2Na (s) + 2H₂O (1) → 2NaOH (aq) + H₂ (g) AH = -367.5 kJ Calculate mass of Na(s) produced during an enthalpy change +100 kJ ​

Without additional information about the specific scenario of the gas particles, it is difficult to determine the best description of their kinetic energy before and after a collision.

What is Kinetic Energy?

It is a scalar quantity that is proportional to the mass of the object and the square of its velocity. The formula for kinetic energy is:

Kinetic energy = 0.5 x mass x velocity^2

where m is the mass of the object, and v is its velocity.

The kinetic energy of two gas particles before and after a collision depends on the mass, velocity, and direction of motion of the particles. If the particles are moving at the same speed and in the same direction before and after the collision, their kinetic energy will remain the same. However, if the particles have different masses or velocities, or if they collide at an angle or with a certain degree of elasticity, their kinetic energy may change as a result of the collision.

Therefore, without additional information about the specific scenario of the gas particles, it is difficult to determine the best description of their kinetic energy before and after a collision.

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The structures of D-ribose and D-arabinose are epimers as the configuration of only one asymmetric carbon is different. Therefore, option (D) is correct.

In stereochemistry, an Epimer can be specified as one of a pair of stereoisomers. Two isomers present in the molecule differ at the stereogenic center while the rest remain identical.

A molecule contains many stereocenters leading to different stereoisomers. Stereoisomers are molecules that contain the same constitution and molecular formula in the three-dimensional orientations of their atoms in space.

D-ribose and D-arabinose epimers create a single difference at the C-2  asymmetric carbon atom. Therefore, the given two molecules are epimers.

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Answer:

Bacteria can use sexual and asexual repoduction

Explanation:

Answer:

KK+reacts with water NaNa+reacts with water LiLi+reacts with water

At equilibrium, the number of moles of Cl2(g) present is approximately 347.37 mol.

To determine the number of moles of Cl2(g) at equilibrium, we need to use the given equilibrium constant (Kc) and set up an ICE table to track the changes in the reactants and products.

The balanced equation for the reaction is:

CO(g) + Cl2(g) ⇌ COCl2(g)

Let's set up the ICE table:

  CO(g)     +     Cl2(g)     ⇌     COCl2(g)

Initial: 0.3500 0.05500 0

Change: -x -x +x

Equilibrium: 0.3500 - x 0.05500 - x x

Using the equilibrium concentrations in the ICE table, we can write the expression for the equilibrium constant (Kc) as:

Kc = [COCl2(g)] / [CO(g)][Cl2(g)]

Substituting the values into the equation, we have:

1.2 × 10^3 = (0.05500 - x) / [(0.3500 - x)(0.05500 - x)]

Simplifying the equation, we can cross-multiply and rearrange:

1.2 × 10^3 × (0.3500 - x)(0.05500 - x) = 0.05500 - x

Expanding and rearranging, we get:

0 = (1.2 × 10^3 × 0.05500 - 1.2 × 10^3x + 0.05500x) - x

Simplifying further:

0 = 66 - 1.245x + 0.05500x - x

0 = 66 - 0.19x

0.19x = 66

x = 66 / 0.19

x ≈ 347.37

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Ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)) mixture could be a useful buffer in a solution. Option C

A buffer is a solution that can resist changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base or a weak base and its conjugate acid. The buffer system works by the principle of Le Chatelier's principle, where the equilibrium is shifted to counteract the changes caused by the addition of an acid or a base.

In option A, acetic acid (\(CH_3CO_2H\)) is a weak acid, but hydrochloric acid (HCl) is a strong acid. This combination does not form a buffer because HCl is completely dissociated in water and cannot provide a significant concentration of its conjugate base.

Option B consists of sodium hydroxide (NaOH), which is a strong base, and elemental sodium (Na), which is a metal. This combination does not form a buffer as there is no weak acid-base pair involved.

Option D contains acetic acid (\(CH_3CO_2H\)), a weak acid, and ammonia (\(NH_3\)), a weak base. Although they are weak acid and base, they do not form a buffer system together as they are both weak acids or bases and lack the required conjugate acid-base pair.

Option C, ammonia (\(NH_3\)), is a weak base, and ammonium chloride (\(NH_4Cl\)) is its conjugate acid. This combination can form a buffer system. When ammonia reacts with water, it forms ammonium ions (NH4+) and hydroxide ions (OH-).

The ammonium ions act as the weak acid, while the ammonia acts as the weak base. The addition of a small amount of acid will be counteracted by the ammonium ions, and the addition of a small amount of base will be counteracted by the ammonia, thus maintaining the pH of the solution relatively stable.

Therefore, option C, consisting of ammonia (\(NH_3\)) and ammonium chloride (\(NH_4Cl\)), is the suitable mixture that could be a useful buffer in a solution.

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There are 3.44 x 10^{22} particles in 0.057 moles of lithium bromide.

The lithium bromide formula also known as the lithium monobromide formula or Bromo lithium formula is explored. It is a counterion bromide-based salt of lithium.

we have to use Avogadro's constant,

Avogadro's constant, is approximately equal to 6.022 x 10^{22} particles per mole.

we can use the following formula:

number of particles = moles x Avogadro's constant

Substitute the values,

number of particles = 0.057 moles x 6.022 x 10^{23} particles/mol

Simplifying the equation

number of particles = 3.44 x 10^{22} particles

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Answer:

def CO2

Explanation:

Answer:

the emperical formula if histame is

Explanation:

C5H9N3

The candle with a 750 mL beaker as the cover would burn longer than the rest.

The burning of the candle requires oxygen and the volume of oxygen each beaker can hold depends on the volume of the beakers.

Thus, a beaker of 750 mL capacity will hold more oxygen than beakers of 250 mL or 150 mL capacity.

Since the higher volume beaker has the capacity to hold more oxygen than the rest, it means that the candle will be able to burn longer in it than the rest.

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Answer:

200.0lg

Explanation:

please give a brainliest

Answer:

methane

Explanation: methane is obtained from the decaying of flora and fauna mostlyunder damp

The final temperature of the combined metals is 34.9°C.

Specific heat is the amount of heat energy required to raise the temperature of one unit of mass of a substance by one degree Celsius (or one Kelvin). It is a physical property that is unique to each substance and is often denoted by the symbol "c."

To solve this problem, we can use the principle of heat transfer, which states that when two objects with different temperatures come into contact, up until they reach the same temperature, heat will transfer from the hotter to the cooler object.

We can start by using the formula for heat transfer:

Q = m * c * ΔT

where Q is the amount of heat transferred, m is the mass of the object, c is the specific heat capacity of the material, and ΔT is the change in temperature.

Let's assume that the final temperature of the combined metals is T. We can set up two equations based on the heat transferred from the gold and the heat transferred to the iron:

Heat transferred from the gold: Q1 = (11.1 g) * (0.129 J/g°C) * (T - 15.7°C)

Heat transferred to the iron: Q2 = (18.7 g) * (0.449 J/g°C) * (T - 52.6°C)

Since the total amount of heat transferred must be equal, we can set Q1 = Q2:

(11.1 g) * (0.129 J/g°C) * (T - 15.7°C) = (18.7 g) * (0.449 J/g°C) * (T - 52.6°C)

Simplifying and solving for T, we get:

T = [ (11.1 g) * (0.129 J/g°C) * 15.7°C + (18.7 g) * (0.449 J/g°C) * 52.6°C ] / [ (11.1 g) * (0.129 J/g°C) + (18.7 g) * (0.449 J/g°C) ]

T = 34.9°C

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Answer:

1.45 x 10⁻² g CO₂

Explanation:

To find the mass of carbon dioxide, you need to (1) convert grams CH₄ to moles CH₄ (via molar mass), then (2) convert moles CH₄ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). The final answer should have 3 sig figs to reflect the given value (5.30 x 10⁻³ g).

Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)

Molar Mass (CH₄): 16.043 g/mol

Combustion of Methane:

1 CH₄ + 2 O₂ ---> 2 H₂O + 1 CO₂

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

5.30 x 10⁻³ g CH₄          1 mole              1 mole CO₂          44.007 g
---------------------------  x  ----------------  x  ---------------------  x  -----------------  =
                                      16.043 g            1 mole CH₄            1 mole

=  0.0145 g CO₂

=  1.45 x 10⁻² g CO₂

Answer:

10.66  grams

Explanation:

Answer:

CoSO₄.7H₂O

Explanation:

Some salts are in its hydrate form to stabilize them. The hydrated form of CoSO₄ is:

CoSO₄.XH₂O

To find the hydration of the COSO₄ you must find the ratio of H₂O and CoSO₄ that is:

0.068921 moles H₂O / 0.0098459 moles CoSO₄ = 7

That means you have 7 moles of water per mol of CoSO₄ and the formula is:

Answer:

H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)

Explanation:

H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq) + SO²⁻₄(aq) + H₂O(l)

A careful observation of the equation above, shows that the equation is already balanced.

To obtain the net ionic equation, we simply cancel Mg²⁺ from both side of the equation as shown below:

H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)

Answer: The equation is \(_{29}^{63}\textrm{Cu}\rightarrow _{30}^{63}\textrm{Zn}+_{-1}^0e\)

Explanation:

Neutron capture is a process where a neutron is converted into a proton and an electron. The released particle is known as beta particle and it carries a charge of -1 units and has a mass of 0 units. It is also known as an electron. The general equation for this process is:

\(_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+ _{-1}^0\e\)

The nuclear equation for the formation of zinc via neutron capture of copper follows:

\(_{29}^{63}\textrm{Cu}\rightarrow _{30}^{63}\textrm{Zn}+_{-1}^0e\)

Answer:

The volume of Component 1 is 36 cubic inches.

Explanation:

To calculate the volume of a composite solid, we need to determine the individual volumes of the different components and then add them together.

In this case, the composite solid consists of multiple components with the following dimensions:

Component 1:

Length: 4 inches

Width: 3 inches

Height: 3 inches

To find the volume of Component 1, we multiply the length, width, and height together:

Volume of Component 1 = Length x Width x Height = 4 in x 3 in x 3 in = 36 cubic inches

Therefore, the volume of Component 1 is 36 cubic inches.

Please provide the dimensions of the remaining components of the composite solid, and I will calculate the total volume by summing up the individual volumes.

Answer:

Environmental cleanup includes removal of heavy metals and other toxic contaminants from the environment and the process is called remediation.

Approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M.

To determine the mass of glucose (C6H12O6) required to prepare a 0.215 M solution in 5000 mL, we need to use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

First, let's convert the volume of the solution from milliliters (mL) to liters (L):

5000 mL = 5000/1000 = 5 L

Now, we can rearrange the formula to solve for moles of solute:

moles of solute = Molarity (M) x volume of solution (L)

moles of solute = 0.215 M x 5 Lmoles of solute = 1.075 mol

Since glucose (C6H12O6) has a molar mass of approximately 180.16 g/mol, we can calculate the mass of glucose using the equation:

mass of solute = moles of solute x molar mass of solute

mass of glucose = 1.075 mol x 180.16 g/mol

mass of glucose = 194.0 g (rounded to three significant figures)

Therefore, approximately 194.0 grams of glucose (C6H12O6) would be required to prepare a 5000 mL solution with a concentration of 0.215 M. It's important to note that the molar mass of glucose used in this calculation may vary slightly depending on the level of precision required.

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In between conductors, which are typically metals, and not-conductors or insulators, such as ceramics, exist materials known as semiconductors. Semiconductors can be pure elements like germanium or silicon or compounds like gallium arsenide.

In the given pictures, 'As' is a N-type semiconductor whereas 'Ga' is a P-type semiconductor.

When pentavalent impurities (P, As, Sb, and Bi) are added to a pure semiconductor (germanium or silicon), four of the five valence electrons form a bond with the four electrons of the pure semiconductor.

The dopant's fifth electron is liberated and used for conduction in the lattice are called N-type semiconductors.

When a trivalent impurity (B, Al, In, or Ga) is added into a pure semiconductor, three of the semiconductor's four valence electrons form a bond with the impurity's three valence electrons.

In the impurity, this results in an electron (hole) being missing called P -type semiconductors.

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Compared to 2,3,4-trifluoropentane, the nonane's carbon chains are longer.

A substance's molar mass is defined as its molecular weight in grams. By adding the molar masses of a substance's constituent atoms, we may get the substance's molar mass. Then, to convert between mass and the quantity of moles of the material, we may utilize the computed molar mass.

Adding the atomic masses of a particular substance results in the calculation of molar mass. Below each element's symbol on the periodic table is a designation of the mass of that specific element. The molar mass is obtained by averaging the atomic masses obtained from the periodic table.

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Answer:

Here are a few more examples:

Smoke and fog (Smog)

Dirt and water (Mud)

Sand, water and gravel (Cement)

Water and salt (Sea water)

Potassium nitrate, sulfur, and carbon (Gunpowder)

Oxygen and water (Sea foam)

Petroleum, hydrocarbons, and fuel additives (Gasoline)

Heterogeneous mixtures possess different properties and compositions in various parts i.e. the properties are not uniform throughout the mixture.

Examples of Heterogeneous mixtures – air, oil, and water, etc.

Examples of Homogeneous mixtures – alloys, salt, and water, alcohol in water, etc.

Explanation:

Answer:

smog,mud, cement?

Explanation:

Smoke and fog (Smog)

Dirt and water (Mud)

Sand, water and gravel (Cement)

Answer: D. baking muffins

Explanation:

A chemical change occurs once some reactants/ingredients are mixed together to create a new product.

In the case of cutting a banana = you still get a banana;

pouring milk = you still have milk;

tossing a salad = cut up veggies = just cut up veggies next to each other;

whereas baking muffins: you begin with flour, sugar, butter, etc. (individual ingredients), and then you bake them (add energy to the system), which in turn gives you a new product = muffins.

The moles of Cl⁻ ions present in 1.8 L of the intravenous solution is 0.063 moles.

The moles of Cl⁻ ions in the intravenous solution are calculated using the conversion factor between mEq/L to mmoles

The conversion between milliequivalents per Liter and millimoles per Liter is:

1 Millimole per Liter = 1 Milliequivalents per Liter.

Hence 35 mEq/L = 35 mmol/L

The mmoles of Cl⁻ ions in 1.8 liters of the intravenous solution will be:

moles of Cl⁻ ions = 35 mmol/L * 1.8 L

moles of Cl⁻ions = 63 mmoles

Converting to moles by dividing by 1000:

moles of Cl⁻ ions = 63 / 1000

moles of Cl⁻ ions = 0.063 moles.

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Answer: 455.26 mL

Explanation:

combined gas law

V2=P1V1T2/ P2T1

V2= 800 X 480 X273.15 /( 760 x 303.15) =

The energy in a single Carbon - Hydrogen (C-H) bond of methane is 413 kJ

From the question, we are to determine the Carbon - Hydrogen bond energy in methane

From the given information,

The total bond energy in a molecule of methane is 1652 kJ

C-H bond energy = \(\frac{Total\ bond\ energy\ in\ methane}{number\ of \ C-H \ bonds}\)

Number of C-H bonds in methane = 4

Therefore,

C-H bond energy = \(\frac{1652}{4}\)

C-H bond energy = 413 kJ

Hence, the energy in a single Carbon - Hydrogen (C-H) bond of methane is 413 kJ

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