A 2.0 mL sample of carbon tetrachloride (CCI4 ) masses at 3.16 g. What is the density of this substance in g/mL?
A. 1.5 g/mL
B. 1.58 g/mL
C. 1.6 g/ml
D. 0.63 g/ml

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:

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:

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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#SPJ2

Answer:

I ratio is

47 : 43 ratio

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:

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:

[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.

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:

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:

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:

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:

[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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