Calculate the specific heat capacity of the unknown metal given the
following information; the mass of the metal is 225 g, the temperature
raises by 13C and the energy absorbed was 1363 joules.

A. 0.466 j/gC
B.1.02j/gC
C.0.96 j/gC
D.2.14 j/gC

Answer:

Sulphate ion

Explanation:

Zinc will undergoes oxidation. Zn2+ - - > Zn + 2e-

Copper will undergoes reduction. Cu2+ + 2e- - - > Cu

Answer:

c) 9.03 x 10^23

Explanation:

find the molar mass of Al

Al is 27.0 grams

Then use that, to find the number of moles in Aluminum.

Then use Avogadro's number which is 6.02 * 10^23

After that, write all of that down with dimensional analysis.

40.5 g * 1 mol/ 27.0 g of Al  * 6.02 x 10^23 / 1mol

As your final answer, you will get 9.03 * 10^23 atoms with sig figs.

Hope it helped!

Hygiene factors are necessary to avoid dissatisfaction, while motivators are essential for promoting job satisfaction and motivation in employees.

Herzberg's two-factor theory proposes two key elements of motivation: hygiene factors and motivators.

1. Hygiene factors: Hygiene factors, such as salary, working conditions, and job security, are work environment-related factors that can cause job dissatisfaction if lacking. They are necessary to prevent dissatisfaction but do not directly motivate employees. Hygiene factors alone may not drive motivation, but their absence or inadequacy can lead to dissatisfaction.

2. Motivators: Motivators are job-related factors that directly contribute to employee motivation and satisfaction. They include opportunities for personal growth, achievement, recognition, responsibility, and advancement. Motivators drive job satisfaction, productivity, and engagement, according to Herzberg's theory.

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A cell membrane is more flexible than a brick wall. Why might many cells benefit from a flexible cell membrane because it allows them to survive in and out of the cell.

The cell is surrounded by a membrane called as cell membrane . cell membrane is a semipermeable membrane. semipermeable membrane is the type of membrane which allows solvent particles to pass through it . or some molecules or ion . Therefore , many cells benefited by cell membrane as they can pass through it.

Thus, A cell membrane is more flexible than a brick wall. Why might many cells benefit from a flexible cell membrane because it allows them to survive in and out of the cell.

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The empirical formula of the compound made up of 70.0% iron and 30.1% oxygen is Fe₂O₃, and the molecular formula can be the same as the empirical formula, Fe₂O₃.

To determine the empirical formula, we need to find the simplest whole-number ratio of the elements present in the compound. Given the percentages of iron (Fe) and oxygen (O), we can assume a 100 g sample of the compound.

The mass of iron is 70.0 g (70.0% of 100 g), and the mass of oxygen is 30.1 g (30.1% of 100 g).

Next, we calculate the moles of each element using their respective molar masses:

moles of Fe = 70.0 g / 55.85 g/mol ≈ 1.252

moles of O = 30.1 g / 16.00 g/mol ≈ 1.881

Then, we divide the number of moles of each element by the smallest number of moles to obtain the simplest whole-number ratio:

1.252 / 1.252 ≈ 1 (for Fe)

1.881 / 1.252 ≈ 1.5 (for O)

Since we need to express the empirical formula using whole numbers, we multiply the ratio by 2 to obtain the simplest ratio:

1 × 2 = 2 (for Fe)

1.5 × 2 = 3 (for O)

Therefore, the empirical formula of the compound is Fe₂O₃, and the molecular formula can be the same as the empirical formula, Fe₂O₃.

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Answer:

The energy needed is 45.7659 kJ

Explanation:

The given mass of the ice, m = 15 g

The initial temperature of the ice, T₁ = -4°C

The temperature

The final temperature, T₂ = 115°C

The specific heat capacity of ice, c₁ = 2.09 J/(g·°C)

The latent heat of ice, l₁ = 334 J/g

The heat capacity of water, c₂ = 4.184 J/(g·°C)

The latent heat of vaporization, l₂ = 2260 J/g

The specific heat of steam, c₃ = 2.02 J/g

Therefore, we get the energy needed, ΔQ, as follows;

ΔQ = m·c₁·(T₂ - Tₐ) + m·l₁ + m·c₂(Tₓ - Tₐ) + m·l₂ + m·c₃·(T₂ - Tₓ)

Where;

Tₐ = The melting point temperature of water = 0°C

Tₓ = The boiling point temperature of water = 100°C

∴ ΔQ = 15×2.09×(0 - (-4)) + 15 × 334 + 15×4.184×(100 - 0) + 15×2260 + 15×2.02×(115 - 100) = 45,765.9

The energy needed, ΔQ = 45,765.9 J = 45.7659 kJ.

The number of atoms in 0.77 mole of Kr is 4.64×10²³ atoms.

From Avogadro's hypothesis,

Thus,

1 mole of Kr = 6.02×10²³ atoms

With the above information in mind, we can obtain the number of atoms in 0.77 mole of Kr as follow:

1 mole of Kr = 6.02×10²³ atoms

Therefore,

0.77 mole of Kr = 0.77 × 6.02×10²³  

Therefore, 0.77 mole of Kr contains 4.64×10²³ atoms.

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This statement is true "A catalyst increases the rate of a reaction without being consumed. It accomplishes this by providing another mechanism that has a lower activation energy".

A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy. Activation energy is the energy required for a chemical reaction to occur. By lowering the activation energy, a catalyst allows the reaction to proceed more readily and at a faster rate.

The catalyst itself remains unchanged at the end of the reaction, meaning it is not consumed or permanently altered by the reaction. It can participate in multiple reaction cycles, continuously facilitating the conversion of reactants into products.

The role of a catalyst is to provide an alternative reaction pathway by forming an intermediate compound with the reactants. This intermediate compound is typically more reactive and has a lower energy barrier compared to the original pathway. As a result, more reactant molecules can overcome the lower activation energy and effectively undergo the reaction, leading to an increased reaction rate.

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Answer:

C. amps

Explanation:

HOPE IT HELPS U...

Answer:

they spread out waiting until each orbital has one electron before doubling up (the last bullet point)

Explanation:

This is because electrons are negatively charged so if they doubled up together in the begging, they would repel each other.  

Answer:

The answer is D.-2

Explanation:

CO2+4H2 -> CH4 +2H2O

An equal number of atoms of each element in both sides of the chemical equation is required to have an equation in a balanced state.

There are a few possible reasons why no observable change took place after adding 10.00 mL of NaOH to the KHP solution:

The KHP may not have completely dissolved in water: It is important that the KHP is completely dissolved in water before adding NaOH. If the KHP did not dissolve completely, the reaction between KHP and NaOH will not occur, and no change will be observed.

The NaOH solution may not have been standardized properly: If the NaOH solution was not standardized properly, its true concentration may be different from the expected concentration. This can lead to inaccurate results and can explain why no change was observed.

The indicator used may not be appropriate for this reaction: The choice of indicator is important in acid-base titrations. The indicator should change color around the equivalence point of the titration. If the indicator used is not appropriate, it may not change color even if the equivalence point has been reached.

The titration may not have been performed accurately: The titration should be performed carefully and accurately to ensure that the correct amount of NaOH is added to the KHP solution. If the titration was not performed accurately, it may not have reached the equivalence point, and no change would be observed.

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There are 0.25 moles of CO₂ (g) in 11.0 grams of CO2 at STP.

Assuming the CO₂ gas in the cylinder is at STP, the total number of moles of gas is calculated as follows:

STP (standard temperature and pressure) is defined as 0°C (273.15 K) and 1 atmosphere (1 atm) of pressure. At STP, 1 mole of any ideal gas occupies a volume of 22.4 liters.

The molar mass of CO₂ is approximately 44.01 g/mol.

To determine the number of moles of CO2 in 11.0 grams at STP, we can use the following formula:

moles = mass / molar mass

Substituting the values we have:

moles = 11.0 g / 44.01 g/mol = 0.25 mol

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Answer:

B

Explanation:

There was time for the gas to escape from the rock bubbling at the surface

The hole in igneous rocks indicate that the magma cooled too quickly for gas to escape.

Igneous rocks are one of the three main rock types the other two including sedimentary and metamorphic rocks.These rocks are formed by cooling and solidification of lava or magma.

The magma is derived from partial melts in the planet's mantle or crust.Melting is caused by increase in temperature or decrease in pressure or even change in composition.Solidification occurs either below surface which is called as intrusive rocks or on the surface as extrusive rocks.

They are classified according to their texture,mode of occurrence,chemical composition,geometry and mineralogy of rock.The two important variables for classification are mineral composition and cooling history.

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The ways to identify the components necessary for a combustion reaction include:

This is referred to as burning and involves high-temperature exothermic redox chemical reaction.

This type of reaction involves the use of a gas such as oxygen which supports it and other flammable substances.

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A): The closest orbital to the nucleus, called the 1s orbital, can hold up to two electrons. This orbital is equivalent to the innermost electron shell of the Bohr model of the atom. It is called the 1s orbital because it is spherical around the nucleus. The 1s orbital is always filled before any other orbital.

Answer:

3750 cm.

Explanation:

You multiply the three side measurements to find the volume.

25cm·10cm·15cm

375cm·10cm

3750 cm.

Hope this helps!

Answer:3750cm

Explanation: 25cm by 10cm is 250cm. Then 250cm by 15cm is 3750cm. To find volume, first you multiply length by width, then the answer of that you multiply by height.

The answer is true, confectioners sugar can be substituted for granulated sugar but it may not always be an exact 1:1 substitution.

Confectioners sugar (also known as powdered sugar or icing sugar) can be substituted for granulated sugar in some recipes, but it may not always be an exact 1:1 ratio substitution.

Confectioners sugar is a very fine powder that contains cornstarch, which is added to prevent caking. Due to its fine texture, it can dissolve quickly in recipes and can add a slightly different texture to baked goods. It can also affect the sweetness level of the recipe as it usually contains cornstarch or other additives, which can also influence the texture of the final product.

In general, substituting confectioners sugar for granulated sugar works best in recipes that don't require the sugar to dissolve completely, such as frosting or glazes. For recipes that require the sugar to dissolve, such as in baking or making meringues, it's best to use granulated sugar.

So, the answer is "True", but the substitution may not be appropriate for all recipes.

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7.89 x 10^24 formula units of copper (II) chloride (CuCl2) weigh more than 7.65 moles of sulfuric acid (H2SO4).

In order to determine the weight of given chemicals, the molar-mass conversion is used which is given by:

Number of moles = mass/molar mass

Where Molar mass = Atomic mass of element x number of atoms

For 7.65 moles of sulfuric acid (H2SO4)

Molar mass of H2SO4 = 2 x 1.0079 g/mol + 1 x 32.06 g/mol + 4 x 16.00 g/mol = 98.0785 g/mol

Hence, Mass = 7.65 x 98.0785 = 750.3 g or 0.75kg

For 7.89 x 10^24 formula units of copper (II) chloride (CuCl2)

The formula unit is converted to moles and mass is determined. A formula unit refers to the chemical formula of an ionic compound that lists the ions in the lowest ratio that equals a neutral electrical charge. The formula units of any compound are determined by multiplying the moles of the given compound by the Avogadro number. Hence, formula unit is converted to moles by dividing the formula unit of the chemical by Avogadro number i.e., 6.022 × 10^23, hence 7.89 x 10^24/6.022 × 10^23 = 13.1.

Molar mass of CuCl2 is = 1 x 63.546 + 2 x 35.453 = 134.452 g/mole

Hence, Mass = 13.1 x 134.452 = 1761.3 g or 1.76kg

Based on the mass, 7.89 x 10^24 formula units of copper (II) chloride (CuCl2) weigh more than 7.65 moles of sulfuric acid (H2SO4).

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The followings are the standard notation of the given numbers.

(a) \(5.55 * 10^3 = 5550\)

(b) \(1.0070 * 10^4 = 10070\)

(c) \(8.85 * 10^{-7} = 0.000000885\)

(d) \(3.004 * 10^{-3} = 0.003004\)

Standard notation, commonly referred to as standard form or scientific notation, is a means of employing powers of 10 to represent extremely big or extremely small values. A number is expressed using standard notation by combining a power of 10 and a decimal number that is more than or equal to 1 but less than 10.

In standard notation, the number 3,500,000, for instance, can be expressed as \(3.5 * 10^6\), where \(10^6\) denotes the quantity of zeros in the initial value (6 zeros in this case). Similar to how the number 0.0000023 can be expressed in standard notation as\(2.3 * 10^{-6\), where -6 denotes the quantity of zeros added to the original value after the decimal point (6 zeros in this case).

Now, in the given question,

(a) \(5.55 * 10^3 = 5.55*1000 = 5550\)

(b) \(1.0070 * 10^4 = 1.0070*10000 = 10070\)

(c) \(8.85 * 10^{-7} = \frac{8.85}{10000000} = 0.000000885\)

(d) \(3.004 * 10^{-3} = \frac{3.004}{1000} = 0.003004\)

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The comparison between the two groups of pennies reveals distinct differences in their composition, appearance, and weight.

Upon separating the pennies into two groups based on their dates, I observed notable differences between the two groups. Pennies dated before 1982 are primarily composed of copper, while those dated after 1982 are made of zinc with a thin copper plating.

In terms of appearance, the pre-1982 pennies have a reddish-brown color due to their high copper content. They often show signs of aging, such as discoloration, tarnish, and wear. In contrast, the post-1982 pennies have a brighter and shinier appearance, resembling a silver-like hue due to the copper coating.

In terms of size, both groups of pennies have the same diameter and thickness. However, the pre-1982 pennies tend to be slightly heavier due to the higher density of copper compared to zinc, which is used in post-1982 pennies.

Pennies dated before 1982 are made of copper, have a reddish-brown color, and are slightly heavier. Pennies dated after 1982, on the other hand, are made of zinc with a copper coating, appear brighter and more silver-like, and are slightly lighter.

These differences arose from the change in materials used by the U.S. Mint in 1982 to reduce production costs.

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Answer:

B) protons and neutrons are shared by two atoms so as to satisfy the requirements of both atoms

Hydrogen peroxide (H2O2) is a pure substance with two atoms, exhibiting multiple oxidation numbers and bonded through charge attraction.

One example of a pure substance with a chemical formula that consists of two atoms and exhibits multiple oxidation numbers is hydrogen peroxide (H2O2).

Hydrogen peroxide is composed of two hydrogen atoms and two oxygen atoms. The oxygen atoms in hydrogen peroxide can have different oxidation states, namely -1 and -2, depending on the reaction conditions.

In hydrogen peroxide, the oxygen atoms have a partial negative charge, while the hydrogen atoms possess a partial positive charge. This electrostatic attraction between the positive and negative charges holds the atoms together.

The oxygen atoms, due to their higher electronegativity, tend to attract electrons more strongly, leading to the formation of peroxide bonds.

Hydrogen peroxide demonstrates a range of redox reactions, which involve the transfer of electrons. It can act as both an oxidizing and reducing agent.

For example, in acidic conditions, hydrogen peroxide can be reduced to water while oxidizing another substance. Conversely, in alkaline conditions, it can be oxidized while reducing another compound.

In summary, hydrogen peroxide is a pure substance with a chemical formula containing two atoms, with the oxygen atoms displaying different oxidation numbers and bonded together through positive/negative charge attraction.

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Answer:

Here's what I get  

Explanation:

(i) Voltaic cell

A voltaic cell is a device that uses a chemical reaction to produce electrical energy.

(ii) Overall Cell Potential

The standard reduction potentials for the half-reactions are

                               ℰ°/V

Cu²⁺ + 2e⁻ ⇌ Cu    0.34

Zn²⁺ + 2e⁻ ⇌ Zn   -0.76

The half-reaction with the more positive potential is the reduction half-reaction. It is the reaction that occurs at the cathode.

The half-reaction with the more negative potential is the oxidation half-reaction. It is the  reaction that occurs at the anode.

We reverse that half-reaction and subtract the voltages to get the cell reaction.

                                                          ℰ°/V

Cathode:  Cu²⁺ + 2e⁻ ⇌ Cu              0.34

Anode:     Zn ⇌ Zn²⁺ + 2e⁻              -0.76

Cell:          Zn +  Cu²⁺ ⇌ Zn²⁺ + Cu    1.10

\(\mathcal{E}_{\text{cell}}^{\circ} = \mathcal{E}_{\text{cat}}^{\circ} - \mathcal{E}_{\text{an}}^{\circ} = \text{0.34 V} - \text{(-0.76 V)} = \text{0.34 V} + \text{0.76 V} = \textbf{1.10 V}\)

(iii) Diagram

The specific labels will depend on your textbook.

They are often as follows.

a. Electron flow

b. Voltmeter or lightbulb

c. Electron flow

d. Cathode or Cu

e. Cu²⁺(aq) and NO₃⁻(aq)

f. Salt bridge

g. Zn²⁺(aq) and NO₃⁻(aq)

h. Anode or Zn

The salt bridge enables ions to flow in the internal circuit and to maintain electrical neutrality in the two compartments.

It often consists of a saturated solution of KCl.

As Zn²⁺ ions form in the anode compartment, Cl⁻ ions move in to provide partners for them.

As Cu²⁺ ions are removed from the cathode compartment, K⁺ ions move in to replace them.

Answer:a. Electron flow

b. Voltmeter or lightbulb

c. Electron flow

d. Cathode or Cu

e. Cu²⁺(aq) and NO₃⁻(aq)

f. Salt bridge

g. Zn²⁺(aq) and NO₃⁻(aq)

h. Anode or Zn

Explanation:

Explanation:

is used for making ropes, used for climbing rocks and for making parachutes. Their usage shows that nylon fibres have high tensile strength