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The chemical formula for lithium acetate is [tex]LiC_2H_3O_2[/tex]. It consists of one lithium ion [tex](Li^+)[/tex] and one acetate ion [tex](C_2H_3O_2^-)[/tex].

Lithium acetate is composed of lithium ions [tex](Li^+)[/tex] and acetate ions [tex](C_2H_3O_2^-)[/tex].

The formula indicates that for each lithium ion, there is one acetate ion. In chemical formulas, the subscripts indicate the number of atoms of each element present in the compound.

In this case, "Li" represents lithium, "C" represents carbon, "H" represents hydrogen, and "O" represents oxygen.

So, the formula [tex]LiC_2H_3O_2[/tex] indicates that there is one lithium atom, two carbon atoms, three hydrogen atoms, and two oxygen atoms in lithium acetate.

Answer:

[tex]2.82\cdot 10^9 y[/tex]

Explanation:

A radioactive isotope is an isotope that undergoes nuclear decay, breaking apart into a smaller nucleus and emitting radiation during the process.

The half-life of an isotope is the amount of time it takes for a certain quantity of a radioactive isotope to halve.

For a radioactive isotope, the amount of substance left after a certain time t is:

[tex]m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}[/tex] (1)

where

[tex]m_0[/tex] is the mass of the substance at time t = 0

m(t) is the mass of the substance at time t

[tex]\tau[/tex] is the half-life of the isotope

In this problem, the isotope is uranium-235, which has a half-life of

[tex]\tau=7.04\cdot 10^8 y[/tex]

We also know that the amount of uranium left in the rock sample is 6.25% of its original value, this means that

[tex]\frac{m(t)}{m_0}=\frac{6.25}{100}[/tex]

Substituting into (1) and solving for t, we can find how much time has passed:

[tex]t=-\tau log_2 (\frac{m(t)}{m_0})=-(7.04\cdot 10^8) log_2 (\frac{6.25}{100})=2.82\cdot 10^9 y[/tex]

Answer: 2500 ml

Explanation:

To dilute a 500 mL solution of 5 M HCl to a 1 M HCl solution, 2000 mL of water must be added, using the formula M1V1 = M2V2 with the values substituted respectively.

The question involves diluting a concentrated solution to a lower concentration, specifically diluting a 500 mL solution of 5 M HCl to obtain a 1 M HCl solution. To solve this, the dilution equation, M1V1 = M2V2, is used. In this equation, M represents molarity, and V represents volume. Initially, we have M1=5 M, V1=500 mL, and M2=1 M. We need to find V2, the final volume after dilution.

Therefore, to dilute the 500 mL solution of 5 M HCl to a 1 M solution, 2000 mL of water needs to be added.

The mass of water that should be added to 22.0 g of KCl to make a 5.50% by mass solution is 488.9 g.

To calculate the mass of water to be added, we can use the concept of mass percent. The mass percent is the mass of the solute divided by the mass of the solution, multiplied by 100%. We can set up the following equation:
Mass of KCl / (Mass of KCl + Mass of Water) = 5.50%
Let's solve for the mass of water:
Mass of Water = Mass of KCl / (0.055 - 1)
Mass of Water = 22.0 g / 0.045
Mass of Water = 488.9 g
Therefore,

488.9 g of water

should be added to 22.0 g of KCl to make a 5.50% by mass solution.

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Temperature change would be 112.6° C.

Explanation:

We can find the amount or heat absorbed or emitted during any reaction by finding the product of their mass, specific heat, and change in temperature of the metal.

Mass of the iron, m = 50.0 g

Amount of heat absorbed, q = 2500 J

Change in temperature, ΔT = ?

Specific heat of Iron, C = 0.444 J/g °C

[tex]\boldsymbol{q}=\boldsymbol{m} \times \boldsymbol{C} \times \boldsymbol{\Delta} \mathbf{T}[/tex]

Plugin the values and rearrange the equation to get the change in temperature as,

[tex]\Delta \mathbf{T}=\frac{\mathbf{q}}{c \times m}[/tex]

[tex]\Delta \mathrm{T}=\frac{2500 \mathrm{J}}{0.444 \frac{J}{\mathrm{g}^{\circ} \mathrm{C}} \times 50 \mathrm{g}}=112.6^{\circ} \mathrm{C}[/tex]

The temperature of the iron sample would change by approximately [tex]\( 111.11 \, ^\circ \text{C} \)[/tex] as a result of absorbing 2500 J of thermal energy.

To determine the change in temperature of the iron sample, we need to use the specific heat capacity formula [tex]\[ q = m \cdot c \cdot \Delta T \][/tex]

where:

[tex]- \( q \)[/tex] is the amount of heat absorbed or released (in joules),

[tex]- \( m \)[/tex] is the mass of the substance (in grams),

[tex]- \( c \)[/tex] is the specific heat capacity of the substance (in joules per gram per degree Celsius),

[tex]- \( \Delta T \)[/tex] is the change in temperature (in degrees Celsius).

For iron, the specific heat capacity [tex]\( c \)[/tex] is approximately [tex]\( 0.450 \, \text{J/g}^\circ \text{C} \)[/tex].

Given:

[tex]- \( q = 2500 \, \text{J} \), - \( m = 50.0 \, \text{g} \), - \( c = 0.450 \, \text{J/g}^\circ \text{C} \).[/tex]

We can rearrange the formula to solve for [tex]\( \Delta T \)[/tex]:

[tex]\[ \Delta T = \frac{q}{m \cdot c} \][/tex]

Now, plug in the values:

[tex]\[ \Delta T = \frac{2500 \, \text{J}}{50.0 \, \text{g} \cdot 0.450 \, \text{J/g}^\circ \text{C}} \] \[ \Delta T = \frac{2500}{50.0 \times 0.450} \] \[ \Delta T = \frac{2500}{22.5} \] \[ \Delta T \approx 111.11 \, ^\circ \text{C} \][/tex]

The answer is: [tex]111.11 \, ^\circ \text{C}[/tex].

Answer:

Valence Electrons are transferred/exchanged

In ionic bonds, valence electrons are transferred from one atom to another, leading to the formation of oppositely charged ions that attract each other to create the bond.

In ionic bonds, valence electrons are transferred from one atom to another. This process involves the movement of electrons from an atom that has a relatively low number of electrons in its valence shell (often a metal) to an atom with a higher affinity for electrons (often a nonmetal), which has a relatively high number of electrons in its valence shell or a few electrons short of completing its shell. Through this transfer, the atom losing electrons becomes a positively charged ion (cation), and the atom gaining electrons becomes a negatively charged ion (anion). This electrostatic attraction between the oppositely charged ions forms the ionic bond. For example, in the formation of sodium chloride (NaCl), sodium (Na) loses one electron to become Na⁺, while chlorine (Cl) gains that electron to become Cl⁻, resulting in a stable ionic compound.

The complete question is:

Fill in the blanks:

In ionic bonds, valence electrons are _______.

Answer:

8L

Explanation:

Using Boyle's law which states that the volume of a given mass of gas is inversely proportional to the pressure, provided temperature remains constant

P1V1= P2V2

P1 =  2atm, V1 = 12L ,

P2 = 3atm , V2 =

12 × 2 = V2 × 3

Divide both sides by 3

V2 = 24 ÷ 3

V2 = 8L

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1 mole of lead has the greatest mass because its molar mass of approximately 207.2 g/mol is higher than that of silver (107.87 g/mol), copper (63.55 g/mol), and tungsten (183.84 g/mol), and all samples contain Avogadro's number of atoms.

To determine which substance has the greatest mass, we need to compare the molar masses of silver, copper, lead, and tungsten. The molar mass is the mass in grams of one mole of a substance, and it is numerically equivalent to the atomic mass for elements. Let's examine the atomic masses of these elements:

Given that 1 mole of each substance contains the same number of atoms (6.022 × 10²23 atoms, Avogadro's number), the substance with the highest molar mass will have the greatest mass. In this case, lead has the highest molar mass at approximately 207.2 g/mol, and therefore 1 mole of lead will have the greatest mass compared to 1 mole of silver, copper, or tungsten.

Final answer:

Lead (Pb) has the greatest mass.

Explanation:

The substance with the greatest mass in this case would be lead (Pb).

To determine which substance has the greatest mass, we need to compare their molar masses. The molar mass of silver (Ag) is approximately 107.87 g/mol, copper (Cu) is approximately 63.55 g/mol, lead (Pb) is approximately 207.2 g/mol, and tungsten (W) is approximately 183.84 g/mol.

Since lead (Pb) has the highest molar mass of the four substances, it has the greatest mass.

The density of ethanol is 0.78 g/mL and the density of benzene is 8.8 g/mL. Benzene is denser than ethanol.

The density of a substance is calculated by dividing its mass by its volume. To find the density of ethanol, divide its mass (3.9g) by its volume (5.0 mL). The density of ethanol is therefore 0.78 g/mL. To find the density of benzene, divide its mass (44g) by its volume (5.0 mL). The density of benzene is 8.8 g/mL. Since the density of benzene is higher than the density of ethanol, benzene is denser than ethanol.

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The density of ethanol is 0.78 g/mL and the density of benzene is 8.8 g/mL. Benzene is denser than ethanol.

To find the density of each liquid, the formula for density will be used:

[tex]\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \][/tex]

Given:

Ethanol:

Volume [tex](\(V\))[/tex] = 5.0 mL

Mass [tex](\(m\))[/tex] = 3.9 g

Benzene:

Volume [tex](\(V\))[/tex] = 5.0 mL

Mass [tex](\(m\))[/tex]  = 44 g

Density of Ethanol:

[tex]\[ \text{Density of Ethanol} = \frac{3.9 \, \text{g}}{5.0 \, \text{mL}} \][/tex]

[tex]\[ \text{Density of Ethanol} = 0.78 \, \text{g/mL} \][/tex]

Density of Benzene:

[tex]\[ \text{Density of Benzene} = \frac{44 \, \text{g}}{5.0 \, \text{mL}} \][/tex]

[tex]\[ \text{Density of Benzene} = 8.8 \, \text{g/mL} \][/tex]

Answer:

20m

Explanation:

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Answer:

The equation is balanced

Explanation:

Before a equation can be said to be balanced the number of mole at the reaction side must be equal to number mole at the product side

Let's start with Zn at the reactant side 4 moles of Zn reacts and 4 moles of Zn was obtained from the product side which makes it balanced

Let's take HNO3

For H,10 moles react and 10 moles was obtained in the product side

For N,10 moles react and 8 moles was obtained initially and 2 moles was obtained as well which makes it 10 moles and it makes it balanced

For O,30 moles react and after calculating the total amount of O at the product side it was observed to be 30 which indicates that the equation is balanced...

The equation is absolutely well balanced

Answer:

The right field is oceanography

Answer:

The new volume will be 42, 7 L.

Explanation:

We use the gas formula, which results from the combination of the Boyle, Charles and Gay-Lussac laws. According to which at a constant mass, temperature, pressure and volume vary, keeping constant PV / T. The conditions STP are: 1 atm of pressure and 273 K of temperature.

P1xV1/T1 =P2xV2/T2

1 atmx 22,4 L/273K = 0,5atmx V2/260K

V2=((1 atmx 22,4 L/273K )x 260K)/0,5 atm= 42, 67L

Answer:

NaClo3 is the reactant here.

Explanation:

In the reaction below:

                        [tex]H_{2} + O_{2}[/tex] ----------------------> 2[tex]H_{2} O[/tex]

Here       [tex]H_{2}[/tex] and [tex]O_{2}[/tex] are reactant, they take part in chemical reaction to form water as a product.

To calculate the total heat absorbed by an 18g ice cube going from -10°C to water at 20°C, you sum the heat needed to raise the temperature of the ice to 0°C, the heat of fusion to melt the ice, and the heat needed to raise the temperature of the resulting water to 20°C.

The student is asking about the heat absorption process when an ice cube melts and warms up to ambient temperature. To find the total heat absorbed, you would calculate the heat needed to warm the ice from -10°C to 0°C, the heat required for the phase change from ice to water at 0°C, and then the heat needed to warm the water from 0°C to 20°C.

First, calculate the heat to warm the ice (Q1) using the formula Q = mcΔT, where m is the mass, c is the specific heat of ice, and ΔT is the temperature change. Then, calculate the heat to melt the ice (Q2) using the molar heat of fusion, given that the molar mass of water is approximately 18 g/mol. Finally, calculate the heat to warm the water (Q3) using the formula Q = mcΔT again, but this time with the specific heat of water.

The total heat (Qtotal) absorbed is the sum of Q1, Q2, and Q3. Note that in this example, the mass of ice and molar mass of water allows us to use a direct conversion from grams to moles for the heat of fusion calculation.

Answer:

There are 70 grams of KOH

Explanation:

First, we calculate the weight of 1 mol of KOH:

Weight 1 mol KOH: Weight K + Weight 0 + Weight H=  39g+ 1g+ 16g= 56 g/mol

1 mol-------56 g KOH

1,25 mol----x= (1,25 molx56 g KOH)/1 mol= 70 g KOH

The correct option is C. Some of the absorbed energy is converted to potential energy and some is converted to kinetic energy

Explanation:

Thus by considering this above explanation it is clear that Option C can be consider as the opt answer

Answer:

aluminum

Explanation:

i took the test

Answer:

Explanation:

Any species that undergoes oxidation is itself a reducing agent or reductant.

The species that undergoes reduction in the course of a chemical reaction is an oxidant or reducing agent.

Any specie that loses electrons to another undergoes oxidation and is a reductant.

Species that gains electrons are called oxidant.

Answer

cation

Explanation:

If an atom losses electrons it forms a positively charged ion called the cation.

cation has more protons than electrons, thereby giving it a net positive charge.

To form a cation one or more electrons must be lost, typically pulled away by atoms with a stronger affinity for them.

The number of electrons lost, and so the charge of the ion, is indicated after the chemical symbol, e.g. zinc (Zn) loses two electrons to become Zn2+.

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Answer:

Empirical formula is CH₂O.

Molecular formula = C₆H₁₂O₆

Explanation:

Given data:

Mass of hydrogen = 3.36 g

Mass of carbon = 20.00 g

Mass of oxygen = 26.64 g

Molar mass of compound = 180.156 g/mol

Empirical formula = ?

Molecular formula = ?

Solution:

Empirical formula:

It is the simplest formula gives the ratio of atoms of different elements in small whole number

Number of gram atoms of H = 3.36 / 1.01 = 3.3

Number of gram atoms of O = 26.64 / 16 = 1.7

Number of gram atoms of C = 20 / 12 = 1.7

Atomic ratio:

            C                      :        H            :         O

           1.7/1.7                :     3.3/1.7       :       1.7/1.7

              1                     :           2          :        1

C : H : O = 1 : 2 : 1

Empirical formula is CH₂O.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

Empirical formula mass = CH₂O = 12×1 + 2× + 16

Empirical formula mass = 30

n = 180.156 / 30

n = 6

Molecular formula = n (empirical formula)

Molecular formula = 6 (CH₂O)

Molecular formula = C₆H₁₂O₆

Answer:

147.473 g/mol

Explanation:

According to the Arrhenius definition, Ca(OH)₂ is a Arrhenius base.

According to the Arrhenius theory, acids are those substance which gives H⁺ ion in the aqueous solution and bases are those substances which gives OH⁻ ion in the aqueous solution.

Chemical equation for the dissociation of Calcium hydroxide is:

Ca(OH)₂ → Ca²⁺ + 2OH⁻

From the above equation it is clear that Ca(OH)₂ is an Arrhenius base as it gives hydroxide ion.

Hence, Ca(OH)₂ is a Arrhenius base.

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Calcium hydroxide, Ca(OH)₂, is a base according to the Arrhenius definition, as it produces hydroxide ions (OH-) when it dissolves in water. The equation representing this is Ca(OH)₂ (aq) → Ca²⁺ (aq) + 2OH- (aq). It is considered a strong base because it dissociates completely in water.

According to the Arrhenius definition, a base is a substance that releases hydroxyl ions (OH-) when it is dissolved in water. Calcium hydroxide, Ca(OH)₂, is considered to be a base because it dissociates in water to form calcium ions (Ca²⁺) and hydroxide ions (OH-). The chemical equation representing this process is: Ca(OH)₂ (aq) → Ca²⁺ (aq) + 2OH- (aq).

This depicts that calcium hydroxide is a base since it generates hydroxide ions (OH-) upon dissolution in water, which aligns with the countenance of a base as defined by Arrhenius. Because it dissociates completely in water solutions, it is acknowledged as a strong base.

The hydroxide ions (OH-) released into the solution are capable of combining with hydrogen ions (H+) to create water molecules. This reduces the solution's acidity and the concentration of free H+ ions, further signifying the behavior of a base.

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Final answer:

The missing coefficient in the reaction Cl2O3 + H2O → HCIO3 is 2 in front of HCIO3, resulting in a balanced equation Cl2O3 + H2O → 2 HCIO3, which is a combination reaction.

Explanation:

The missing coefficient in the chemical equation Cl2O3 + H2O → HCIO3 should balance the number of atoms of each element on both sides of the equation. In this case, to balance chlorine and oxygen atoms, we should place a coefficient of 2 in front of HCIO3, resulting in Cl2O3 + H2O → 2 HCIO3. This is an example of a combination reaction, where two or more substances combine to form a single product. Below are several examples of different reaction types for context:

a. Combustion reaction: C6H5CH3 + 9O2 → 7CO2 + 4H2Ob. Decomposition reaction: 2NaHCO3 → Na2CO3 + H2O + CO2c. Synthesis reaction: C + 2H2 → CH4

By integrating two or even more materials whose properties are quite distinctive is composite materials. The different materials work together to build special properties for the composite, but within the composite one can easily tell the various materials apart that they don't dissolve or mix together.

It is also recognized as Fiber-Reinforced Polymer (FRP) composites, which are constructed from a polymer matrix strengthened with engineered, man-made or natural fiber like glass, steel or aramid or other strengthening material.

Magnesium and calcium are more reactive with oxygen in the air than aluminum.

Magnesium and calcium are more reactive with oxygen in the air than aluminum.

Magnesium reacts with oxygen to form magnesium oxide (MgO), while calcium reacts with oxygen to form calcium oxide (CaO). Aluminum also reacts with oxygen, but it forms a thin layer of aluminum oxide (Al2O3) on its surface, which acts as a protective barrier to further reaction. This layer prevents aluminum from reacting as readily with oxygen compared to magnesium and calcium. Therefore, magnesium and calcium are more reactive with oxygen in the air than aluminum.

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Answer:

0.45M

Explanation:

The following were obtained from the question:

C1 = 0.75M

V1 = 100mL

V2 = 165mL

C2 =?

Applying the dilution formula C1V1 = C2V2, the concentration of the diluted solution can be calculated for as follows:

0.75 x 100 = C2 x 165

Divide both side by 165

C2 = (0.75 x 100) /165

C2 = 0.45M

The concentration of the diluted solution is 0.45M

Answer:

74.09 atm

Explanation:

Using the gas laws ( Charles and Boyle's law). We have the formula ,

P1/T1 = P2/T2

Where P1 = 30.3atm

T1 = -100 degree Celsius

to kelvin = -100+ 273 = 173K

T2 = 150 degree Celsius

To Kelvin = 150 = 150+273 = 423K

Imputing values

P1/T1 = P2/T2

30.3/173 = P2/ 423

Cross multiply

173×P2 = 30.3 ×423

173P2 = 12816.9

Divide both sides by 173

P2 = 12816.9/173

P2 = 74.09 atm

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Answer:

P2=74.086atm

Explanation:

Using the formula p1/t1=P2/t2

P1=30.3atm

P2=?

T1=-100°c to Kelvin=273-100=173k

T2=150°c to Kelvin=273+150=423k

30.3\173=x/423

Cross multiply

173x=423×30.3

173x=12816.9

Divide both sides by 173

X=74.086atm

Answer:

(A) Removed and solid

Explanation:

The new volume is 38 ml.

Explanation:

At NTP,

P1  = 760 mm,

P2 = 800 mm,

V1 = 40 ml

Boyle's law states that when a gas is held at a constant temperature and mass in a closed container, the volume and pressure vary inversely.

By applying Boyle's law,

                       P1  V 1  = P2  V2

where P1, P2 represents the pressure of a gas,

           V1, V2 represents the volume of a gas.

                        P1  V 1  = P2  V2

​              760×40 = 800×V  2

                          V 2   =  (760×40)  / 800

                           V2  = 38 mL.

The new volume of the gas at the increased pressure, while maintaining the same temperature, is 38 ml.

The correct formula to use in this situation is Boyle's Law, which states that for a given mass of gas at constant temperature, the pressure of the gas is inversely proportional to its volume. Mathematically, this can be expressed as:

[tex]\[ P_1V_1 = P_2V_2 \][/tex]

where [tex]\( P_1 \)[/tex] and [tex]\( V_1 \)[/tex] are the initial pressure and volume, and [tex]\( P_2 \)[/tex] and [tex]\( V_2 \)[/tex] are the final pressure and volume.

Given:

- Initial pressure, [tex]\( P_1 = 760 \)[/tex] mm Hg

- Final pressure, [tex]\( P_2 = 800 \)[/tex] mm Hg

- Initial volume, [tex]\( V_1 = 40 \)[/tex] ml

We want to find the final volume [tex]\( V_2 \)[/tex].

Using Boyle's Law:

[tex]\[ P_1V_1 = P_2V_2 \][/tex]

[tex]\[ 760 \times 40 = 800 \times V_2 \][/tex]

[tex]\[ V_2 = \frac{760 \times 40}{800} \][/tex]

[tex]\[ V_2 = \frac{30400}{800} \][/tex]

[tex]\[ V_2 = 38 \text{ ml} \][/tex]