Calculate the pH of a buffer solution prepared by mixing 200 mL of 0.10 M NaF and 100 mL of 0.050 M HF.

Answer:

Explanation:

As the 50mL of water are in a big temperature will give heat to the other sample of water.

With the equation:

Q = m*C*T

Where Q is heat, m  is mass of the sample, C is specific heat of the sample (4.184J/molK for pure water) and T is temperature in K

We can determine the energy of each sample of water:

Sample 1:

Q = 30.0g*4.184J/molK*282K

Mass is 30.0g because density of water is 1g/mL

Q = 35396.6J

Sample 2:

Q = 50.0g*4.184J/molK*328K

Q = 68617.6J

Total energy is:

Q = 104014.2J

As total mass is 30+50 = 80g, final temperature is:

104014.2J = 80.0g*4.184J*T

310.7K is final temperature of the mixture

Answer:tell the instructor of the lab

Explanation:

Answer:

Explanation:

The density of a substance can be found by using the formula

[tex]Density = \frac{mass}{volume} [/tex]

From the question

mass of CCl4 = 3.16 g

volume of CCl4 = 2.0 mL

Substitute the values into the above formula and solve for the Density

That's

[tex]Density = \frac{3.16}{2} [/tex]

We have the final answer as

Hope this helps you

Answer:

Four Four

Explanation:

A molecule is formed by the overlap of orbitals. The combination of atomic orbitals during bond formation leads to the formation of bonding and anti bonding molecular orbitals.

When a molecule is formed, the number of bonding molecular orbitals must be equal to the number of anti bonding molecular orbitals.

Hence, when methane is formed, four bonding and four anti bonding orbitals are formed because methane forms four bonds per molecule.

Answer:

12 this is your answer

hope you like it

Answer:

(c) Nitric acid [tex]\rm HNO_3\, (aq)[/tex] is the limiting reactant.

(d) Approximately [tex]6.6\; \rm g[/tex] of [tex]\rm CuO\, (s)[/tex] will be in excess.

(e) The percentage yield of [tex]\rm Cu(NO_3)_2[/tex] is approximately [tex]91\%[/tex]. (Rounded to two significant figures, as in other quantities in the question.)

Explanation:

Start with the balanced chemical equation:

[tex]\rm CuO\, (s) + 2\; HNO_3\, (aq) \to Cu(NO_3)_2\, (aq) + H_2O\, (l)[/tex].

Look up relevant relative atomic mass data on a modern periodic table:

Calculate the formula mass of the species:

There are two reactants in this reaction: [tex]\rm CuO[/tex] and [tex]\rm HNO_3\, (aq)[/tex]. Assume that [tex]\rm CuO\![/tex] is the limiting one. In other words, assume that all the [tex]\rm CuO\!\![/tex] is consumed before [tex]\rm HNO_3\, (aq)\![/tex] was.

Consider: how many moles of [tex]\rm HNO_3\, (aq)\!\![/tex] would be required to convert all that [tex]7.0\; \rm g[/tex] of [tex]\rm CuO\!\!\![/tex] to [tex]\rm Cu(NO_3)_2[/tex]?

Calculate the number of moles of [tex]\rm CuO[/tex] formula units in [tex]7.0\;\rm g[/tex] of

[tex]\displaystyle n({\rm CuO}) = \frac{m}{M} = \frac{7.0\; \rm g}{79.545\; \rm g \cdot mol^{-1}} \approx 0.088001\; \rm mol[/tex].

Note the ratio between the coefficients of [tex]\rm CuO\, (s)[/tex] and [tex]\rm HNO_3\, (aq)[/tex]:

[tex]\displaystyle \frac{n(\mathrm{HNO_3\, (aq)})}{n(\mathrm{CuO\, (s)})} = \frac{2}{1} = 2[/tex].

Therefore:

[tex]\begin{aligned}&n(\text{$\mathrm{HNO_3\, (aq)}$, consumed (under assumption)})\\ &= \frac{n(\text{$\mathrm{HNO_3\, (aq)}$, consumed})}{n(\mathrm{CuO\, (s)})}\cdot n(\mathrm{CuO\, (s)}) \\ &\approx 2 \times 0.088001\; \rm mol \approx 0.17600\; \rm mol\end{aligned}[/tex].

On the other hand, how many moles of [tex]\rm HNO_3\, (aq)[/tex] are actually available?

Convert the volume of that [tex]\rm HNO_3\, (aq)[/tex] solution to the standard unit (liter.)

[tex]V(\rm HNO_3\, (aq)) = 50\; \rm mL = 0.050\; \rm L[/tex].

Calculate the number of moles of [tex]\rm HNO_3\, (aq)[/tex] in that [tex]0.20\; \rm M[/tex] solution:

[tex]n({\rm HNO_3\, (aq)}) = c\cdot V = 0.050\; \rm L \times 0.20\; \rm mol \cdot L^{-1} = 0.010\; \rm mol[/tex].

Apparently, the quantity of [tex]\rm HNO_3\, (aq)[/tex] required exceeded the quantity that is available. Therefore, the assumption is invalid, and [tex]\rm CuO[/tex] cannot be the limiting reactant. At the same time,

Since it is now known that all that [tex]0.010\; \rm mol[/tex] of [tex]\rm HNO_3\, (aq)[/tex] will be consumed, apply that coefficient ratio again to obtain the quantity of [tex]\rm CuO[/tex] consumed in this reaction:

[tex]\begin{aligned}&n(\text{$\mathrm{CuO}$, consumed})\\ &= n(\text{$\mathrm{HNO_3\, (aq)}$, consumed}) \left/\frac{n(\text{$\mathrm{HNO_3\, (aq)}$, consumed})}{n(\mathrm{CuO\, (s)})}\right. \\ &\approx (0.010 \; \rm mol) / 2 \approx 0.0050\; \rm mol\end{aligned}[/tex].

It was already shown that the formula mass of [tex]\rm CuO[/tex] is (approximately) [tex]79.545\; \rm g \cdot mol^{-1}[/tex]. Therefore, the mass of that [tex]0.0050\; \rm mol[/tex] formula units of [tex]\rm CuO\![/tex] would be:

[tex]\begin{aligned}& m(\text{$\rm CuO$, consumed}) \\&= n \cdot M \approx 0.0050\; \rm mol \times 79.545\; \rm g\cdot mol^{-1} \\&\approx 0.39773\; \rm g\end{aligned}[/tex].

Before the reaction, [tex]7.0\; \rm g[/tex] of [tex]\rm CuO[/tex] is available. Therefore, all that [tex]7.0\; \rm g - 0.39773\; \rm g \approx 6.6\; \rm g[/tex] of [tex]\rm CuO\![/tex] would be in excess.

Similarly:

[tex]\displaystyle \frac{n(\mathrm{HNO_3\, (aq)})}{n(\mathrm{Cu(NO_3)_2\, (aq}))} = \frac{2}{1} = 2[/tex].

Apply this ratio to find the theoretical yield of [tex]\rm Cu(NO_3)_2[/tex]:

[tex]n(\text{$\mathrm{Cu(NO_3)_2\, (aq)}$, theoretical yield}) \approx (0.010 \; \rm mol) / 2 \approx 0.0050\; \rm mol[/tex].

Find the mass of that [tex]0.0050\; \rm mol[/tex] of [tex]\rm Cu(NO_3)_2[/tex] formula units using its formula mass:

[tex]m(\text{$\rm Cu(NO_3)_2\, (aq)$, theoretical yield}) \approx 0.93777\; \rm g[/tex].

Calculate the percentage yield given that the actual yield is [tex]0.85\; \rm g[/tex]:

[tex]\begin{aligned}&\text{Percentage Yield} \\ &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\% \\ &= \frac{0.85\; \rm g}{0.93777\; \rm g} \times 100\% \approx 0.91\% \end{aligned}[/tex].

(Rounded to two significant figures.)

(c) The limiting reactant is HNO₃

(d) The mass of the excess reactant after the reaction is approximately 6.6 grams

(e) The percentage yield of copper (II) nitrate from the reaction is approximately 90.64%

The reason the above values are the correct values is as follows;

The given parameters are;

The mass of copper(II)oxide in the reaction = 7.0 g

The volume of the 0.20 M nitric acid, HNO₃ = 50 mL

(c) Concentration of the reactants

The molar mass of CuO = 79.545 g/mol

Number of moles = Mass/(Molar mass)

The number of moles of CuO = (7 g)/79.545 g/mol ≈ 0.088 moles

50 mL of 0.20 M HNO₃, contains 50/1000 × 0.2 = 0.01 moles of HNO₃

The chemical equation for the reaction is CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O

Therefore;

One mole of CuO reacts with two moles of HNO₃ to produce one mole of Cu(NO₃)₂ and one mole of H₂O

Therefore, 0.088 moles of CuO reacts with 2 × 0.088 = 0.176 moles of HNO₃

Given that there is only 0.01 moles of HNO₃, the limiting reactant is the HNO₃, which is not enough to completely react with the CuO which is the excess reactant

(d) The mass of the CuO that reacts with the 0.01 moles of HNO₃ is given as follows;

1 mole of CuO reacts with 2 moles HNO₃

0.01 moles of HNO₃ will react with 0.01/2 = 0.005 moles of CuO

Mass = Number of moles × Molar mass

The mass of 0.005 moles of CuO = 0.005 moles × 79.545 g/mol = 0.397725 grams

The mass of the CuO left = Initial mass - Reacting mass

∴ The mass of the CuO left = 7 grams - 0.397725 grams = 6.602275 grams

The mass of the excess reactant (CuO) after the reaction ≈ 6.6 grams

(e) The theoretical number of moles of copper (II) nitrate, Cu(NO₃)₂ produced = Half the number of moles of HNO₃ in the reaction

The number of moles of HNO₃ in the reaction = 0.01 moles

∴ The theoretical number of moles of copper (II) nitrate, Cu(NO₃)₂ produced = (1/2) × 0.01 moles = 0.005 moles

The molar mass of Cu(NO₃)₂ = 187.56 g/mol

The theoretical mass of Cu(NO₃)₂ produced = 0.005 moles × 187.56 g/mol = 0.9378 grams

The actual yield of copper (II) nitrate is 0.84 g

[tex]Percentage \ yield = \mathbf{\dfrac{Actual \ yield}{Theoretical \ yield} \times 100}[/tex]

Therefore;

[tex]\% \ yield \ of \ Cu(NO_3)_2 = \dfrac{0.85 \, g}{0.9378 \, g} \times 100 \approx 90.64 \%[/tex]

The percentage yield of copper (II) nitrate, %yield ≈ 90.64%

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Answer:

0.15 (M) of NaCl            

Explanation:

An isotonic solution is defined as the solution having the same osmolarity, or the same solute concentration, as the another solution. If the two solutions are to be separated by any semipermeable membrane, then water will start flowing in equal parts from each solution and also into the other.  

In the context, since it is given that the red blood cells exerts the same osmotic pressure as a 0.15 (M) of NaCl solution.

Thus, 0.15 (M) of NaCl is a isotonic solution.  

Answer:

Explanation:

The density of a substance can be found by using the formula

To calculate the Density in g/mL we must first convert the mass from kg to g and the volume from L to mL

For the mass

That's

1.34 kg

1kg = 1000 g

1.34 kg = 1000 × 1.34 = 1340 g

For the volume

That's

2.0 L

1 L = 1000 mL

2.0 L = 2 × 1000 = 2000 mL

So we have

mass = 1340 g

volume = 2000 mL

Substitute the values into the above formula and solve for the Density

That's

We have the final answer as

Hope this helps you

Answer:

B

Explanation:

Soil structure affects water and air movement in a soil, nutrient availability for plants, root growth, and microorganism activity. ... This allows for greater air and water movement and better root growth.

Answer:

b

Explanation:

Answer:

4.07 × 10⁹ mL

Explanation:

Step 1: Given data

Length of the pool (L): 60.0 m

Width of the pool (W): 35.0 m

Depth of water in the pool (D): 6.35 ft

Step 2: Convert "D" to m

We will use the relationship 1 m = 3.28 ft

[tex]6.35ft \times \frac{1m}{3.28ft} = 1.94m[/tex]

Step 3: Calculate the volume of water (V)

We will use the following expression.

[tex]V = L \times W \times D = 60.0m \times 35.0m \times 1.94 m = 4.07 \times 10^{3} m^{3}[/tex]

Step 4: Convert "V" to mL

We will use the relationship 1 m³ = 10⁶ mL.

[tex]4.07 \times 10^{3} m^{3} \times \frac{10^{6}mL }{1m^{3} } = 4.07 \times 10^{9} mL[/tex]

Answer:

a. Physical Change

b. Chemical Change

c. Physical Change

d. Chemical Change

e. Physical Change

Physical and chemical change has been differentiated by the formation of new substances. Silver pin tarnishing and digestion of food are considered chemical changes. Thus, options b and d are correct.

Chemical changes are characterized by alterations in the properties of the substances that results in the formation of new substances that differs in their composition and properties. These changes are said to be permanent changes that cannot be reversed back into their main constituents.

Physical changes are said to be temporary changes that are reversible as the substance produced has the same structures and includes changes in texture, color, pattern, temperature, state of matter, etc.

Silver tarnishes under air is a chemical change as it undergoes a chemical reaction under oxygen and produces a new product. Digestion of food is also a chemical change that breaks the food into smaller pieces under the action of the enzymes.

Rolling of the pie dough, cutting the trees into boards, and melting the chocolate bars are physical changes as no new substances with different chemical compositions are produced.

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Answer:

0.275

Explanation:

Given that:

Initial number of miles of radioactive isotope = 4.4 moles

How much if the sample remains after 4 half life's

Half-life refers to the time taken by a sample to reduce to half of it's initial value.

After the first half life :

Initial / 2 = 4.4 / 2 = 2.2 moles

Second half life :

2.2 / 2 = 1.1 moles

Third half life :

1.1 / 2 = 0.55 Moles

Fourth half life :

0.55 / 2 = 0.275 moles

Answer:

-891kj

Explanation:

We have the chemical equation as:

2SO2(g) + O2(g) ----2SO3; ∆H° = 198kj

If sulphur trioxide decomposes we have opposite reaction. Enthalpy of decomposition will not be positive for forward reaction.

2 moles of SO3 ----∆H° = -198Kj

1 mole = -198/2

9 moles = -198/2 x 9

= -891kj

Therefore change in enthalpy when 9 moles is decomposes is -891kj.

Answer:

C

Explanation:

Answer:

4 sig fis so look at the next number which is "4"

answer is 19.97

all numbers not zeros are significant

Answer:

The speed is 7 every second.

Equation:

14/2 = 7

A student graphed distance versus time for an object that moves 14 every 2 s. The speed of the object is 7 m/sec.

The speed is the distance traveled by an object with a specific unit of time. Speed is calculated by dividing distance by time. The SI unit of speed is a meter per second.

S = d / t

Distance is the space or extent between two things. It is measured in meters and kilometers.

Time is a variable that equals the distance divided by time.

The slope of a graph is the rate at which the y and x axis change together, the "rise over the run." Slope can be simply calculated using this equation:

Given, that distance = 14 meters

The time is 2 sec.

By the formula

d = 14 / 2 = 7 m /sec.

Thus, the speed of the object is 7 m/sec.

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Explanation:

a chemical nomenclature is a set of rules to generate systematic names for chemical compounds.

Answer:

0.72 g

Explanation:

Note that, the mass in gram of any substance is obtained from the relationship;

Mass= number of moles × molar mass

From the sequence of the reaction, the number of moles of Lithium diisopropylamide is 6.75mmol which is the same as 6.75 × 10^-3 moles

Molar mass of base= 107.1233 g/mol

Hence mass of the base is given by;

6.75 × 10^-3 moles × 107.1233 g/mol = 0.72 g

Answer:

Valence of iron in FeS2

Let the valency of iron be x

x+2(oxidation number of Sulphur)=0 as their is no charge on FeS2

As sulphur belongs to Group VI (chalcogens) so the valency/oxidation number of Sulphur is -2

x+2(-2)=0

x-4=0

x=4

So the valency/oxidation number of iron in FeS2 is +4

Explanation:

Hope it will help you monkey ^_^ it's crab

Answer:

Hope you are better now your parents must be worried about you get well soon  drink plenty of milk it will make your bones strong.

NOW TIME FOR WISHES ^_^

Answer:

The weak base is hydroxylamine

Explanation:

The general equilibrium of a weak base (A), is:

A(aq) + H₂O(l) ⇄ HA⁺ + OH⁻

Where Kb is:

Kb = [HA⁺] [OH⁻] / [A]

As both HA⁺ and OH⁻ comes from the equilibrium of A; HA⁺ = OH⁻. So, you can take:

Kb = [OH⁻]² / [A]

OH⁻ is determined from pH, thus:

pOH = -log [OH⁻]

pOH = 14 - pH

pOH = 14 - 9.44

pOH = 4.56

[OH⁻] = 10^{-4.56}

[OH⁻] = 2.75x10⁻⁵

As the concentration of the weak base is 0.0678M:

Kb = [OH⁻]² / [A]

Kb = [2.75x10⁻⁵]² / [0.0678M]

Kb = 1.1x10⁻⁸ = Kb of Hydroxylamine.

Answer:

The mass of FeSO₄.7H₂O  needed  [tex]\simeq[/tex] 225.89 grams

Explanation:

From the given question

The molecular weight of iron(ii) sulfate heptahydrate FeSO₄.7H₂O = 278.02 g/mol

Molarity = 3.25

Volume = 250.0 mL

The mass of FeSO₄.7H₂O  needed = [tex]\dfrac{moles \times molecular weight \times 250\ mL}{1000 \ mL}[/tex]

The mass of FeSO₄.7H₂O  needed  = [tex]\dfrac{3.25 \times 278.02 \times 250}{1000}[/tex]

The mass of FeSO₄.7H₂O  needed  = 225.89125 grams

The mass of FeSO₄.7H₂O  needed  [tex]\simeq[/tex] 225.89 grams

Answer:

66.441

I hope this helps!

Answer:

66.441

Explanation:

(5.391 × 10-7) + (2.5531 × 10-6)

(46.91) + (19.531)

66.441

Answer:  85.7 mL

Explanation:

Given the information from the question as plotted in the graph i will be uploading along side this answer,

Average of total volume of DCPIP used is

= (1.21 + 1.11 + 1.06)mL / 3

= 1.12 mL

and corresponding ( ascorbic acid ) is 0.70 g/L

Two parameter given as volume of DCPIP in final syringe and total volume of DCPIP are quite ambiguous

700mg ⇒ 1 L

THEREFORE volume that contains 60mg =  (1000/700) × 60 = 85.7 mL

Answer:

I ratio is

47 : 43 ratio

Answer:

HELLO YOUR QUESTION IS INCOMPLETE ATTACHED BELOW IS THE COMPLETE QUESTION

answer : 0.033 mol  ( A )

Explanation:

To determine the number of moles of AI3 that will be reduced when 0.1 Faraday passes through a cell during the production of AI

AI3+  + 3 e -------- AI  this means that it will take 3 mol to reduce AI3 to AI

1 mol AI = 3 F  electrons

therefore : 1 F electrons  = 1/3 mol of AI

therefore : 0.1 F ( Faraday ) = 0.1 * 1/3  = 0.033 mol of AI is needed

Answer:

The correct answer is 6.65 mg.

Explanation:

Based on the given information, the number of days given is 45.6 days. The original mass of sodium phosphate give is 175 mg. The half-life of phosphorus-32 is 14.3 days, therefore, the n or the number of half life will be,

n = 45.6/14.3 = 3.19

So, after 45.6 days, the mass of sodium phosphate sample left will be,

= Original mass × 1/2ⁿ

= 175 mg × 1/2 ^3.19

= 175 mg × 0.1096

= 34.3 mg of sodium phosphate left after 45.6 days

The molecular mass of Na₃³²PO₄ is 3 (23) + 32 + 4 (16) amu = 165 amu

Therefore, 165 grams of sodium phosphate comprise 32 grams of phosphorus.

So, 34.3 mg or 0.343 gram of sodium phosphate will contain,

= 0.343 × 32/165

= 6.65 mg of P³².

Answer:

Two (2) types of ions

Explanation:

Metal ions (cations) and negatively charged ions (anions) of either chloride, carbonate, sulphides or hydroxides are present during the selective precipitation.

Majorly, the process determines the metal ions present by initiating precipitation as certain conditions.

I hope this was helpful.

Answer:

Helps in accepting and defending the conclusion.

Explanation:

The main advantages of having proof before you make any conclusion is that it gives a reason for accepting the answer given by you because you have the proof to prove your answer so no one questions on your answer. The proof give you strength to defend your conclusion and make your conclusion right in front of the world. Without proof no one can accept your answer because every answer of a scientific question wants a proof for its validation.

Answer:

[tex]t_{1/2}=3.10x10^{-3}min=0.186s[/tex]

Explanation:

Hello,

In this case, for the calculation of the half-life for the mentioned reaction we first must realize that considering the units of the rate constant and the form of the rate law, it is a second-order reaction, therefore, the half-life expression is:

[tex]t_{1/2}=\frac{1}{k[A]_0}[/tex]

Therefore, we obtain:

[tex]t_{1/2}=\frac{1}{1291\frac{1}{M*min}*0.250M}\\\\t_{1/2}=3.10x10^{-3}min=0.186s[/tex]

Regards.

The half-life for the reaction is 0.003 mins

From the question given above, the following data were obtained:

Rate law = (1291 M⁻¹min⁻¹)[A]²

Initial concentration of A = 0.250 M

From the rate law given, we can see that the reaction is indicating a 2nd order reaction.

Thus, the half-life can be obtained as follow:

t½ = 1 / K[A₀]

t½ = 1 / (1291 × 0.25)

Therefore, the half-life for the reaction is 0.003 mins

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Answer:

[tex]0.564[/tex]

Explanation:

Hello,

In this case, by performing the given division we obtain:

[tex]\frac{9.15}{16.23}=0.563770795[/tex]

Nevertheless since 9.15 has three significant figures and 16.23 four, we must show the result with three significant figures taking into account that it must be shown with the least significant figures from the initial data:

[tex]0.564[/tex]

Regards.