For the standard cell, the Cu2 half-cell was made with 1. 0L of 1. 0MCu(NO3)2 and the Zn2 half-cell was made with 1. 0L of 1. 0MZn(NO3)2. The experiment was repeated, but this time the Cu2 half-cell was made with 0. 50L of 2. 0MCu(NO3)2 and the Zn2 half-cell was made with 1. 0L of 1. 0MZn(NO3)2. Is the cell potential for the nonstandard cell greater than, less than, or equal to the value calculated in part (b)

Answer:

A. The color change from blue to green happens in nature.

Explanation:

The color change is an indicator of a chemical reaction that has occurred. Although color changes are not necessarily good diagnostic tool to measure if a chemical reaction has occurred or not.

The evidence that shows that a chemical reaction has taken place is the disappearance of the steel wool shows that a new substance has formed.

The electrochemical series is arrangement of elements in order of decreasing reactivity. It is important to note that the elements that are high up in the electrochemical series can displace the elements that are lower in the series from their aqueous solution.

Steel contains iron which lies above copper in the electrochemical series hence a chemical change occurs when steel dissolves in the copper solution.

Hence, the evidence that shows that a chemical reaction has taken place is the disappearance of the steel wool shows that a new substance has formed.

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Answer: biology is the study of life

Source:trust me bro

Answer:

Answer:

The mole ratio of C₄H₁₀ and CO₂ is 2 : 8, which  simplifies to 1 : 4.

Explanation:

The mole ratio is the relative proportion of the moles of products or reactants that participate in the reaction according to the chemical equation.

The chemical equation given is:

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Once you check that the equation is balanced, you can set the mole ratios for all the reactants and products. The coefficients used in front of each reactant and product, in the balanced chemical equation, tells the mole ratios.

In this case, they are: 2 mol C₄H₁₀ :  13 mol O₂ : 8 mol CO₂ : 10 mol H₂O

Since you are asked about the mole ratio of C₄H₁₀ and CO₂ it is:

2 mol C₄H₁₀ : 8 mol CO₂ , which dividing by 2, simplifies to

1 mol C₄H₁₀ : 4 mol CO₂, or

1 : 2.

Explanation:

Litmus paper and Phenolphthalein are both indicators. This means they will change colour in the presence of an acid or a base.

Litmus paper turns blue in a base and remains red when it is in contact with an acid or neutral solution. Phenolphthalein turns pink in a base, but is coluorless in an acid or neutral solution.

Answer:

c. 1.0 mole C5H5N and 1.5 mole C5H5NHCl

Explanation:

We can determine pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] is concentration of conjugate base and [HA] concentration of weak acid. These concentrations can be computed as moles of each species.

We need to determine pKa of both NH₃ and C₅H₅N buffers, thus:

pKb = -log Kb

NH₃ pKb = -log 1.8x10⁻⁵ = 4.74

C₅H₅N pKb = -log 1.7x10⁻⁹ = 8.77

And pKa = 14 - pKb:

NH₃ pKa = 14 - 4.74 = 9.26

C₅H₅N pKb = 14 - 8.77 = 5.23

A buffer works only under pH's between pKa-1 and pKa + 1. As pKa NH₃ buffer is 9.23 is not possible to produce a buffer with pH 5.05 for this system.

Thus, we only will compute the buffers made with C₅H₅N:

c. 1.0 mole C5H5N (Weak base) and 1.5 mole C5H5NHCl (Conjugate acid)

pH = pKa + log [A⁻] / [HA]

pH = 5.23+ log [1.0 moles] / 1.5 moles]

d. 1.5 mole C5H5N and 1.0 mole C5H5NHCl

pH = pKa + log [A⁻] / [HA]

pH = 5.23+ log [1.5 moles] / 1.0 moles]

Right solution is:

When the element comes near an atom having seven valence electrons, no reaction will occur.

The name of the element is Neon.

The electron configuration of elements shows how the total number of electrons present in the atoms of the elements are arranged in the orbitals around the nucleus.

In other words, the electron configuration of elements indicates the total number of electrons on its neutral atom, and hence the atomic number of elements.

In this case, the electron configuration is \(1s^2 2s^2 2p^6\). The total number of electrons is 10. Hence, the atomic number of the element is 10. The element with atomic number 10 is Neon (Ne).

Neon represents one of the inert elements on the periodic table. They possess octet electron configuration and as such, are not reactive ordinarily. Thus, if an atom of neon comes near an atom with seven valence electrons, no reaction will occur.

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A competitive inhibitor is a type of enzyme inhibitor that binds to the active site of the enzyme and competes with the substrate for binding.

As a result, the competitive inhibitor can increase the apparent Michaelis-Menten constant (Km) of the enzyme.

Km is a measure of the affinity of the enzyme for its substrate. A higher value of Km indicates a lower affinity of the enzyme for the substrate, which means that the enzyme requires a higher concentration of the substrate to reach half of its maximum reaction rate (Vmax).

When a competitive inhibitor is present, it decreases the affinity of the enzyme for its substrate by occupying the active site and preventing the substrate from binding.

This means that a higher concentration of the substrate is required to overcome the inhibition caused by the competitive inhibitor, which leads to an increase in the apparent Km.

In other words, a competitive inhibitor increases the Km because it makes it more difficult for the substrate to bind to the enzyme.

The effect of the competitive inhibitor on Vmax depends on the concentration of the inhibitor relative to the concentration of the substrate.

If the concentration of the inhibitor is high enough to completely saturate the active site, it can decrease the Vmax by preventing the substrate from binding to the enzyme.

However, if the concentration of the inhibitor is low, the effect on Vmax may be minimal, as the substrate can still bind to the enzyme and reach the maximum reaction rate.

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Answer:

Divide the distance the object traveled by the time it took to get there.

Explanation:

To calculate the speed on an object, start by determining how far the object has traveled. Next, figure out the amount of time that the object took to cover that distance. Finally, divide the distance the object traveled by the time it took to get there. Don't forget to label the speed with the correct units of measurement.

The standard enthalpy change for reaction (3) is 117.1 kJ.

The standard enthalpy change for reaction (3) can be calculated by using the enthalpy changes of reactions (1) and (2) and applying Hess's Law.

To do this, we need to manipulate the given equations so that the desired reaction (3) can be obtained.

First, we reverse reaction (1) to get the formation of C2H4(g) from C2H6(g):

C2H4(g)C2H6(g) ΔH° = -52.3 kJ

Next, we multiply reaction (2) by 2 and reverse it to obtain 2 moles of C2H6(g) reacting to form 3 moles of H2(g):

2C2H6(g)2C(s) + 3H2(g) ΔH° = 169.4 kJ

Now, we add the two modified equations together:

C2H4(g)C2H6(g) ΔH° = -52.3 kJ
2C2H6(g)2C(s) + 3H2(g) ΔH° = 169.4 kJ

When adding these equations, the C2H6(g) on the left side cancels out with the C2H6(g) on the right side, leaving us with the desired reaction (3):

C2H4(g) + H2(g)C2H6(g) ΔH° = -52.3 kJ + 169.4 kJ = 117.1 kJ

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Using  hypothetical scenario, the Separation of  the Pennie’s are given below

Pennies dated after 1982 are lighter and thinner because zinc is less dense than copper. However, they have a similar diameter of 19.05mm to pennies dated before 1982. The difference in composition between the two groups is important because copper is a valuable metal, and the cost of producing a penny exceeded its value. By reducing the amount of copper in pennies, the U.S. Mint was able to save money on production costs.

Pennies dated before 1982 are made of 95% copper and 5% zinc and weigh 3.11 grams. They have a diameter of 19.05mm and a thickness of 1.55mm. Pennies dated after 1982 are made of 97.5% zinc and 2.5% copper and weigh 2.5 grams. They have a diameter of 19.05mm and a thickness of 1.52mm.

Therefore, the main difference between the two groups is their composition, weight, and thickness. The pennies dated before 1982 are larger, heavier, and thicker than those dated after 1982.

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Answer:2

Explanation:

Answer:

1 ppm :)

Explanation: i was guessing and 1 ppm was the answer :)

Hazard signs such as "Flammable," "Oxidizer," and "Corrosive" National Fire Protection Association (NFPA) diamond Gas Cylinder sign posted signs convey information about chemical storage.

Pictograms are pictorial symbols that are used to express particular information about a chemical's risks. OSHA mandates pictograms on principal labels to indicate chemical dangers. The exact OSHA danger categorization determines each pictogram(s). There are several forms of risks in the workplace, including chemical, ergonomic, physical, and psychological hazards, to mention a few. Dangerous situations. Employees who operate with machinery or on construction sites are more likely to be exposed to safety concerns.  Biological dangers. Biological dangers are exceedingly hazardous  Physical dangers Ergonomic risks Chemical risks Workload risks.

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It is found that a pH of 8 and a temperature of 37°C were most effective for the protease's maximum caseinolytic activity.

With plant leaf extracts at various stages of purification (crude soup is the initial supernatant after homogenization and centrifugation, 40% ammonium soup is the phosphate-dissolved pellet after 40% ammonium sulphate fractionation, and pooled soup is the final collection of pure fractions came from DEAE cellulose column), we have tested the protease activity in terms of caseinolytic activity. Prior to the protease experiment, all three samples were dialyzed to remove EDTA and protease inhibitor. Pure protein's protease activity was tested at various pH values (4–9) and temperatures (4–70°C). Different amounts of -casein were used as substrate for the enzyme activity assay for protease, which was carried out at pH-8 in 37°C according to the ideal conditions established by the preceding studies (ideal temperature and pH).

Here, the enzyme concentration was held constant while the substrate concentration (-casein) changed in the range (0.81, 1.6, 2.4, 4.03, 5.2) mg/ml.

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The balanced chemical equation for the reaction between copper (II) nitrate and sodium hydroxide, Cu(NO3)2(aq) + 2NaOH(aq) → Cu(OH)2(s) + 2NaNO3(aq)

Chemical equations make use of symbols to represent factors such as the direction of the reaction and the physical states of the reacting entities. Chemical equations were first formulated by the French chemist Jean Beguin in the year 1615

In this reaction, copper (II) nitrate (Cu(NO3)2) reacts with sodium hydroxide (NaOH) to form solid copper (II) hydroxide (Cu(OH)2) and soluble sodium nitrate (NaNO3). The (aq) label indicates that the species is in aqueous solution, while the (s) label indicates that the species is a solid.

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Answer:

The molarity is 0.4 \(\frac{moles}{liter}\)

Explanation:

The Molarity (M) or Molar Concentration is a concentration unit that indicates the number of moles of the solute per liter of solution. In other words, molarity is defined as the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by the expression:

\(Molarity (M)=\frac{number of moles of solute}{volume}\)

Molarity is expressed in units (\(\frac{moles}{liter}\)).

In this case:

Replacing:

\(Molarity (M)=\frac{0.2 moles}{0.5 L}\)

Molarity= 0.4 \(\frac{moles}{liter}\)

The molarity is 0.4 \(\frac{moles}{liter}\)

The given representation of one unit of C₆H₁₂O₆ in water is incomplete as it does not include the water molecules that are essential for the dissolution process.

In the given representation, only the C₆H₁₂O₆ molecules are shown, while the water molecules are intentionally not depicted. However, when C₆H₁₂O₆  dissolves in water, it forms a solution where C₆H₁₂O₆ molecules are surrounded by water molecules, resulting in a hydrated state.

Therefore, the representation is incomplete and inaccurate since it neglects the presence of water molecules, which play a crucial role in the dissolution and formation of a hydrated C₆H₁₂O₆ complex in water.

The question should be:

A representation of one unit of C₆H₁₂O₆ in water is shown below. (The water 12 molecules are intentionally not shown.)

(a) What is wrong with this representation?

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Sample of helium gas initially at 37.0 c, 785 torr and 2 l was heated to 58.0c while the volume expanded to 3.24 l .The final pressure of the helium gas is 8.66 atm.

Helium is a chemical element with the symbol He and atomic number 2. It is the second lightest and second most abundant element in the universe. Helium is a colourless, odourless, tasteless, non-toxic, inert monatomic gas. It is the most common element in the universe, making up about 24% of its mass.

The ideal gas law can be used to solve this problem. The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature.We can rearrange the equation to P = (nRT)/V.

We know the values of n, R, V, and T. n is 1 mol of helium, R is 0.0821 atm∙L/mol∙K, V is 3.24 L, and T is 58.0 °C, which is equal to 331.15 K.

Plugging these values into the equation, we get:

P = \((1 mol*0.0821 atm∙L/mol∙K*331.15 K)/3.24 L = 8.66 atm\)

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Answer:

the  specific heat capacity of the sample is 1011.05 J /kg k

Explanation:

The computation of the  specific heat capacity of the sample is shown below:

Given that

Q=mcΔθ

Now Heat gain by the calorimeter is

= (0.200)×  (4180) × (26.1 - 19) + (0.15) × (390) ×(26.1 - 19)

=6 350.95J

Now

6350.95=(0.0850) × c ×(100-26.1)

c= 6350.95 ÷ ((0.085) × (100 - 26.1))

= 1011.05 J /kg k

Hence, the  specific heat capacity of the sample is 1011.05 J /kg k

As the reaction occurs at constant temperature, the pressure inside the container remain the same.

Pressure is defined as the force was applied perpendicular to an object's surface per unit area across which that force is dispersed. 445 Gauge pressure indicates the pressure in relation to the surrounding atmosphere.

Pressure is expressed using a variety of units. Some of these stem from a unit of forces exerted by an area unit. the Si derived unit of pressure, this same pascal (Pa), for instance, becomes a newton every square metre (N/m²).

P×V = n×R×T

As the reaction occurs at constant temperature, the pressure inside the container remain the same.

Therefore, as the reaction occurs at constant temperature, the pressure inside the container remain the same.

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Answer:

Density (ρ) = 5 kilogram/cubic meter

Explanation:

Steps:

ρ =  

m

V

=  

10 kilogram

2 cubic meter

=  5 kilogram/cubic meter

Answer:

68.81ml

Explanation:

Using the formula:

CaVa = CbVb

Where;

Ca = concentration of acid (M)

Cb = concentration of base (M)

Va = volume of acid (ml)

Vb = volume of base (ml)

Based on the information given in this question;

Ca = 0.545M

Cb = 0.150M

Va = 250ml

Vb = ?

Using CaVa = CbVb

0.545/250 = 0.150/Vb

Cross multiply

250 × 0.150 = 0.545Vb

37.5 = 0.545Vb

Vb = 37.5/0.545

Vb = 68.81ml

Answer:

5,596.8

Explanation:

159×2.2= 349.8×16= 5,596.8

The equation for cellular respiration is :

Glucose + oxygen ⇒ water + energy + carbon dioxide

HOW CELLULAR RESPIRATION WORKS:

The process through which living things receive the energy (ATP) required for metabolic activity is called cellular respiration.

It entails the breakdown of sugar molecules (with or without oxygen) to release carbon dioxide and produce energy in the form of ATP (CO2).

The reactants of cellular respiration, or the substances that are used for the beginning of the chemical process, are sugar (glucose) and oxygen. The products, or the results of cellular respiration, are water, carbon dioxide, and ATP energy. Thus it is the answer.

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For the specified reversible reaction, the equilibrium constant expression is:

Kc = ([A]a × [B]b) / ([C]c × [D]d)

The ratio of the product concentrations (C and D) increased to their respective stoichiometric coefficients (C and d) divided by the reactant concentrations (A and B) elevated to their respective stoichiometric coefficients (A and B) is represented by this expression.

A species' population density is indicated by square brackets ([]). The exponents in the expression for the equilibrium constant are determined by the stoichiometric coefficients (a, b, c, and d) in the balanced equation.

The equilibrium position for the specified reaction is quantified by the equilibrium constant (Kc). It is clear that the equilibrium favors the products (C and D) if Kc is bigger than 1. If Kc is less than 1, it means that the reactants (A and B) are favored in the equilibrium. When Kc is equal to 1, the reactants and products are present at equilibrium in about equal amounts.

It's important to note that the expression [A]×[B] represents the product of the concentrations of A and B, but it does not accurately describe the equilibrium constant expression for the given system. The correct expression is Kc = C × [D]d / ([A]a × [B]b).

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--The question is incomplete, the given complete question is:

"Consider the general reversible reaction.

aA + bB

C + dD

What is the equilibrium constant expression for the given system?

0

c[C]d[D]

a[A]b[B]

[C][D]

[A][B]

0

[A][B]

0

[C] [D]

[C]°[

DO

0

[A]×[B]"--

Since the breakdown of hydrogen peroxide can cause a reduction in concentration and potency over time, it is a good idea to check the concentration of hydrogen peroxide that has been exposed to air and light.

The chemical student employed titration with potassium permanganate to measure the quantity of hydrogen peroxide.

Using this technique, the volume of potassium permanganate solution needed to completely react with the hydrogen peroxide sample is measured after adding a known quantity of a standard potassium permanganate solution to the hydrogen peroxide sample until the reaction is complete.

Hydrogen peroxide and potassium permanganate react in the following way: 5 H2O2 + 2 KMnO4 + 3 H2SO4 5 O2 + 2 MnSO4 + K2SO4 + 8 H2O

The amount of hydrogen peroxide in the sample may be determined from the volume and concentration of the potassium permanganate solution used in the titration because the reaction uses 5 moles of hydrogen peroxide for every 2 moles of potassium permanganate.

The chemical student should carry out the following procedures to examine the hydrogen peroxide samples in dark and light vials that have been exposed to air for 10, 20, and 30 hours:

1. Make a standard potassium permanganate solution with a known concentration.

2. Pour a known volume of each test sample of hydrogen peroxide into a flask.

3. To serve as a catalyst for the reaction, add a tiny quantity of diluted sulfuric acid to the hydrogen peroxide sample.

4. Continue titrating the hydrogen peroxide sample with the potassium permanganate standard solution until the reaction is finished, which is shown by a lingering pink hue of the potassium permanganate solution.

5. To confirm the correctness of the results, repeat the titration using the same sample.

6. Using the volume and concentration of the potassium permanganate solution used in the titration, determine the concentration of hydrogen peroxide in the sample.

The chemical student can find out if the exposure to air and light has changed the concentration of hydrogen peroxide over time by analysing the concentration of hydrogen peroxide in the light and dark vials after 10, 20, and 30 hours.

The findings may be calculated by comparing them to the original concentration of 3% by weight.

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One of the most commonly known catalysed reaction is that of Hydrogen Peroxide turning into Water

In this reaction, we can use Potassium Permanganate (KMnO₄) or Palladium (Pd) as a catalyst.

These catalysts helps Hydrogen Peroxide turn into water and oxygen

Catalyst: Substances that change the rate of a chemical reaction and themselves remain chemically unchanged after the reaction are called catalysts.

Example:- In the process of manufacturing of ammonia Fe is used as Catalyst.

\(N _{2} (g)+ 3H _{2} (g) \huge \rightarrow {}^{Fe}\rightarrow \small{3NH _{3}(g)}\)

Note:- there is only one arrow. I was unable to put iron over the arrow ..