Suppose you need to prepare 117.3 mL of a 0.434 M aqueous solution of NaCl. What mass, in grams, of NaCl do you need to use to make the solution

Answer: A beta particle is being emitted, not an alpha particle.

More unstable an electron configuration , the more reactive an atom will    become.

How electron configuration influences the chemical behavior of an atom?

This is happen generally, If we look at the Group 1 elements in the periodic table, they are all highly reactive as they have 1 electron in their outermost shells - an unstable configuration in terms of energy.

Also, the noble gases in Group 8 in the periodic table are 'inert' that means they don't react (or more correctly, have an incredibly low reactivity). This is because they have 8 electrons in their outermost shell and thus have no need to acquire or lose electrons to possess a stable electron configuration.

Hence, electron configuration influences the chemical behavior of an atom.

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Answer:

6FCRXCTV

Explanation:

Answer:

b

Explanation:

a p e x

When considering the electrophilic addition of Br2 to isoprene (2-methyl-1,3-butadiene), the product mixture predominantly contains 3,4-dibromo-3-methyl-1-butene (21%) over 3,4-dibromo-2-methyl-1-butene (3%) due to the following reasons:

1. Regioselectivity: During the electrophilic addition reaction, the initial attack of the electrophile (Br+) to isoprene's double bond leads to the formation of two possible carbocation intermediates.
2. Carbocation stability: The intermediate carbocation that leads to the formation of 3,4-dibromo-3-methyl-1-butene is a tertiary carbocation (3°), which is more stable than the secondary carbocation (2°) leading to 3,4-dibromo-2-methyl-1-butene.
3. Hyperconjugation: The greater of the tertiary carbocation is due to hyperconjugation, which is the interaction between the adjacent C-H σ-bonds stability with the empty p-orbital of the carbocation. The tertiary carbocation has more of these stabilizing interactions than the secondary carbocation.

In conclusion, the predominance of 3,4-dibromo-3-methyl-1-butene (21%) over 3,4-dibromo-2-methyl-1-butene (3%) in the product mixture results from the greater stability of the tertiary carbocation intermediate formed during the electrophilic addition reaction of Br2 to isoprene (2-methyl-1,3-butadiene).

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Answer: d. energy must transfer to the carbon dioxide from its surroundings.

Explanation:

Sublimation is the phase change from a solid to a gas, meaning that the particles will have more kinetic energy (since particles in the solid phase are more ordered than particles in the gas phase, and thus particles in the gas phase have more energy). Thus, for sublimation to occur, the particles need more energy, which must come from its surroundings.

Answer:

Two hydrogen atoms and one oxygen atom (water) was removed.

Explanation:

yw:))

The moles of HCl required for the reaction with 1.4g of Zn can be determined by stoichiometry and subtracting that amount from the total moles of HCl available.

The balanced equation for the reaction between zinc (Zn) and hydrochloric acid (HCl) is given as:

Zn + 2HCl → ZnCl₂ + H₂

From the balanced equation, we can see that 1 mole of Zn reacts with 2 moles of HCl. To determine the moles of HCl required for the reaction with 1.4g of Zn, we need to convert the mass of Zn to moles.

Using the molar mass of Zn (65.38 g/mol):

Moles of Zn = Mass of Zn / Molar mass of Zn

Moles of Zn = 1.4 g / 65.38 g/mol ≈ 0.0214 mol

According to the balanced equation, the mole ratio between Zn and HCl is 1:2. Therefore, 0.0214 mol of Zn would react with 2 × 0.0214 mol = 0.0428 mol of HCl.

To find the amount of HCl available, you would subtract the moles of HCl required (0.0428 mol) from the total moles of HCl available.

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Cooling the sugar cube-water mixture: Decrease in temperature decreases the kinetic energy of the water molecules, which in turn reduces their ability to interact with and dissolve the sugar molecules. Therefore, cooling the sugar cube-water mixture will decrease the rate of dissolving of the sugar cube in water.

To prepare a 2.75% (m/m) solution of sucrose, we need 2.75 grams of sucrose per 100 grams of water. Therefore, the mass of sucrose required in 375 g of water can be calculated as follows:

2.75 g of sucrose per 100 g of water

x g of sucrose per 375 g of water

Cross-multiplying, we get:

\(100x = 2.75 x 375\)

\(x = (2.75 x 375)/100\)

\(x = 10.31 g\)

Therefore, we need 10.31 grams of sucrose to prepare a 2.75% (m/m) solution in 375 grams of water.

To calculate the number of grams of sucrose present in 185 mL of a 2.50 M sucrose solution, we can use the following formula:

Molarity = moles of solute/volume of solution in liters

We can first calculate the number of moles of sucrose present in the solution as follows:

2.50 M = moles of sucrose/1 L of solution

moles of sucrose = 2.50 mol/L x 0.185 L

moles of sucrose = 0.4625 mol

The mass of sucrose can be calculated from the number of moles as follows:

mass = moles x molar mass

mass = 0.4625 mol x 342.34 g/mol

mass = 158.50 g

Therefore, 185 mL of a 2.50 M sucrose solution contains 158.50 grams of sucrose.

To prepare a 1M silver nitrate solution from 24 mL of a 3M stock solution of silver nitrate, we can use the formula:

M1V1 = M2V2

where M1 is the initial molarity (3M), V1 is the initial volume (24 mL), M2 is the final molarity (1M), and V2 is the final volume (unknown).

Rearranging the formula to solve for V2, we get:

V2 = (M1V1)/M2

V2 = (3M x 24 mL)/1M

V2 = 72 mL

Therefore, 48 mL of water should be added to 24 mL of the stock solution to prepare a 1M silver nitrate solution.

To prepare a 6.75% (m/m) solution of NaCl, we need 6.75 grams of NaCl per 100 grams of solution. Therefore, the mass of NaCl required in 100 grams of solution can be calculated as follows:

6.75 g of NaCl per 100 g of solution

x g of NaCl per 20.0 g of solution

Cross-multiplying, we get:

100x = 6.75 x 20.0

x = (6.75 x 20.0)/100

x = 1.35 g

Therefore, 1.35 grams of NaCl should be added to 18.65 grams of water to prepare a 6.75% (m/m) solution.

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The structure and function of a bat's wings and a butterfly's wings are similar.

The use of both wings—those of a bat and a butterfly—for flight is a similarity between them. From the foregoing, we can deduce that these organs are analogous because, despite having a different origin, they carry out similar tasks. Both a butterfly's and a bat's wings are used for flight, making them similar in function. Convergent evolution produced these two. A butterfly's wing and a bat's wing are similar in structure. Both have wings and can fly, but they come from different places. Bat wings are made of bones, whereas butterfly wings are made of other materials and have a very distinct structure. No, the wings of a bat and a butterfly should not be regarded as homologous organs since they have different origins and structures but serve the same purpose in flight, making them analogous organs.

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Explanation:

Density  =  mass  ÷  volume

              =    65g  ÷  (15 cm × 5 cm × 3 cm)

              =    65g ÷ 225 cm³

Density  =    0.289 g/cm³

Answer: The specific heat of given metal is \(1.195 J/^{o}C\).

Explanation:

Given: Mass of sample = 22.8 g

Heat energy = 1450 J

Initial temperature = \(21.8^{o}C\)

Final temperature = \(75^{o}C\)

Formula used to calculate the specific heat is as follows.

\(q = m \times C \times (T_{2} - T_{1})\)

where,

q = heat energy

m = mass of substance or sample

C = specific heat

\(T_{1}\) = initial temperature

\(T_{2}\) = final temperature

Substitute the values into above formula as follows.

\(q = m \times C \times (T_{2} - T_{1})\\1450 J = 22.8 g \times C \times (75 - 21.8)^{o}C\\C = \frac{1450 J}{22.8 \times 53.2^{o}C}\\= 1.195 J/^{o}C\)

Thus, we can conclude that the specific heat of given metal is \(1.195 J/^{o}C\).

ANSWER

The formula of strontium oxide is SrO

EXPLANATION

Strontium oxide is formed when strontium reacts with oxygen

Below is the reaction between strontium and oxygen

Hence, the formula of strontium oxide is SrO

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There are approximately 9.03 x 10^25 molecules of SO2 in the sample.


First, we need to determine the molecular formula of SO2. The atomic mass of sulfur (S) is 32.06 g/mol, while the atomic mass of oxygen (O) is 16.00 g/mol. So, the molecular mass of SO2 is:

molecular mass of SO2 = (atomic mass of S x 1) + (atomic mass of O x 2)
= (32.06 g/mol x 1) + (16.00 g/mol x 2)
= 64.06 g/mol

Next, we can use Avogadro's number to convert moles of SO2 to molecules. Avogadro's number is approximately 6.02 x 10^23 molecules/mol.

Number of molecules of SO2 = Number of moles of SO2 x Avogadro's number
= 150 mol x 6.02 x 10^23 molecules/mol
= 9.03 x 10^25 molecules

Therefore, there are approximately 9.03 x 10^25 molecules of SO2 in the sample.

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There are approximately 9.033 x 10²⁵ molecules of SO₂ in the sample of gas.

To find out how many molecules of SO₂ are in the sample of gas, we need to use Avogadro's number, which states that one mole of any substance contains 6.022 x 10²³ molecules.

First, we need to determine the number of moles of SO₂ in the sample. Assuming that all of the gas is composed of SO₂, we can use the mole ratio of SO₂ to total gas to find out:

150 moles total gas x (1 mole SO₂ / 1 mole total gas) = 150 moles SO₂

Next, we can use Avogadro's number to convert the number of moles of SO₂ to the number of molecules:

150 moles SO₂ x (6.022 x 10²³ molecules / 1 mole SO₂) = 9.033 x 10²⁵ molecules SO₂

Therefore, there are approximately 9.033 x 10²⁵ molecules of SO₂ in the sample of gas.

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HCl>H2S>PH3

Explanation:

Acidity is directly proportional to non metallic character. as we move in period (i.e. from left to right) non metallic character increases hence acidic strength increases.

The order of the acidic strength from strongest to weakest: HCl > H₂S > PH₃. Therefore, option (a) is correct.

The acidic strength measures the ability of the acid to lose its H⁺ ion. In general dissociation of acid can be shown as:

HA  →  H⁺ + A⁻

Some examples of strong acids are hydrochloric acid, perchloric acid  (HClO4), sulfuric acid (H2SO4), etc.

For the given acids, with the increase in the electronegativity from phosphorous to chlorine in the period. The bond pair of electrons attract more strongly by the electronegative atom.

Due to this, the weaker bond can easily donate the proton. Hence, HCl has a greater tendency to liberate proton (H⁺) than phosphine (PH₃). After dissociation,  the anion can be easily stabilized by a more electronegative Chlorine atom.  

Therefore, the order of acidic strength for the given acids is HCl> H₂S > PH₃.

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The reaction is double displacement reaction of potassium phosphate and magnesium sulfate between balanced equation for the precipitation reaction is

2 K3PO4 + 3 MgSO4 → Mg3(PO4)2 + 3 K2SO4

Here reaction is  between potassium phosphate and magnesium sulfate we have to understand the balance chemical equation. precipitation reactions are double displacement reactions in which the conversion of a solution into solid by converting the substance into insoluble form or by making the solution a super saturated one. here the reaction is two mole of  aqueous tribasic potassium phosphate react with three mole of aqua magnesium sulphate to form one mole of solid farringtonite and three mole of aqueous potassium sulphate this is balanced chemical reaction of potassium phosphate and magnesium sulfate.

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Answer: C

Explanation: none of the others make sense if you think about it

Answer:

Gold

Explanation:

Gold(I) chloride is a compound of gold and chlorine with the chemical formula AuCl

Answer:

P and Q

Q and R

Explanation:

To begin, we must bear in mind the meaning of isotope.

Isotopes are atoms having the same atomic number (i.e proton) but different mass number due to the difference in the neutron number present in the atoms involved.

Thus, we shall write out the number of protons, neutrons and electrons present in the model given above. This is given:

Model P:

Proton = 12

Neutron = 13

Electron = 13

Model Q:

Proton = 12

Neutron = 12

Electron = 13

Model R:

Proton = 12

Neutron = 13

Electron = 12

From the above data, we can say that:

1. P and Q are isotope since they have different Neutron number.

2. P and R are not isotope since they have the same Neutron number.

3. Q and R are isotope since they have different Neutron number.

Answer:

I agree with D

Explanation:

Took the test on Edge

The characteristic do all parts of the electromagnetic spectrum share is that they travel at the same speed through empty space. The speed of these waves are the speed of light in vacuum.

Electromagnetic spectrum is the arrangement of different types of radiations in the order of their increasing frequency or decreasing wavelength.

The order of the waves in the increasing order of frequency is radio waves, microwaves, infrared rays, visible light, ultraviolet rays, x-rays and gamma rays.

The frequency and wavelength of all these waves are different. However, they travel at the same speed in  empty space equal to the speed we say for the light.

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Answer:

Answer Option A) To attain a noble-gas electron configuration.

Explanation:

Correct answer from test

Taking into account the definition of covalent bond and octet rule, the correct answer is option A.  Atoms share electrons in covalent bonds to achieve a noble gas electron configuration.

The covalent bond is the chemical bond that occurs between atoms of non-metallic elements. These atoms have many electrons at their outermost level and have a tendency to share them to acquire the stability of the noble gas electronic structure. In this way a molecule is formed.

The tendency of the elements to reach a stable configuration is known as the octet rule. For this, the octet rule explains that there must be eight electrons in their last energy levels, thus acquiring the structure of a noble gas.

So, a covalent bond is a force that joins two atoms of non-metallic elements to form a molecule in order to comply with the octet rule.

Finally, the correct answer is option A.  Atoms share electrons in covalent bonds to achieve a noble gas electron configuration.

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Answer:

B

Explanation:

the hydronium ion concentration is 0.0064 mol/L and the ph of the solution is 2.19 that results when 75.0 ml of 0.405 m ch3cooh is mixed with 104 ml of 0.210 m naoh. acetic acid's ka is 1.70 ✕ 10−5

First, we need to determine the amount of acid and base that reacts with each other. To do this, we use the following equation:

n(CH3COOH) = C(CH3COOH) x V(CH3COOH) = (0.405 mol/L) x (0.075 L) = 0.0304 mol

n(NAOH) = C(NAOH) x V(NAOH) = (0.210 mol/L) x (0.104 L) = 0.0218 mol

Since the acid and base react in a 1:1 ratio, we see that the limiting reagent is the NaOH. Therefore, all of the NaOH will react, leaving us with 0.0086 mol of CH3COOH.

Next, we need to calculate the concentration of the remaining CH3COOH:

[CH3COOH] = n(CH3COOH) / V(total) = (0.0086 mol) / (0.179 L) = 0.048 mol/L

Using the Ka expression for acetic acid, we can solve for the hydronium ion concentration:

Ka = [H3O+][CH3COO-] / [CH3COOH]

[H3O+] = sqrt(Ka x [CH3COOH] / [CH3COO-]) = sqrt((1.70E-5)(0.048)/(0.0218)) = 0.0064 mol/L

Finally, we can calculate the pH:

pH = -log[H3O+] = -log(0.0064) = 2.19

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The hydronium ion concentration is 0.0237 M and the pH is 1.63. This is found by calculating the moles of acid and base, determining the limiting reactant, and then using the balanced equation to calculate the excess reactant. The excess OH- concentration is used to calculate the hydronium ion concentration and pH using the Kw expression and the definition of p H.

To calculate the hydronium ion concentration and pH, we first determine the moles of acid and base using their respective concentrations and volumes. Then, we determine the limiting reactant, which is acetic acid in this case. The balanced equation for the reaction is CH3COOH + OH- → CH3COO- + H2O. We can use the stoichiometry of the balanced equation to determine the excess OH- concentration. The concentration of hydronium ions can be calculated using the Kw expression, and the pH is found using the definition of pH. The resulting values indicate that the solution is acidic.

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Answer:

B

Explanation:

weight , w= mg  where m is the mass of the object (unit is kg ) and g is the gravitational force acting on the object ( unit is m/s^2 )

hope it helps

An acid base reaction always produces salt and water.

In neutralization reaction, an acid and a base react together to produce salt and water.

In this chemical reaction hydrogen ions (H+ ions) come form an acid while hydroxide ions (OH- ions) are released by a base which combine together to form water.

For example, let's take the reaction between sodium hydroxide and hydrochloric acid.

NaOH + HCl -------> NaCl + H2O

the above reaction produces sodium chloride salt and water.

Overall, it is a neutral reaction because no excess ions are left behind in the solution.

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Answer: The specimens appears larger with a higher magnifications because a smaller area of the object is spread out to cover the filed view of your eye.

Explanation: