.8g[/tex]

Regards.

Answer:

The new volume is less than the initial volume.

The new volume is 322mL

Explanation:

Based on Boyle's law, the pressure of a gas is inversely proportional to its volume under constant temperature. That means if the pressure of a gas is increased, the volume decrease and vice versa. The formula is:

P₁V₁ = P₂V₂

Where P is pressure and V is volume of 1, initial state and 2, final states.

In the problem, the pressure of the gas increased from 4.04atm to 7.17atm, That means the new volume is less  than initial volume because the gas is compressed occupying less volume.

Replacing in Boyle's equation:

4.04atm*571mL = 7.17atmV₂

Beeing the new volume of the compressed gas 322mL

Answer:

C

Explanation:

A- incorrect, not in alphabetical order

B- incorrect, symbol for salt in NA not SA

C- correct

D- incorrect, not based on discovery

Each element represented in the periodic table is represented by a one or two-letter symbol.Hence , Option (C) is Correct.

The periodic table, also known as the periodic table of the elements, is a tabular display of the chemical elements.

It is widely used in chemistry, physics, and other sciences, and is generally seen as an icon of chemistry.

Each element represented in the periodic table is represented by a one or two-letter symbol.Hence , Option (C) is Correct.

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Answer:

5H3AsO3(aq) + 2MnO4-(aq) + 6H+(aq) → 5H3AsO4(aq) + 2Mn2+(aq) + 3H2O(aq)

Explanation:

Every net balanced ionic equation is composed of a union of two half equations;

The oxidation half equation (indicating electron loss) and the reduction half equation (indicating electron gain). Remember that redox reactions is a process in which electrons are lost and gained by chemical species simultaneously. One specie looses electrons in the oxidation half equation while the other specie gains electrons in the reduction half equation.

The balanced redox reaction equation shows the overall redox process and shows at a glance the total number of elect tribe lost or gained in the redox process. The overall redox reaction equation for the titration described in the question is;

5H3AsO3(aq) + 2MnO4-(aq) + 6H+(aq) → 5H3AsO4(aq) + 2Mn2+(aq) + 3H2O(aq)

Answer:

1. [tex]10.6\; \rm m[/tex] (one decimal place.)

2.[tex]0.79\; \rm g[/tex] (two decimal places.)

3. [tex]16.5\;\rm cm^2[/tex] (three significant figures.)

Explanation:

The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.

For example, in the first expression:

Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: [tex]\rm m[/tex].)

Therefore:

[tex]\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m[/tex]. (Rounded to one decimal place.)

Similarly:

Therefore, the result should be rounded to two decimal places. Its unit should be [tex]\rm g[/tex] (same as the unit of the two inputs.)

[tex]\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g[/tex]. (Rounded to two decimal places.)

When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:

Therefore, the result should have only three significant figures.

The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be [tex]\rm cm \cdot cm[/tex], which is occasionally written as [tex]\rm cm^2[/tex].

[tex]\rm 4.51 \; cm \times 3.6666 \; cm \approx 16.5\; \rm cm^2[/tex]. (Rounded to three significant figures.)

Answer:

[tex]M_{acid}=0.0444M[/tex]

Explanation:

Hello,

In this case, for the given reaction:

[tex]H _2 S O _4 + 2 N a O H \rightarrow 2 _H 2 O + N a _2 S O_ 4[/tex]

We find a 1:2 molar ratio between the acid and the base respectively, for that reason, at the equivalence point we find:

[tex]2*n_{acid}=n_{base}[/tex]

That in terms of concentrations and volumes we can compute the concentration of the acid solution:

[tex]2*M_{acid}V_{acid}=M_{base}V_{base}\\\\M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}}=\frac{0.697M*28.07mL}{2*220.1mL}\\ \\M_{acid}=0.0444M[/tex]

Best regards.

The concentration of the sulfuric acid solution (H₂SO₄) is 0.0444 M

From the question,

We are to determine the concentration of the H₂SO₄ solution

Using the titration formula

[tex]\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}[/tex]

Where

[tex]C_{A}[/tex] is the concentration of acid

[tex]C_{B}[/tex] is the concentration of base

[tex]V_{A}[/tex] is the volume of acid

[tex]V_{B}[/tex] is the volume of base

[tex]n_{A}[/tex] is the mole ratio of acid

and [tex]n_{B}[/tex] is the mole ratio of base

The given balanced chemical equation for the reaction is

H₂SO₄ + 2NaOH ⟶ 2H₂O + Na₂SO₄

∴ [tex]n_{A} = 1[/tex]

and [tex]n_{B} = 2[/tex]

From the given information

[tex]V_{A} = 220.1 \ mL[/tex]

[tex]C_{B} = 0.697 \ M[/tex]

[tex]V_{B} = 28.07 \ mL[/tex]

Putting the parameters into the equation, we get

[tex]\frac{C_{A} \times 220.1}{0.697 \times 28.07}= \frac{1}{2}[/tex]

∴ [tex]C_{A} = \frac{1 \times 0.697 \times28.07}{2\times 220.1}[/tex]

[tex]C_{A} = \frac{19.56479}{440.2}[/tex]

[tex]C_{A} = 0.0444 \ M[/tex]

Hence, the concentration of the sulfuric acid solution (H₂SO₄) is 0.0444 M

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The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.  

The no. of moles can be determined by using the formula,  

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

= 0.0144 moles

The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.  

moles = 0.461 g /16 g/mol

= 0.0288 moles

The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,  

moles = 9.89 g / 207.2 g/mol

= 0.0477 moles

The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,  

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.  

Answer:

sulfur-35

Explanation:

Sulfur-35 is a radioactive isotope that contains 19 neutrons.

Isotopes are represented with mass numbers. Mass number is the addition of number of proton and number of neutrons.

The number of proton in sulfur = 16

Number of neutron = 19

So, mass number = no. of protons + no. of neutrons

                              = 16 + 19

                              = 35

Hence, the correct answer is sulfur-35.

Answer:

sulfur -35

Answer:

The amount of solute that will dissolve

Explanation:

A P E X

Covalent compounds are held together with an intra molecular attraction which is weaker than metallic bond

hence covalent compounds exist as liquids, gases and soft solids