State whether the error introduced by each of the following problems would result in a high or a low value for the Cu recovery or would not affect the results. Explain. a. Some of the copper nitrate solution splashes out of the beaker in step 1. _____________

Answer:

я не знаю ответа :(

Explanation:

Answer:

Se

Explanation:

First of all, we must note that any element that we must choose is an element that is in group sixteen.

Elements of groups 16 have six electrons in their outermost shell which can be used for bond formation thereby yielding a total of twelve electrons on the valence shell.

However, this is only possible for the heavier members of the group 16 (from sulphur downwards) which are able to expand their octet.

Oxygen can not expand its octet hence it is not the answer.

Answer:

5.25 moles of protons. Option e

Explanation:

Reaction between phosphoric acid and sodium hydroxide is neutralization.

We can also say, we have an acid base equilibrium right here:

H₃PO₄  +  3NaOH →  Na₃PO₄  +  3H₂O

Initially we have 5.25 moles of base.

We have data from the acid, to state its moles:

M = mol/L, so mol = M . L

mol = 1.75 moles of acid

If we think in the acid we know:

H₃PO₄  →  3H⁺  +  PO₄⁻³

We know that 1 mol of acid can give 3 moles of protons (hydrogen ions)

If we have 1.75 moles of acid, we may have

(1.75 . 3) /1 = 5.25 moles of protons

These moles will be neutralized by the 5.25 moles of base

H₃O⁺  +  OH⁻  ⇄  2H₂O     Kw

In a titration of a weak acid and a strong base, we have a basic pH

Answer:

Explanation:

Can you provide a picture? I can outline the reactions though. a) will make a Z double bond from a triple bond and then peroxyacid can do epoxidation. b) will make the Z double bond then ozonolysis to double bond will create to aldehyde compounds. c) is essentially useless unless there is a ketone or aldehyde in the compound already since H2/Pd will fully reduce the alkyne (which I am assuming is present) and so the peroxyacid can't do epoxidation and can only do baeyer villiger oxidation, and d) reduces the alkyne to an E alkene and then do epoxidation to give an epoxide (with trans steroechemistry)

Answer:

Molecular formula => CH₅N

Explanation:

We'll begin by calculating the number of mole of the compound. This can be obtained as follow:

Pressure (P) = 0.982 atm

Volume (V) = 125 mL = 125 / 1000 = 0.125 L

Temperature (T) = 22 °C = 22 + 273 = 295 K

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =?

PV = nRT

0.982 × 0.125 = n × 0.0821 × 295

0.12275 = n × 24.2195

Divide both side by 24.2195

n = 0.12275 / 24.2195

n = 0.0051 mole

Next, we shall determine the molar mass of the compound.

Mass of compound = 0.158 g

Mole of compound = 0.0051 mole

Molar mass of compound =?

Mole = mass /molar mass

0.0051 = 0.158 / molar mass

Cross multiply

0.0051 × molar mass = 0.158

Divide both side by 0.0051

Molar mass = 0.158 / 0.0051

Molar mass of compound = 31 g/mol

Next, we shall determine the empirical formula of the compound. This can be obtained as follow:

C = 38.6%

H = 16.4%

N = 45%

Divide by their molar mass

C = 38.6 / 12 = 3.217

H = 16.4 / 1 = 16.4

N = 45 / 14 = 3.214

Divide by the smallest

C = 3.217 / 3.214 = 1

H = 16.4 / 3.214 = 5

N = 3.214 / 3.214 = 1

Empirical formula => CH₅N

Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:

Empirical formula = CH₅N

Molar mass of compound = 31 g/mol

Molecular formula = Empirical × n = molar mass of compound

[CH₅N]ₙ = 31

[12 + (5×1) + 14]n = 31

[12 + 5 + 14]n = 31

31n = 31

Divide both side by 31

n = 31 / 31

n = 1

Molecular formula => [CH₅N]ₙ

Molecular formula => [CH₅N]₁

Molecular formula => CH₅N

Answer:

See explanation and image attached

Explanation:

The acid-catalyzed hydrolysis of a nitrile to give a carboxylic acid occurs by initial protonation of the nitrogen atom. This step is shown in the image attached.

The next step is the nucleophilic addition of water. The task is to show the movement of electrons in this step of the reaction mechanism. This was clearly shown in the image attached to this answer.

Answer:

Average speed of helium is higher than argon.

Explanation:

The average speed of helium is higher than argon atom under the conditions specified because of lower mass of the helium atom as compared to argon atom. Average speed of an atom is inversely proportional to mass of the atom. If mass of an atom decreases, the atom moves with higher speed while on the other hand, if the mass of an atom increases the average speed of an atom decreases.

Answer:

true

Explanation:

Answer:

[tex]k_2=0.504s^{-1}[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the rate constant at 55 °C by using the temperature-variable version of the Arrhenius equation:

[tex]ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )[/tex]

Thus, we plug in the temperatures, activation energy and universal constant of gases in consistent units to obtain:

[tex]ln(\frac{k_2}{0.0215s^{-1}} )=-\frac{72000\frac{J}{mol}}{8.3145\frac{J}{mol*K}}(\frac{1}{55+273} -\frac{1}{20+273} ) \\\\ln(\frac{k_2}{0.0215s^{-1}} )=3.154\\\\k_2=0.0215s^{-1}exp(3.154)\\\\k_2=0.504s^{-1}[/tex]

Regards!

Answer:

c......................

Answer:

electrons will fill the lowest energy orbitals first and then move up to higher energy orbitals only after the lower energy orbitals are full

Answer:

1.8 g

Explanation:

Step 1: Write the balanced equation

CH₃CH₃(g) + 3.5 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(g)

Step 2: Determine the limiting reactant

The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:112.0 = 0.2684:1.

The experimental mass ratio of CH₃CH₃ to O₂ is 0.60:3.52 = 0.17:1.

Thus, the limiting reactant is CH₃CH₃

Step 3: Calculate the mass of CO₂ produced

The theoretical mass ratio of CH₃CH₃ to O₂ is 30.06:88.02.

0.60 g CH₃CH₃ × 88.02 g CO₂/30.06 g CH₃CH₃ = 1.8 g

Answer:

After the solution is heated, but before additional solute is added

Explanation:

An unsaturated solution is a solution that contains less solute than it can normally hold at a given temperature. Hence an unsaturated solution can still dissolve more solute.

When the solution is heated, the saturated cold solution becomes an unsaturated hot solution which is capable of dissolving more solute at this point.

Once more solute is dissolved, the solution becomes saturated again just before it begins to cool since no more solute dissolves in the solution at some point before cooling and addition of seed crystals.

After the solution is heated and more solute is added, but before it is cooled is the point the solution was unsaturated.

This is referred to as the point in which a solution can still dissolve more of a substance when added to it.

In this case, the solution was still unsaturated when the solution was heated and more solute is added, but before cooling took place.

Read more about Unsaturation here https://brainly.com/question/490217

Answer:

a. 6mL of ethanol and 35mL of water: the solute is ethanol (smallest volume) and the solvent is water (greater volume).

b. 300 g of water containing 8g of NaHCO3: the solute is NaHCO3 (smallest mass) and the solvent is water (greater mass).

c. 0.005L of CO2 and 2L of O2: the solute is CO2 (smallest volume) and the solvent is O2 (greater volume).

Explanation:

Hello there!

In this case, according to the given problem, it turns out possible for us to solve these questions by bearing to mind the fact that in a solution, we can find two substances, solute and solvent, whereas the former is in a smaller proportion in comparison to the latter; in such a way, we infer the following:

a. 6mL of ethanol and 35mL of water: the solute is ethanol (smallest volume) and the solvent is water (greater volume).

b. 300 g of water containing 8g of NaHCO3: the solute is NaHCO3 (smallest mass) and the solvent is water (greater mass).

c. 0.005L of CO2 and 2L of O2: the solute is CO2 (smallest volume) and the solvent is O2 (greater volume).

Regards!

Answer:

1)  [tex]molality = 0.19[/tex]

2)  [tex]Mole\ fraction= 0.003486[/tex]

3)[tex]Mass\ percent = 2.8%[/tex]

Explanation:

Concentration of Tartaric acid=0.190mole /l

1)

Generally

[tex]Mass\ of\ tartaric\ acid = 150.087 *0.190[/tex]

[tex]Mass\ of\ tartaric\ acid = 28.5 g[/tex]

Since 1L of solution tartartic acid is

[tex]T_{1l}= density * volume[/tex]

[tex]T_{1l}= 1.016Kg / L X 1 L[/tex]

[tex]T_{1l}= 1016[/tex]

Therefore

[tex]Mass of solvent = 1016-28.5[/tex]

[tex]Mass of solvent = 987.5 g[/tex]

Generally the equation for molality is mathematically given by

[tex]molality = \frac{moles}{Kg}[/tex]

[tex]molality = \frac{0.190 * 1000}{987.5}[/tex]

[tex]molality = 0.19[/tex]

2.

Generally the equation for Moles of water  is mathematically given by

[tex]Moles\ of\ water = \frac{mass}{mol wt}[/tex]

[tex]Moles\ of\ water = \frac{987.5}{18 }[/tex]

[tex]Moles\ of\ water= 54.86[/tex]

Therefore

[tex]Mole\ fraction = \frac{Moles\ of\ solute}{total\ moles}[/tex]

[tex]Mole\ fraction= \frac{0.190}{54.5}[/tex]

[tex]Mole\ fraction= 0.003486[/tex]

3

Generally the equation for Mass Percent is mathematically given by

[tex]Mass\ percent = \frac{mass\ of\ tartaric\ acid}{total mass}[/tex]

[tex]Mass\ percent = \frac{28.5* 100}{1016}[/tex]

[tex]Mass\ percent = 2.8%[/tex]

Answer:

The postabsorptive state (also called the fasting state) occurs when the food is already digested and absorbed, and it usually occurs overnight, when you sleep (if you skip meals for some days, you will enter in this state).

The catabolic state is the metabolic breakdown of molecules into simpler ones, releasing energy (heat) and utilizable resources.

Now, when you are in a postabsorptive state, the glucose levels start to drop, then the body starts to depend on the glycogen stores, which are catabolized into glucose, this is defined as the start of the postabsorptive state.

So yes, as the postabsorptive states, catabolic processes start to happen, so the statement is true.

Answer:

The right answer is "60.56 atm".

Explanation:

As we know,

Vander wall's equation is:

⇒ [tex](P+\frac{n^2 a}{v^2} )(v-nb)=nRT[/tex]

or,

⇒ [tex]P=\frac{nRT}{(v-nb)}-\frac{n^2 a}{v^2}[/tex]

Here,

a = 3.59 L² atm mol⁻²

b = 0.0427 L mol⁻¹

By putting the values in the above equation, we get

⇒ [tex]P=\frac{0.467\times 0.0821\times 432}{0.245-0.467\times 0.0427}[/tex]

       [tex]=\frac{(0.467)^2\times 3.49}{(0.245)^2}[/tex]

       [tex]=73.61-13.05[/tex]

       [tex]=60.56 \ atm[/tex]

Solution :

1 lb = 453.592 g

1 gallon = 3785 g

For every 5 gallon gasoline = 5 lb of C is found

                                              = 5 x 453.592 g of C atoms

                                              = 2267.96 g of C atoms

Assume the consumption of car = 15 miles per kg of gasoline

The amount of gasoline used per year  [tex]$=\frac{12000}{15}$[/tex]

                                                                   = 800 kg

In gallons = [tex]$\frac{800}{3.785} = 211.36\text{ gallons}$[/tex]    

5 gallons will produce = 2267.96 g of C atoms

Therefore,  

211.36 gallons will produce = [tex]$\frac{211.36 \times 2267.96}{5}$[/tex]

                                             = 95871.21 g

                                             = 95.87 kg

or                                           = 25.32 gallon

Answer:

Atoms are similar in the way that their nuclei contain only protons and neutrons.

Answer:

All things are made of atoms, and all atoms are made of the same three basic particles - protons, neutrons, and electrons. But, all atoms are not the same. The difference in the number of protons and neutrons in atoms account for many of the different properties of elements.

I hope this helps :)

Answer: The pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.

Explanation:

Given: Initial concentration of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] = 0.10 M

[tex]K_{a} = 1.0 \times 10^{-8}[/tex]

Let us assume that amount of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] dissociates is x. So, ICE table for dissociation of  [tex]Cu(H_{2}O)^{2+}_{6}[/tex]  is as follows.

                               [tex]Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}[/tex]

Initial:                       0.10 M                       0                       0

Change:                    -x                              +x                      +x

Equilibrium:           (0.10 - x) M                    x                        x

As the value of [tex]K_{a}[/tex] is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.

And, (0.10 - x) will be approximately equal to 0.10 M.

The expression for [tex]K_{a}[/tex] value is as follows.

[tex]K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}[/tex]

Hence, [tex][H_{3}O^{+}] = 3.2 \times 10^{-5}[/tex]

Formula to calculate pH is as follows.

[tex]pH = -log [H^{+}][/tex]

Substitute the values into above formula as follows.

[tex]pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49[/tex]

Thus, we can conclude that the pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.

Answer:

Likely [tex]\rm In[/tex] (indium.)

Explanation:

Number of atoms: [tex]N = 2.241 \times 10^{21}[/tex].

Dividing, [tex]N[/tex], the number of atoms by the Avogadro constant, [tex]N_A \approx 6.023 \times 10^{23} \; \rm mol^{-1}[/tex], would give the number of moles of atoms in this sample:

[tex]\begin{aligned} n &= \frac{N}{N_{A}} \\ &\approx \frac{2.241 \times 10^{21}}{6.023 \times 10^{23}\; \rm mol^{-1}} \approx 3.72 \times 10^{-3}\; \rm mol \end{aligned}[/tex].

The mass of that many atom is [tex]m = 0.4272\; \rm g[/tex]. Estimate the average mass of one mole of atoms in this sample:

[tex]\begin{aligned}M &= \frac{m}{n} \\ &\approx \frac{0.4272\; \rm g}{3.72 \times 10^{-3}\; \rm mol} \approx 114.82\; \rm g \cdot mol^{-1}\end{aligned}[/tex].

The average mass of one mole of atoms of an element ([tex]114.82\; \rm g \cdot mol^{-1}[/tex] in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass [tex]114.82[/tex]. Indium, [tex]\rm In[/tex], is the closest match.

The symbol of the element is In

From the question, we are to determine the identity of the element

First, we will determine the number of moles of sample present

Using the formula

[tex]Number \ of\ moles = \frac{Number\ of \ atoms }{Avogadro's\ constant}[/tex]

Number of moles of the sample = [tex]\frac{2.241\times 10^{21} }{6.022\times 10^{23} }[/tex]

Number of moles of the sample = 0.003721355 mole

Now, we will determine the Atomic mass of the sample

From the formula,

[tex]Atomic\ mass = \frac{Mass}{Number\ of\ moles}[/tex]

Therefore,

Atomic mass of the sample = [tex]\frac{0.4272}{0.003721355}[/tex]

Atomic mass of the sample = 114.8 amu

The element that has an atomic mass of 114.8 amu is Indium. The symbol of Indium is In.

Hence, the symbol of the element is In.

Learn more on Stoichiometry here: https://brainly.com/question/22495545

Answer:

Explanation:

CCl4 => C + 2Cl2

Answer:

Oxidation half-reaction: 2 Fe (s) ----> 2 Fe³+ (aq) + 6e-

Reduction half-reaction: 3 Pb²+ (aq) + 6 e- ----> 3 Pb (s)

Explanation:

A redox reaction reaction is one in which oxidation and reduction occur simultaneously and to the same extent.

Oxidation involves a loss of electron, hence, a positive increase in the oxidation number of the atom or ion. The oxidation half-reaction is as follows:

2 Fe (s) ----> 2 Fe³+ (aq)

The metallic element iron, Fe , having an oxidation number of zero, loses three electrons to form the Fe³+ ion with a charge of +3. Since each atom loses three electrons each, The number of moles of electrons lost is six.

2 Fe (s) ----> 2 Fe³+ (aq) + 6e-

Reduction involves a gain of electrons, hence, a decrease in the oxidation number of the atom or ion. The reduction half-reaction is given below:

3 Pb²+ (aq) ---> 3 Pb (s)

The lead (ii) ion, Pb²+ having a charge of +2 gains two electrons each to become the neutral metallic lead atom, Pb, with oxidation number of zero. Since 3 moles of Pb²+ are reacting, 6 moles of electrons are gained.

3 Pb²+ (aq) + 6 e- ----> 3 Pb (s)

Answer: The purpose of washing the organic layer with sodium bicarbonate solution is to

(to remove any of the unreacted benzoic acid).

Explanation:

Methyl benzoate is a typical example of an ester in organic chemistry. It is formed by the action of benzoic acid and methanol. Sulphuric acid is often used as a catalyst in this reaction.

The procedure for the preparation of the ester includes:

--> the acid and the alcohol are added into a round bottom flask,

--> concentrated sulphuric acid is added down the side of the flask,

--> a separatory funnel is used to extract the ester into an organic layer where sodium bicarbonate is added.

--> when the aqueous layer from the sodium bicarbonate wash is acidified with concentrated HCl, the unreacted benzoic acid should precipitate. This is how the unreacted benzoic acid is removed.

Answer:

See explanation

Explanation:

Let us remember that Sodium bicarbonate is basic in nature. It is one of the common bases used to neutralize acids.

In the preparation of methyl benzoate, H2SO4 is used as the catalyst for the esterification reaction.

Hence, when we wash the organic layer with sodium bicarbonate we are trying to remove the H2SO4 thereby neutralizing the organic layer.

Answer:

True

Explanation:

When two atoms are at infinite distance from each other, the both atoms posses high energy.

However, as they begin to approach each other, the distance between them gradually decreases and so does their energy.

A point is eventually reached when the potential energy curve reaches its minimum value. The internuclear distance between the two atoms at this point is called the bond length of the system.

Answer:

I do not speak Spanish.

Explanation:

Answer:

a. increase.

Explanation:

Hello!

In this case, according to the given information, it turns out possible for us to tell that the common ion effect decreases the solubility of the ionic solid by firstly increasing the concentration of the common ion, which is further solved for the solubility in order to evidence the aforementioned decrease.

As an example, we can consider the solubility equilibrium for silver chloride:

[tex]Ksp=[Ag^+][Cl^-][/tex]

Which goes to:

[tex]Ksp=s^2[/tex]

Whereas s is the solubility to be solved for. However, when a silver- or chloride-containing solution is added, say 0.1 sodium chloride, the equilibrium expression changes to:

[tex]Ksp=(s)(s+0.1)[/tex]

Which turns out into a smaller value for s.

Regards!

Answer:

4.77mol is the correct answer

Answer:

Option D: Carbohydrates

Explanation:

Carbohydrates are a group of macromolecules that are a vital energy source for the cell, provide structural support to many organisms, and can be found on the surface of the cell as receptors or for cell recognition.

Answer:

The mass of silver carbonate precipitated is 5.18 grams.

Explanation:

Molarity of the silver nitrate solution = 0.671 M

Volume of the silver nitrate solution = 56.0 mL

[tex]1 mL = 0.001 L\\56.0 mL = 56.0\times 0.001 L=0.0560 L[/tex]

Moles of silver nitrate = n

[tex]Molarity=\frac{\text{Moles of compound}}{\text{Volume of solution in Liters}}\\\\0.671 M=\frac{n}{0.0560 L}\\n=0.671 M\times 0.0560 L=0.0376 mol[/tex]

Moles of silver nitrate used = 0.0376 mol

[tex]K_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2KNO_3[/tex]

According to the reaction, 2 moles of silver nitrate gives 1 mole of silver carbonate, then 0.0376 moles of silver nitrate:

[tex]=\frac{1}{2}\times 0.0376 mol=0.0188 \text{mol of }Ag_2CO_3[/tex]

Moles of the silver carbonate formed = 0.0188 mol

Molar mass of silver carbonate = 275.7453 g/mol

Mass of silver carbonate :

[tex]=275.7453 g/mol\times 0.0188 mol=5.1840 g\approx 5.18 g[/tex]

The mass of silver carbonate precipitated is 5.18 grams.