Which metal(s) is most likely to accommodate Hatoms (radius 37.0 pm) with the least distortion to the crystalline lattice, given their atomic radii? Choose one or more: titanium (atomic radii - 147 pm) zirconium (atomic radii - 160 pm) hafnium (atomic radii - 159 pm)

The mass of 2.30 x 10²⁴ particles of water (H₂O) is 60.84 grams.

The amount of matter in a particle or object is represented by its mass, which is denoted by the symbol m. In the International System (SI), the kilograms serves as the default unit of mass (kg).

The formula N = n L describes the relationship between a substance's number of particles and number of moles.

n also defined using the equation n = m/M

N = n L

N = m/M L

Then substitute N and L values,

2.30×10²⁴ = m/M × 6.02×10²³

Molar mass of water = 18

2.30×10²⁴ = m/18  × 6.02×10²³

2.30×10²⁴/6.02×10²³ = m/18

3.38 = m/18

m = 3.38 × 18

m = 60.84grams.

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150 mL of 0.100 M NaOH is needed to completely neutralize 50.0 mL of 0.300 M HCl. To determine the amount of NaOH needed to neutralize the HCl, we must first balance the chemical equation. HCl + NaOH → NaCl + H2O. The balanced equation tells us that one mole of NaOH reacts with one mole of HCl.

We can use the formula M1V1 = M2V2 to calculate the amount of NaOH needed. First, we determine the number of moles of HCl present in 50.0 mL of 0.300 M HCl:

0.300 mol/L x 0.0500 L = 0.0150 moles HCl

Since one mole of NaOH is needed to neutralize one mole of HCl, we need 0.0150 moles of NaOH.

Now, we can use the concentration of the NaOH solution to calculate the volume needed:

0.100 mol/L x V = 0.0150 moles

V = 0.0150 moles / 0.100 mol/L = 0.150 L = 150 mL

Therefore, 150 mL of 0.100 M NaOH is needed to completely neutralize 50.0 mL of 0.300 M HCl.

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Answer: 2.04

Explanation:

Answer:

Approximately \(0.821\; \rm mol\).

Explanation:

Look up the relative atomic mass of \(\rm Ba\), \(\rm O\), and \(\rm H\) on a modern periodic table:

Calculate the formula mass of \({\rm Ba(OH)_2}\):

\(\begin{aligned}& M({\rm Ba(OH)_2}) \\ &= 137.327 + 2\times(15.999 + 1.008) \\ &\approx 171.334\; \rm g \cdot mol^{-1}\end{aligned}\).

Calculate the number of moles of \({\rm Ba(OH)_2}\) formula units in that \(235\; \rm g\) of this compound:

\(\begin{aligned}& n({\rm Ba(OH)_2}) \\ &= \frac{m({\rm Ba(OH)_2})}{M({\rm Ba(OH)_2})} \\ &= \frac{235\; \rm g}{171.334\; \rm g \cdot mol^{-1}} \approx 1.37159\; \rm mol \end{aligned}\).

Calculate the volume of a \(c({\rm Ba(OH)_2}) = 1.67\; \rm mol \cdot L^{-1}\) with approximately \(n({\rm Ba(OH)_2}) = 1.37159\; \rm mol\) of the solute:

\(\begin{aligned}& V({\rm Ba(OH)_2}) \\ &= \frac{n({\rm Ba(OH)_2})}{c({\rm Ba(OH)_2})} \\ &= \frac{1.37159\; \rm mol}{1.67\; \rm mol \cdot L^{-1}} \approx 0.821\; \rm L \end{aligned}\).

Answer:

5.2 x 10^8

Explanation:

Answer:

5.2x10^8

Explanation:

hope this helped

Answer:

who do I submit my questions to there is no submit or done sign I download this app for help not to waste my money and time. I need help with math please don't let us start off on the wrong foot

The equilibrium for the hydration of alkenes strongly favors the reverse reaction due to thermodynamic factors.

When an alkene is hydrated, an alcohol is formed by adding water across the double bond. This reaction is exothermic and releases energy. As a result, the entropy of the system decreases since fewer molecules are present in the products than in the reactants.

In contrast, the reverse reaction is endothermic and requires energy to proceed. This results in an increase in the entropy of the system since more molecules are present in the products than in the reactants.

Therefore, the equilibrium for the hydration of alkenes strongly favors the reverse reaction due to the thermodynamic factors of energy and entropy.

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The actual formula for volume for a cube is the length multiplied by the width and then multiplied by the height. Since all three measurements are the same, the formula results in the measurement of one side cubed. For the example, 5^3 is 125 cm^3. Multiply the volume by the known density, which is the mass per volume.

The first step in this calculation is to know how many liters is equal to 1 gallon, and the value is 3.785 liters, so now we have to make the following calculation:

1 gal = 3.785 Liters

30.0 gal = x Liters

x = 3.785 * 30.0

x = 114 Liters

The wavelength of the neutron is 4.65 n. Wavelength is typically measured in units of meters (m), centimeters (cm), or nanometers (nm).

What is Wavelength?

Wavelength is the distance between successive crests, troughs, or any other equivalent points in a wave. It is usually denoted by the Greek letter lambda (λ) and is commonly used to describe electromagnetic waves, such as light, as well as other types of waves, including sound and water waves.

Since the wavelength of a neutron is given by λ = h/p, where h is Planck's constant and p is the momentum of the neutron, we can calculate the wavelength using the equation λ = h/mv, where m is the mass of the neutron, v is its velocity, and h is Planck's constant.

First, we need to convert the velocity from m/s to cm/s, since the units of h are cm^2 g/s:

85.0 m/s × (100 cm/1 m) = 8500 cm/s

Next, we can plug in the values for m, v, and h:

λ = h/mv = (6.626 × 10^-27 cm^2 g/s)/(1.67493 × 10^-24 g × 8500 cm/s)

λ = 4.65 nm

Therefore, the wavelength of the neutron is 4.65 nm.

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Answer:

radiation B species x because short canines are less likely to break.

Answer:

18 * 10^19 atoms

Explanation:

We must first convert 57.8 mg to grams.

If 1000 mg = 1g

  57.8 mg = 57.8/1000 = 57.8 * 10^-3 g

Now;

If 1 gold atom has a mass of 3.27X10^-22 grams

x gold atoms have a mass of 57.8 * 10^-3 g

x = 57.8 * 10^-3 g/3.27X10^-22 g

x = 18 * 10^19 atoms

According to the electronic configuration, Li,C, F belong to same period as they have 2 shells .Among F,S, P it is neither same period or group and for O,S,Se they belong to the same group as they have 6 valence electrons.

Electronic configuration is defined as the distribution of electrons which are present in an atom or molecule in atomic or molecular orbitals.It describes how each electron moves independently in an orbital.

Knowledge of electronic configuration is necessary for understanding the structure of periodic table.It helps in understanding the chemical properties of elements.

Elements undergo chemical reactions in order to achieve stability. Main group elements obey the octet rule in their electronic configuration while the transition elements follow the 18 electron rule. Noble elements have valence shell complete in ground state and hence are said to be stable.

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Answer:

b

Explanation:

water is generally transparent and smooth to the touch unless it is frozen, then it is rough

Explanation:

n=,4.5 moles C2H6

the ratio 2:7

4.5:x

x=15.75 moles

When table salt (sodium chloride which ionizes into Na and Cl) is added to alginate, a gel does not form and spherification does not occur. This happens because the salt only has one positive charge that neutralizes the negative charge in the alginate.

There are various types of Spherification. Spherification is the creation of small spheres with a thin film on the surface and a liquid center. The process of spherification is mostly used in molecular gastronomy to make small, flavorful balls of liquid ingredients that burst in the mouth when bitten into. The method involves a process of encapsulating liquid droplets in a sphere made of a gel-like film. This process requires sodium alginate (E401), a gel-forming ingredient that thickens the liquids.

Sodium alginate gelation occurs as a result of the mixture of an alginate solution with a cation solution that causes the solution to gel. The sodium ions present in the solution swap with calcium ions present in the cation solution, causing a gel to form. This occurs as a result of a chemical reaction known as cross-linking. When table salt is added to the alginate solution, a gel does not form and spherification does not occur since the salt only has one positive charge that neutralizes the negative charge in the alginate. Alginate requires a doubly charged cation to cross-link.

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The moles of water vapor in the mixture are 0.00165 mol.

To find the moles of water vapor in the mixture, we need to consider the total pressure of the gases and the vapor pressure of water.

The total pressure of the gases (P_total) is given as 749.0 mmHg, and the vapor pressure of water (P_water) is given as 21.5 mmHg.

The pressure exerted by the water vapor in the mixture (P_vapor) can be calculated by subtracting the vapor pressure from the total pressure:

P_vapor = P_total - P_water

= 749.0 mmHg - 21.5 mmHg

= 727.5 mmHg

Now, we can use the ideal gas law to calculate the moles of water vapor (n_vapor). The ideal gas law equation is:

PV = nRT

Where:

P is the pressure (in atm or mmHg),

V is the volume (in liters),

n is the number of moles,

R is the ideal gas constant (0.0821 L·atm/(mol·K)),

T is the temperature (in Kelvin).

Since we are given the pressure (P_vapor), volume is not specified, and temperature is assumed to be constant, we can simplify the equation to:

n_vapor = P_vapor / (RT)

To use this equation, we need to convert the pressure from mmHg to atm and the temperature to Kelvin. Assuming the temperature is known and constant, let's use 298 K.

Converting pressure to atm:

P_vapor = 727.5 mmHg * (1 atm / 760 mmHg)

= 0.957 atm

Now we can calculate the moles of water vapor:

n_vapor = 0.957 atm / (0.0821 L·atm/(mol·K) * 298 K)

≈ 0.00165 mol

Therefore, the moles of water vapor in the mixture are approximately 0.00165 mol.

The moles of water vapor in the mixture are approximately 0.00165 mol.

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0.00170mol of H_(2) was collected over water. If the total pressure of the gases was 749.0mmHg and the vapor pressure was 21.5mmHg, find the moles of water vapor in the mixture.

Option C ;(NH4)2SO4 is the formula for ammonium sulfate.

It is an inorganic salt. It appears as fine white crystals. Its molecular weight is 132.14g/mol. Its density is 1.77g/cm³.It has many commercial uses. It is mainly used as fertilizer for alkaline soils because ammonium ions forms small amounts of acid which lowers the pH of the soil.

It is acidic with pH of 5.5. Its melting point is 235 °C. it is soluble in water. It is obtained by the reaction of sulfuric acid with ammonia. Sulfuric acid will react with two equivalents of ammonia.

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Answer:

C. (NH4)2SO4

And the second question is

A. KOH

Explanation:

Answer:liters

Explanation:

Answer: 10=14 1=t6

Explanation:

The binding energy per nucleon for 141Ba is approximately 1.350 x 10^-13 J/nucleon.

To calculate the binding energy per nucleon for 141Ba, we'll follow the steps outlined earlier:

Given:

Nuclear mass of 141Ba = 140.883 amu

Using the atomic masses from the periodic table:

Mass of proton = 1.007276 amu

Mass of neutron = 1.008665 amu

Number of protons (Z) = 56 (atomic number of barium)

Number of neutrons (N) = (140.883 - 56) = 84

Step 1: Calculate the mass defect (Δm):

Sum of proton masses = 56 x 1.007276 amu

Sum of neutron masses = 84 x 1.008665 amu

Δm = (Sum of proton masses + Sum of neutron masses) - Actual nuclear mass

Step 2: Calculate the binding energy (BE):

Binding Energy (BE) = Δm x c^2

where c = 2.998 × 10^8 m/s

Step 3: Calculate the binding energy per nucleon:

Binding Energy per Nucleon = BE / (Z + N)

Performing the calculations:

Sum of proton masses = 56 x 1.007276 amu = 56.446656 amu

Sum of neutron masses = 84 x 1.008665 amu = 84.68586 amu

Δm = (56.446656 amu + 84.68586 amu) - 140.883 amu = 0.249516 amu

BE = (0.249516 amu) x (2.998 × 10^8 m/s)^2 = 2.23899 x 10^-11 J

Binding Energy per Nucleon = (2.23899 x 10^-11 J) / (56 + 84) = 1.350 x 10^-13 J/nucleon.

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Answer:

M = (6.0 M x 67.5 mL) / 450 mL

M = 0.9 M

Answer:

I CANT

Explanation:

Answer:

phisical i think

Explanation:

chemical would be baking powder in a pancake, the pancake would be in turn more brown cause of baking powder mixing with flower.

Considering the law of multiple proportions, they obey the law of multiple proportion.

The law of multiple proportions or Dalton's law establishes that if two elements form more than one compound, establishing the composition of one of them, the other element will be in the ratio of natural numbers (integers and simple).

In other words, when two elements form more than one compound, the masses of one of them that combine with the same mass of the other element are in a simple relationship, expressible by a quotient of whole numbers.

In this case, two samples A and B of chloride of gold contains 15.1% and 35.1% chloride respectively.

This is, A contains 15.1% of Cl and  84.9% of Au . So:

\(\frac{84.9}{15.1}= 5.6\)

This indicates that there is 5.6 g of Au per gram of Cl.

And B contains 35.1% of Cl and  64.9% of Au . So:

\(\frac{64.9}{35.1}= 1.8\)

This indicates that there is 1.8 g of Au per gram of Cl.

Then, the relationship can be expressed as:

\(\frac{5.6}{1.8}=\frac{3}{1}\)

Since the composition of the atoms in the compounds are in simple ratio to each other, they obey the law of multiple proportion.

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The 90% confidence interval for the difference in the incidence of hail between spraying and not spraying the clouds is (-0.141, 0.241).

The experiment aimed to determine if spraying chemicals on clouds for artificial rainfall could reduce hail occurrence. The data collected included the number of days when chemicals were sprayed on the clouds and the number of days when no intervention was done.

Out of the observed cloudy days with hail, 7 days had spraying while 43 days had no spraying. On the other hand, out of the observed cloudy days without hail, 43 days had spraying and 122 days had no spraying.

To calculate the 90% confidence interval for the difference in hail incidence, we use a formula that takes into account the sample sizes and proportions. After performing the calculations, the resulting confidence interval is (-0.141, 0.241). This means that there is not enough evidence to conclude a significant difference in hail occurrence between spraying and not spraying the clouds, as the confidence interval contains zero.

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Answer:

The concentration of the solution is 2 g/dm³

Explanation:

The equation used to find the concentration of a solution is "c = m/v"

C= concentration

M= mass

V= volume

The question asks for the answer to be in decimeters, so you need to 100 cm³ to dm³. To do this, you divide 100 by 10, giving you 10 cm³. Then you just plug in your numbers into the formula: c = 20g/10dm³.

The concentration of the potassium chloride solution has been \(\rm 2\;\times\;10^-^3\;g/dm^3\).

The concentration has been defined as the mass of sample present in a definite volume of the solution. The concentration has been given by the moles, molarity, g/ml, g/L etc.

The concentration attained by the student has been by dissolving 20 g of KCl in \(\rm 100\;cm^3\) water has been given in \(\rm g/cm^3\) as:

\(\rm \dfrac{g}{cm^3}=\dfrac{20\;g}{100\;cm^3}\\ \dfrac{g}{cm^3}=0.2\;g/cm^3\)

The concentration of the solution obtained by the student has been \(\rm 0.2\;g/cm^3\).

The conversion of \(\rm cm^3\;to\;dm^3\) has been given as:

\(\rm 1\;g/cm^3=0.001\;g/dm^3\)

The concentration of \(\rm 0.2\;g/cm^3\) solution has been given as:

\(\rm 1\;g/cm^3=0.001\;g/dm^3\\0.2\;g/cm^3=0.2\;\times\;0.001\;g/dm^3\\0.2\;g/cm^3=2\;\times\;10^-^3\;g/dm^3\)

The concentration of the potassium chloride solution has been \(\rm 2\;\times\;10^-^3\;g/dm^3\).

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Option (d), P=ATC=MR=MC, accurately represents the long-run equilibrium condition for perfect competition, reflecting the balance between price and cost for firms operating in a competitive market.

The long-run equilibrium condition for perfect competition is that price (P) is equal to average total cost (ATC), which is also equal to marginal cost (MC), and marginal revenue (MR).

Option (d), P=ATC=MR=MC, best represents the long-run equilibrium condition for perfect competition. In perfect competition, firms operate at the minimum point of their average total cost curve, where price equals both average total cost and marginal cost. This condition ensures that firms are earning zero economic profit and are producing at an efficient level.

In the long run, if firms are earning economic profit, new firms will enter the market, increasing competition and driving prices down. Conversely, if firms are experiencing losses, some firms may exit the market, reducing competition and causing prices to rise. This process continues until firms reach a state where price equals average total cost, marginal cost, and marginal revenue, ensuring a long-run equilibrium.

Therefore, option (d), P=ATC=MR=MC, accurately represents the long-run equilibrium condition for perfect competition, reflecting the balance between price and cost for firms operating in a competitive market.

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