Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH=1.06×10−10m, into the answer box.

Answers

Answer 1

Answer:

dH = 1.06 x 10⁻¹⁰ m

Explanation:

Scientific notation is a way of minimizing large figures in smaller decimal form. For example, 2300000 is a large figure so, its size can be minimized by converting it into scientific notation form, 2300000 can also be written as 2.3 x 106.

There are number of formats to write dH=1.06×10−10m in scientific notation by shifting the decimal in right or left direction. For example, if we shift the decimal in right direction, the addition of +1 will occur in the power of 10 i.e. 1.06 x 10-10 will become 10.6 x 10-11 or 106 x 10-12 (these formats of scientific notation are also correct).

In some softwares and programming languages, scientific notation are written as 1.06E-10 so, avoid using these kinds of notations since, E indicates as variable in mathematics.

Variables are alphabetical value in an equation. For example, in equation 2a + 3b, a and b are variables while 2 and 3 are constants.

Answer 2
Final answer:

To express the diameter of a hydrogen atom in scientific notation, it would be 1.06×10⁻¹⁰ meters. Scientific notation simplifies the representation of very large or small numbers and is particularly useful for expressing measurements in physics, like the mass of a hydrogen atom, which is 1.67×10⁻¹⁼ kilograms.

Explanation:

The diameter of a hydrogen atom in its ground state is given as dH=1.06×10⁻¹⁰ meters. To enter this number in scientific notation, you place the decimal point such that there is only one non-zero digit to the left of the decimal point. In this case, the diameter would be 1.06×10⁻¹⁰ m.

Scientific notation is a convenient way to express large or small numbers. For instance, the mass of a hydrogen atom can be expressed as 1.67×10⁻¹⁼ kg. This system lets us write numbers as a product of a number between 1 and 10 and a power of 10.

An example of applying scientific notation is a scale model of a hydrogen atom: if one were building a scale model where the atom's diameter is 1.00 m, finding the proportional size of the nucleus would require understanding the actual size relationship, which in scientific notation is a much simpler task than with standard numeral representation.


Related Questions

A student conducted an experiment to determine DHrxn for the reaction between HCl(aq) and NaOH(aq). The student ran two trials using the volumes of HCl(aq) and NaOH(aq) indicated in the table above, and determined the amount of heat released. Which of the following best explains the relationship between X and Y?

Answers

Answer:

Y=X

Explanation:

The number of moles of acid and base reacting with each other is the same in both trials.

Final answer:

In the neutralization reaction between HCl and NaOH, a one-to-one mole ratio leads to a predictable enthalpy change (ΔHrxn). The temperature increase signifies an exothermic reaction, and ΔHrxn can be calculated using the formula q=mcΔT, considering the reaction's exothermic nature.

Explanation:

The relationship between X and Y refers to the amount of heat (ΔH) released during the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH). In this reaction, a one-to-one mole ratio is observed, meaning 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of sodium chloride (NaCl) and water (H₂O)

When calculating the enthalpy change, the student must use the temperature change, the mass of the solution, and the specific heat capacity (which is assumed to be the same as water's for this type of experiment). The amount of heat produced (ΔHrxn) can be calculated using the formula q = mcΔT, where 'm' is the mass of the solution, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature.

In a typical calorimetry experiment, the student would need to carefully measure the temperature changes and use stoichiometry to relate the amounts of the reactants to the heat released. If the temperature increases, as in the example provided, it indicates that the reaction is exothermic, and the value for ΔHrxn would be negative, signifying that heat is released into the surroundings.

If you are involved in a vehicle accident, you should not drink any alcohol up to ___ hours after the accident or you could be legally charged with operating a vehicle while under the influence of alcohol.

Answers

Answer: 6

Explanation:

Alcohol is a psychoactive drug with a high number of side effects that can seriously affect our body. The amount and circumstances of consumption play an important role in determining the duration of intoxication. For example, consuming alcohol after a large meal is less likely to produce visible signs of intoxication than on an empty stomach.

Alcohol has a two-phase effect on the body, meaning that its effects change over time.  Initially, it produces feelings of relaxation and joy, but later consumption can lead to blurred vision and coordination problems.  Cell membranes are highly permeable to alcohol, so once alcohol is in the bloodstream, it can spread to almost all body tissues.  

Unconsciousness can follow excessive consumption, and extreme levels of consumption can lead to alcohol poisoning and death. Death can also be caused by asphyxiation, if vomit - a frequent result of excessive drinking - blocks the windpipe and the individual is too drunk to respond.  An appropriate first-aid response to an unconscious, drunk person is to place him or her in the recovery position.  

When alcohol enters the bloodstream (30-90 minutes after ingestion), there is a decrease in the sugars present in the bloodstream, causing a feeling of weakness and physical exhaustion. This is because alcohol accelerates the transformation of glycogen (a substance that stores sugar in the liver) into glucose, which is then eliminated more quickly.  

After the ingestion of alcohol, a series of effects or symptoms occur in the short term, depending on the dose ingested (although other individual factors are affected).

Euphoria and excitement phaseIntoxication: The nervous system is affected as the loss of the capacity to coordinate movements is caused, producing imbalance and sometimes falls. If abused, locomotor ataxia can occur, which is a paralysis characteristic of alcoholics. Hypnotic or confused phase: Irritability, agitation, drowsiness, headache, ysarthria, ataxia, dysmetry, nausea and vomiting. Anaesthetic phase or stupor and coma, inconsistent language. Bulbar or death phase: Cardiovascular shock, inhibition of the respiratory center, cardio-respiratory arrest and death.

Alcohol is broken down mainly in the liver, which can metabolize about 1 drink per hour for men. Factors such as age, weight, gender, and the amount of food eaten can affect how quickly the body can process alcohol. The amount of time alcohol can be detected in blood is up to 6 hours. Then it is not recommended to drink alcohol until after 6 hours if you have been involved in an accident, since alcohol can be detected in the blood in that time range. Because of the effects alcohol has on the body and how it can affect driving, if alcohol is detected you can be legally charged with driving under its influence.

If an atom of thorium-234 were to undergo beta emission twice, which atom would result?


Rn-226


U-234


Th-230


U-238

Answers

Answer:

The answer to your question is  U-234

Explanation:

Data

Thorium-234

beta emission twice

Definition

Beta emission is when a beta particle (electron) is emitted from an atomic nucleus.

First beta emission

                                            ²³⁴₉₀Th   ⇒    ²³⁴₉₁Pa + e⁻

Second beta emission

                                            ²³⁴₉₁ Pa   ⇒  ²³⁴₉₂U  + e⁻

The atom will be Uranium-324

Answer: U-234

Explanation: no matter how much of beta decay, the mass number of the resulting element is unchanged

A solution is made by dissolving solute B in solvent A. Consider the A-A attractive forces, the B-B attractive forces, and the A-B attractive forces. If the solution process is exothermic, what can you say about the relationships between these attractive forces?

Answers

Final answer:

An exothermic dissolution process suggests that A-B attractive forces are significant enough to overcome A-A and B-B intermolecular forces, resulting in energy being released as heat.

Explanation:

When a solution is made by dissolving solute B in solvent A and the solution process is exothermic, it indicates that the energy released in forming A-B attractive forces is greater than the energy required to overcome both A-A and B-B attractive forces. Since the process is exothermic, this means that the solute-solvent attractions are strong enough to not only break the solute-solute and solvent-solvent interactions but also provide excess energy that is released as heat. If the energy required to separate the solute and solvent was greater than the energy released upon mixing, an endothermic reaction would occur, and the solutes might not dissolve.

ATP labeled with 32P a radioactive isotope of phosphorus at the gamma phosphate is added to a crude extract of a tissue rich in the enzymes of glycolysis along with glucose. What is the first intermediate of glycolysis that will no longer be radioactive?

Answers

Answer:

Piruvate

Explanation:

Glucose + ATP32 --------> Glucose-6-P32 ----->Fructose-6-P32

Fructose-6-P32 + ATP32 ---------> Fructose-1,6 bi-P32

Fructose-1,6 bi-P32 ----------> Glyceraldehyde-3-P32 + Dihydroxiacetone Phosphate32 (that are in equilibrium)

Glyceraldehyde-3-P32 +NADH +Pi (non radioactive) -----> Glyceraldehyde-1,3-P32 (1,3 biphosphoglycerate where only in C3, there is a P32)

After that, the phosphorus in Carbon1 is donated to ADP to for ATP (non radioactive), and we have 3-Phosphoglycerate (radioactive because its P32), then it's converted to 2-Phosphoglycerate (radioactive), then Phosphoenolpiruvate (radioactive), that donates its P32 to ADP to produce ATP, remaining Piruvate as end product at the end of glucolysis

Consider the reaction. Upper H subscript 2 upper o (g) plus upper C l subscript 2 upper O (g) double-headed arrow 2 upper H upper C l upper O (g). At equilibrium, the concentrations of the different species are as follows. [H2O] = 0.077 M [Cl2O] = 0.077 M [HClO] = 0.023 M What is the equilibrium constant for the reaction at this temperature?

Answers

Answer:

0.089

Explanation:

Step 1:

The balanced equation for the reaction is given below:

H2O + Cl2O <=> 2HClO

Step 2:

Data obtained from the question. This includes:

Concentration of H2O, [H2O] = 0.077 M

Concentration of Cl2O, [Cl2O] = 0.077 M

Concentration of HClO, [HClO] = 0.023 M

Equilibrium constant, K =?

Step 3:

Determination of the equilibrium constant. This is illustrated below:

The equilibrium constant for the above reaction is given below:

K = [HClO]^2 / [H2O] [Cl2O]

K = (0.023)^2 / (0.077 x 0.077)

K = 0.089

Therefore, the equilibrium constant for the above reaction is 0.089

Answer:

The answer is 0.089

I just took the test

Explanation:

Consider the reaction.

Upper H subscript 2 upper o (g) plus upper C l subscript 2 upper O (g) double-headed arrow 2 upper H upper C l upper O (g).

At equilibrium, the concentrations of the different species are as follows.

[H2O] = 0.077 M

[Cl2O] = 0.077 M

[HClO] = 0.023 M

What is the equilibrium constant for the reaction at this temperature?

0.089

0.26

3.9

11

Suppose we want to charge a flask with 1.9 g of sugar. We put the empty flask on a balance and it is determined to weigh 450 g. Enter the weight we would expect to see on the balance when we're done adding the sugar.

Answers

Answer: 451.9g

Explanation:

Weight of empty flask= 450g

Weight of sugar= 1.9g

Weight of sugar+flask=450+1.9=451.9g

The sum of the massed of both the sugar and the flask is the mass the balance will read after the addition of the sugar.

A solution is prepared by mixing 93.0 mL of 5.00 M HCl and 37.0 mL of 8.00 M HNO3. Water is then added until the final volume is 1.00 L. Calculate [H ], [OH -], and the pH for this solution.

Answers

Answer:

[tex][H^{+}] = 0.761 \frac{mol}{L}[/tex]

[tex][OH^{-}]=1.33X10^{-14}\frac{mol}{L}[/tex]

[tex]pH = 0.119[/tex]

Explanation:

HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.

[tex]n_{H^{+} } from HCl = [HCl](\frac{mol}{L}). V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)[/tex]

Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows

[tex]n_{H^{+} } from HCl = (5.00)(0.093)[/tex]

[tex]n_{H^{+} } from HCl = 0.465 mol[/tex]

[tex]n_{H^{+} } from HNO_{3} = (8.00)(0.037)[/tex]

[tex]n_{H^{+} } from HNO_{3} = 0.296 mol[/tex]

[tex]n_{H^{+}(total) } = 0.296 + 0.465[/tex]

[tex]n_{H^{+}(total) } = 0.761 mol[/tex]

For molar concentration of hydrogen ions:

[tex][H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}[/tex]

[tex][H^{+}] = \frac{0.761}{1.00}[/tex]

[tex][H^{+}] = 0.761 \frac{mol}{L}[/tex]

From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows

[tex]K_{w} = [H^{+} ][OH^{-} ][/tex]

[tex][OH^{-}]=\frac{Kw}{[H^{+}] }[/tex]

[tex][OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }[/tex]

[tex][OH^{-}]=1.33X10^{-14}\frac{mol}{L}[/tex]

The pH of the solution can be measured by the following formula:

[tex]pH = -log[H^{+} ][/tex]

[tex]pH = -log(0.761)[/tex]

[tex]pH = 0.119[/tex]

sample of copper(II) sulfate pentahydrate, CuSO4⋅5H20 , is heated to remove the water of hydration from the crystals. The mass of the original sample before heating was 50.00 grams. If the mass of the cool, dry sample after heating is 34.95 grams, find the experimental percentage of water in the hydrate.

Answers

Answer:

30.10%

Explanation:

The mass of water is:

50.00 g − 34.95 g = 15.05 g

So the percentage of water is:

15.05 g / 50.00 g × 100% = 30.10%

Final answer:

The experimental percentage of water in copper(II) sulfate pentahydrate, after heating a sample and measuring the mass loss, is calculated to be 30.10%.

Explanation:

Experimental Percentage of Water in a Hydrate

The question involves calculating the percentage of water in a hydrate, specifically copper(II) sulfate pentahydrate, which is a classic chemistry experiment. To determine the percentage of water, you have to compare the mass of water lost upon heating to the original mass of the hydrate.

First, calculate the loss of mass due to heating: 50.00 grams (initial mass) - 34.95 grams (final mass) = 15.05 grams of water lost.

Percentage of water in the hydrate is found by the formula:
Mass% of H₂O = (Mass of water lost / Original mass of hydrate) * 100%.
Substituting the known values gives us: Mass% of H₂O = (15.05 g / 50.00 g) * 100% = 30.10%.

The experimental percentage of water in the copper(II) sulfate pentahydrate is therefore 30.10%, which is determined by this experimental process of heating and weighing.

Convert grams of FeCl2 to moles. Then rearrange M = n/V to solve for V: V = n/M (in Liters) What volume of a 0.01 M solution can be made using 120 grams of FeCl2?

Answers

Answer:

The answer to your question is Volume = 214.3 ml

Explanation:

Data

mass = 120 g of FeCl₂

concentration = 0.01 M

volume = ?

Formula

Molarity = [tex]\frac{number of moles}{volume}[/tex]

Solve for volume

Volume = [tex]\frac{number of moles}{molarity}[/tex]

Process

1.- Convert grams to moles

Atomic weight = 56 g

                            56 g of Fe --------------- 1 mol

                         120 g of Fe  ----------------  x

                               x = (120 x 1) / 56

                               x = 2.14 moles

2.- Calculate the volume

Volume = [tex]\frac{2.14}{0.01}[/tex]

Volume = 214.3 ml

Answer:

We need 94.67 liters if a 0.01 M solution

Explanation:

Step 1: Data given

Molarity of the solution = 0.01 M

Mass of FeCl2 = 120.0 grams

Molar mass FeCl2 = 126.75 g/mol

Step 2: Calculate moles of FeCl2

moles FeCl2 = massFeCl2 / molar mass FeCl2

Moles FeCl2 = 120.0 grams / 126.75 g/mol

Moles FeCl2 = 0.9467 moles

Step 3: Calculate  volumes FeCl2

Molarity = moles / volume

Volume = moles / molarity

Volume = 0.9467 moles /0.01 M

Volume = 94.67 L

We need 94.67 liters if a 0.01 M solution

A tank of hydrogen gas has a volume of 22.9 L and holds 14.0 mol of the gas at 12°C. What is the pressure of the gas in atmospheres?

Answers

Answer: 14.3 atm

Explanation: solution attached

Final answer:

The pressure of hydrogen gas in the tank is approximately 4.92 atm.

Explanation:

To find the pressure of the hydrogen gas, we need to use the ideal gas law equation: PV = nRT. Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.

To convert the temperature from Celsius to Kelvin, we add 273 to the given temperature. Now, let's plug in the values: P(22.9 L) = (14.0 mol)(0.0821 L·atm/mol·K)(12°C + 273). Solving this equation for P, the pressure of the gas is approximately 4.92 atm.

Learn more about Ideal Gas Law here:

https://brainly.com/question/1063475

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An organic compound was extracted into dichloromethane and then the aqueous layer is shaken with saturated sodium chloride solution.What is the purpose of the sodium chloride?

Answers

Answer:

The sodium chloride serves to pull out the water from the organic layer to the aqueous or water layer

Explanation:

Shaking it with sodium chloride also called brine pulls the water away from the organic composition to the water layer due to the affinity of the salt to absorb more water and become less concentrated and dense as the salt has more ability to absorb the water than to absorb organic compounds

spontaneously form membranes when mixed in water and most likely were one of the first organic compounds formed on Earth?

Answers

Answer:

phospholipids

Explanation:

As regards the lipid membrane of the first protocells, it is most likely that it initially consisted of simpler fatty acids than the phospholipids that make up the current membranes (see appendix). If the modern membranes are bilayers of glycerol phospholipids, the primitive membranes would probably be made up of simpler, single-chain molecules, also amphiphilic in nature (with a soluble part and another insoluble in water), such as monocarboxylic acids or alcohols. The origin of these compounds could be multiple. On the one hand, it has been seen that they are very abundant in meteorites of the type of carbonaceous chondrites, so they could have arrived on Earth already formed from outer space. But it is also possible that they were formed abioticly on the primitive Earth by the reaction of CO and hydrogen to give rise to various hydrocarbons, a reaction that would be viable at high temperatures in the presence of ferric catalysts, on the surface of montmorillonite clays and also in hydrothermal conditions. Regardless of the origin, the result would be the presence of fatty acids initially very diluted in an aqueous solution, but which would be concentrated by successive evaporation cycles, or by the formation of small aerosolized drops that would also transfer those vesicles to points distant from the place where The first organic membrane compounds that formed on Earth were formed and would be.

Identify the true statement. Choose one: A. The expansion of ice sheets at the end of the Pleistocene caused glacial rebound in northern Canada. B. Regions covered by glacial ice are termed periglacial environments. C. Glaciers move by basal sliding as friction between the ice and its substrate increases. D. The Antarctic ice sheet has been calving off huge icebergs over the last few decades.

Answers

Answer:

Option (D)

Explanation:

Antarctica is located in the south pole and covers a large area of ice-covered region. It is an extremely cold environment, having a low freezing temperature. The glaciers and ice-bergs of this region has been constantly affected due to the increasing global surface temperature. This rise in the surface temperature of the earth is due to the increasing concentration of CO₂ in the atmosphere. One of the reasons for this temperature increase is also due to the introduction of the industrial revolution, which is responsible for the emission of a large number of harmful gases, including carbon into the atmosphere. Due to this global climate change, these glaciers are melting at a faster rate, resulting in the rise of the sea level.

Thus, the Antarctic ice sheets have been calving off large pieces of icebergs over the last few decades.

Hence, the correct answer is option (D), which is the true statement.

a chemist wishes to mix some pure acid with some water to produce 16L of a solution that is 30% acid how much pure acid and how much water should be mixed?

Answers

Answer: The volume of acid and water that must be mixed will be 4.8 L and 11.2 L

Explanation:

We are given:

Volume of mixture = 16 L

Percent of acid present = 30 %

Calculating the percentage of acid present in the mixture:

[tex]\Rightarrow 16\times \frac{30}{100}=4.8L[/tex]

The mixture is made entirely of acid and water.

Volume of acid in the mixture = 4.8 L

Volume of water in the mixture = 16 - 4.8 = 11.2 L

Hence, the volume of acid and water that must be mixed will be 4.8 L and 11.2 L

Final answer:

The chemist needs 4.8 liters of pure acid and 11.2 liters of water to prepare 16 liters of a solution that is 30% acid.

Explanation:

To prepare 16 liters of a solution that is 30% acid, we start by setting up an equation to represent the mix of pure acid and water. Let the amount of pure acid needed be x liters, and therefore the amount of water would be 16 - x liters. Since the solution is 30% acid, we can write the equation as follows:

0.30 × 16 = x

When we solve for x, we get:

4.8 = x

This means we need 4.8 liters of pure acid and 16 - 4.8 = 11.2 liters of water to make the 30% acid solution.

Antoine Lavoisier, the French scientist credited with first stating the law of conservation of matter, heated a mixture of tin and air in a sealed flask to produce tin oxide. Did the mass of the sealed flask and contents decrease, increase, or remain the same after the heating?

Answers

Answer:

Remain the same

Explanation:

The law of mass conservation states that matter cannot be created nor destroyed but can be converted from one form to another.

Essentially, what Lavoisier was trying to proof is that by heating the mixture, after all the change the mass still remains. That was why he used a sealed flask. If the flask was not sealed, it probably would have been that some of the mass will escape as vapor to the atmosphere which might be difficult to account for

"A sphere of radius 0.50 m, temperature 27oC, and emissivity 0.85 is located in an environment of temperature 77oC. What is the net flow of energy transferred to the environment in 1 second?"

Answers

Explanation:

It is known that formula for area of a sphere is as follows.

                     A = [tex]4 \pi r^{2}[/tex]

                        = [tex]4 \times 3.14 \times (0.50 m)^{2}[/tex]

                        = 3.14 [tex]m^{2}[/tex]

    [tex]T_{a}[/tex] = (27 + 273.15) K = 300.15 K

          T = (77 + 273.15) K = 350.15 K

Formula to calculate the net charge is as follows.

             Q = [tex]esA(T^{4} - T^{4}_{a})[/tex]

where,    e = emissivity = 0.85

               s = stefan-boltzmann constant = [tex]5.6703 \times 10^{-8} Wm^{-2} K^{-4}[/tex]

                A = surface area

Hence, putting the given values into the above formula as follows.

                 Q = [tex]esA(T^{4} - T^{4}_{a})[/tex]

                     = [tex]0.85 \times 5.6703 \times 10^{-8} Wm^{-2} K^{-4} \times 3.14 \times ((350.15)^{4} - (300.15)^{4})[/tex]

                     = 1046.63 W

Therefore, we can conclude that the net flow of energy transferred to the environment in 1 second is 1046.63 W.

How many joules of heat must be absorbed by 500g h2O @ 50CELCIUS to convert to steam @ 120 celcius?

vaporization:40.7 mol
steam; 36.5 j/mol
h2o: 75.3 j/mol

Answers

Answer:

Q = 1267720 J

Explanation:

Qt = QH2O + ΔHv

∴ QH2O = mCpΔT

∴ m H2O = 500 g

∴ Cp H2O = 4.186 J/g°C = 4.183 E-3 KJ/g°C

∴ ΔT = 120 - 50 = 70°C

⇒ QH2O = (500 g)(4.183 E-3 KJ/g°C)(70°C) = 146.51 KJ

∴ ΔHv H2O = 40.7 KJ/mol

moles H2O:

∴ mm H2O = 18.015 g/mol

⇒ moles H2O = (500 g)(mol/18.015 g) = 27.548 mol H2O

⇒ ΔHv H2O = (40.7 KJ/mol)(27.548 mol) = 1121.21 KJ

⇒ Qt = 146.51 KJ + 1121.21 KJ = 1267.72 KJ = 1267720 J

In some vintage science fiction movies, space travelers find themselves on a planet orbiting a distant star in which there are curious forms of life based on silicon instead of carbon. Although the story clearly is sci-fi, there is an aura of plausibility in the choice of silicon, an atom with 14 protons, in place of carbon as this alien life-form's central atom. The reason is that silicon:

Answers

Answer:

yes because it is a great planet

Explanation:

because of the reason for silicon

You wish to prepare 0.13 M HNO3 from a stock solution of nitric acid that is 16.6 M. How many milliliters of the stock solution do you require to make up 1.00 L of 0.13 M HNO3?

Answers

Answer:

0.0078 L of the stock solution is required to make up 1.00 L of 0.13 M [tex]HNO_{3}[/tex]

Explanation:

According to laws of equivalence, [tex]C_{1}V_{1}=C_{2}V_{2}[/tex]

where, [tex]C_{1}[/tex] and [tex]C_{2}[/tex] are initial and final concentration respectively. [tex]V_{1}[/tex] and [tex]V_{2}[/tex] are initial and final volume respectively.

Here, [tex]C_{1}=16.6M[/tex], [tex]C_{2}=0.13M[/tex] and [tex]V_{2}=1.00L[/tex]

So, [tex]V_{1}=\frac{C_{2}V_{2}}{C_{1}}[/tex]

or, [tex]V_{1}=\frac{(0.13M)\times (1.00L)}{(16.6M)}[/tex]

or, [tex]V_{1}=0.0078L[/tex]

Hence 0.0078 L of the stock solution is required to make up 1.00 L of 0.13 M [tex]HNO_{3}[/tex]

A 1.87 L aqueous solution of KOH contains 155 g of KOH . The solution has a density of 1.29 g/mL . Calculate the molarity ( M ), molality ( m ), and mass percent concentration of the solution

Answers

Answer:

[KOH] = 1.47 M

[KOH] = 1.22 m

KOH = 6.86 % m/m

Explanation:

Let's analyse the data

1.87 L is the volume of solution

Density is 1.29 g/mL → Solution density

155 g of KOH → Mass of solute

Moles of solute is (mass / molar mass) = 2.76 moles.

Molarity is mol/L → 2.76 mol / 1.87 L = 1.47 M

Let's determine, the mass of solvent.

Molality is mol of solute / 1kg of solvent

We can use density to find out the mass of solution

Mass of solution - Mass of solute = Mass of solvent

Density = Mass / volume

1.29 g/mL = Mass / 1870 mL

Notice, we had to convert L to mL, cause the units of density.

1.29 g/mL . 1870 mL = Mass → 2412.3 g

2412.3 g - 155 g = 2257.3 g of solvent

Let's convert the mass of solvent to kg

2257.3 g / 1000 = 2.25kg

2.76 mol / 2.25kg = 1.22 m (molality)

% percent by mass = mass of solute in 100g of solution.

(155 g / 2257.3 g) . 100g = 6.86 % m/m

Final answer:

The molarity, molality, and mass percent concentration of the KOH solution are approximately 1.476 M, 1.224 m, and 6.43%, respectively.

Explanation:

The first step in solving this problem is to determine the number of moles of KOH in the solution. The molar mass of KOH is approximately 56.11 g/mol, so you can calculate the number of moles by dividing the mass of the KOH by its molar mass: moles of KOH = 155 g / 56.11 g/mol = 2.762 mol.

Molarity (M) is defined as the number of moles of solute per liter of solution. To find the molarity, divide the number of moles by the volume of the solution in liters: M = 2.762 mol / 1.87 L = 1.476 M.

To calculate molality (m), you need the mass of the solvent in kilograms. First, find the total mass of the solution: mass = volume x density = 1.87 L x 1.29 g/mL = 2411.3 g. Then, subtract the mass of the KOH to find the mass of the solvent: mass of solvent = 2411.3 g - 155 g = 2256.3 g = 2.2563 kg. Now, you can find the molality: m = 2.762 mol / 2.2563 kg = 1.224 m.

Finally, the mass percent concentration of the solution is the mass of the solute divided by the mass of the solution, multiplied by 100: mass percent = (155 g / 2411.3 g) x 100 = 6.43%.

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The cations in an aqueous solution that contains 0.150 m Ba(NO3)2 and 0.0800 m Ca(NO3)2 are to be separated by taking advantage of the difference in the solubilities of their sulfates. Ksp(BaSO4) 5 1.1 3 10210 and Ksp(CaSO4) 5 2.4 3 1025. What should be the concentration of sulfate ion for the best separatio?

Answers

Answer:Ba^2+ is a solid because the concerntration ofSO4 ion for the separation is 3.0×10^4M

Explanation: The equation for the reaction are:

1. BaSO4 ---->Ba^2+ SO4^2-

Ksp=1.1×10^-10

2.CaSO4 ------> Ca^2+ + SO4^2-

Ksp=2.4×10^-5

/Ba^2+/=0.150M

/Ca^2+/=0.080M

Calcuim sulfate is more soluble than barium sulfate, therefore add SO4^2- ion to leaveCa^2+ in the solution and precipitate Ba^2+ ions.

SO4^2-= 2.4×10^-5/0.080

SO4^2-=3.0×10^-4 M

ForCa^2+ ion to be left in the solution,reaction quotient must be less than the Ksp

Q=[Ca^2+][SO4^2-] <ksp

SO4^2- =Ksp/[Ca^2+]

Which response includes all the following that are properties of most metals, and no other properties?
Z1) They tend to form cations.
Z2) They have high first ionization energies.
Z3) They have outer electronic shells that are more than half-filled.
Z4) They tend to form ionic compounds when they combine with the elements Group VIIA.

1. Z3 and Z4 only
2. Z2, Z3, and Z4 only
3. Z1, Z2, and Z3 only
4. Z1 and Z4 only
5. Z1 and Z3 only

Answers

Answer:Z1 and Z4 only

Explanation:

One general property of all metals is that they form positive ions (cations) by electron loss. Secondly, elements of group VIIA have high electron affinities hence they easily accept electrons to form negative ions. The bonding between metals and elements in this group is almost always ionic in nature.

HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)ΔH°=−57.1kJ/molrxn
The chemical equation above represents the reaction between HCl(aq) and NaOH(aq). When equal volumes of 1.00MHCl(aq) and 1.00MNaOH(aq) are mixed, 57.1kJ of heat is released. If the experiment is repeated with 2.00MHCl(aq), how much heat would be released?

Answers

Answer:

57.1 kilo Joules of heat would be released.

Explanation:

[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex] ΔH°=-57.1kJ/mol

Molarity of HCl = 2.00 M

Molarity of NaOH = 1.00 M

According to reaction , 1 M of HCl reacts with 1 M of NaOH. Then 2.00 M of HCl will react with:

[tex]\frac{1}{1}\times 2.00 M= 2M [/tex] of NaOH

But according to question we only have 1.00 M NaOH .So, this means that NaOH is limiting reagent and HCl is an excessive reagent.

Heat evolved will depend upon concentration of NaOH solution :

Heat evolved when 1.00 M of NaOH reacts =

[tex]1.00\times (-57.1 kJ/mol)=-57.1 kJ[/tex]

Negative sign means that heat is released during the reaction.

57.1 kilo Joules of heat would be released.

The heat released during the repeated experiment with 2 mole of HCl is 57.1 kJ.

The given reaction,

[tex]\bold {HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l) \ \ \ \ \ \ \ \ \ \Delta H = - 57.1\ kJ/mol}[/tex]

Concentration of HCl is 2 mole.

So, 2 mole of HCl react with 2 mole of NaOH but only mole of NaOH is available.

So, NaOH is a Limiting factor in the reaction and HCl is excessive factor.

Only, 1mole of HCl will react with available 1 mole of NaOH,

Therefore, the heat released during the repeated experiment with 2 mole of HCl is 57.1 kJ.

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Complete the dissociation reaction and the corresponding Ka equilibrium expression for each of the following acids in water. (Type your answer using the format [NH4]+ for NH4+ and [Ni(CN)4]2- for Ni(CN)42-. Use the lowest possible coefficients.)
(A) HC2H3O2
HC2H3O2(aq) (twosidedarrow) [ ]H+(aq) + [ ] [ ](aq)
Ka = [ ][ ] / [ ]
(B) Co(H2O)63+
Co(H2O)63+(aq) (twosidedarrow) [ ]H+(aq) +[ ][ ](aq)
Ka = [ ][ ] / [ ]
(C) CH3NH3+
CH3NH3+(aq) (twosidedarrow) [ ]H+(aq) +[ ][ ](aq)
Ka = [ ][ ] / [ ]

Answers

Final answer:

The dissociation reactions for acetic acid, hexaaquacobalt(III) ion, and methylammonium along with their Ka equilibrium expressions show how each acid disassociates into its constituent ions in water, and how the concentration of these ions at equilibrium can be represented.

Explanation:

Completing the dissociation reaction and the Ka equilibrium expression for each of the following acids in water:

(A) Acetic acid [tex]($\text{HC}_2\text{H}_3\text{O}_2$)[/tex]

[tex]\[\text{HC}_2\text{H}_3\text{O}_2(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{C}_2\text{H}_3\text{O}_2^-(\text{aq})\\\\K_a = \frac{[\text{H}^+][\text{C}_2\text{H}_3\text{O}_2^-]}{\text{HC}_2\text{H}_3\text{O}_2}\][/tex]

(B) Hexaaquacobalt(III) ion [tex]($\text{Co(H}_2\text{O)}_6^{3+}$)[/tex]

[tex]\[\text{Co(H}_2\text{O)}_6^{3+}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{Co(H}_2\text{O)}_5\text{OH}_2^+(\text{aq})\\\\K_a = \frac{[\text{H}^+][\text{Co(H}_2\text{O)}_5\text{OH}_2^+]}{\text{Co(H}_2\text{O)}_6^{3+}}\][/tex]

(C) Methylammonium[tex]($\text{CH}_3\text{NH}_3^+$)[/tex]

[tex]\[\text{CH}_3\text{NH}_3^+(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{CH}_3\text{NH}_2(\text{aq})\\\\K_a = \frac{[\text{H}^+][\text{CH}_3\text{NH}_2]}{\text{CH}_3\text{NH}_3^+}\][/tex]

A solution surrounding a cell is hypertonic solution if: a. It contains fewer nonpenetrating solute particles than the interior of the cell b. It contains more nonpenetrating solute particles than the interior of the cell c. It contains the same amount of nonpenetrating solute particles as the interior of the cell

Answers

Answer:

The answer is B. It contains more non-penetrating solute particles than the interior of the cell.

Explanation:

This means that it has a greater concentration or number of solute particles outside a membrane than there are inside it.

A typical example is Saline solution.

Final answer:

In a hypertonic solution, the extracellular fluid has a higher concentration of solutes than the cell's cytoplasm. Water will leave the cell, causing it to shrink.

Explanation:

A hypertonic solution is any external solution that has a high solute concentration and low water concentration compared to body fluids. In a hypertonic solution, the net movement of water will be out of the body and into the solution.

This means that the solution contains more nonpenetrating solute particles than the interior of the cell. As a result, water will leave the cell, causing it to shrink.

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Haley's parents bought her a used bicycle for her birthday. She was thrilled until she learned that her best friend received a brand-new bicycle for her birthday. Haley's declining satisfaction illustratesA)relative deprivation.B)the adaptation-level phenomenon.C)the catharsis hypothesis.D)the behavior feedback effect.Page 7

Answers

Answer:

A) relative deprivation

Explanation:

The given example is of the relative deprivation principle of sociology.

The Theory of Relative Deprivation. Throughout sociology, the principle of relative deprivation is a concept of societal change and campaigns, whereby people act for social changes in order to gain something (for example, privileges, prestige, or wealth) that others have and that they also feel they should have the same. Here Haley is feeling deprived of brand new bicycle as has a used one.

How many hydrogen atoms are connected to the indicated carbon atom?

Answers

Answer:

Your question is incomplete, but it could deal with organic structures just like the ones I attached in this answer. So I hope it helps you.

Q1. A) One hydogen atom

Q2. E) None

Q3. A) One Hydrogen atom

Explanation:

Each carbon can make 4 covalent bonds with Hydrogene because of its 4 valencies. If a carbon has only three covalent bonds, the C atom will be charged; in the 3 questions that I´m aswering here, the carbons are charged and they are carbocations (there is no lone pair of electrons and the charge is positive). We can also find carbanions (there is one lone pair of electrons and the charge is negative).

In the molecules on the attachement file, you can notice that the first one has already 2 bonds with adjacent carbons, the next one has 3 bonds and the last one has 2; in every structure, the indicated carbon is a carbocation (no lone pairs of electons).  A cabocation can make 3 covalent bonds, so, for the first and the last molecules, there is one hyrogen atom connected to it, for the second there is none.

Where would you expect to find many hydrophobic amino acids in the 3-D structure of a folded protein that functions in an aqueous solution?

Answers

Answer:

Hydrophobic amino cluster on the inside of the 3-D structure of the folded protein

Explanation:

Hydrophobic interactions, describe the process in which nonpolar, hydrophobic amino acids cluster or aggregate together in the internal part of the protein, to allow the hydrophilic amino acids interact with surrounding water molecules on the outside.

1. Do you think evolution is still taking place in the Galapagos finches? Why or why not?
2. Discuss whether or not human activities lead to speciation. Explain your reasoning.

Answers

The answers are as follows:

1) Evolution is still taking place in the Galapagos finches

Why?

Evolution basically consists on random mutations that affect the population of a species, that are conserved when those mutations help the affected individuals to better adapt to their environment, in a process known as natural selection.

Random mutations are still affecting the finch population in the Galapagos Islands, and when one of those mutations provides an advantage, the species will continue to evolve. However, these changes occur over long periods of time, and its unlikely that we will see the evolved species in our lifetime.

2)  Human activities can lead to speciation

Speciation is the process in which new species are created by natural or artificial selection.

Why?

Human activities can speed up this process by creating external pressure in the form of artificial selection. There are many examples of this in our history, but common cases are the separation of the Dog (Canis familiaris) from the Wolf (Canis lupus). Wolves with desirable traits (non-aggressive, cooperative, small size) were artificially selected by humans and in time, a new species was created. Other examples can be the different species of cattle and sheep.

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