7. When an astronaut travels from the Earth to the Moon, which of their physical properties changes: mass or weight?
Explain why one property changes but the other does not.

Substituents are trans to each other if the appear on opposite sides in the structure. C, E, G, H and K are trans to the bromo substituent.

The substituted cyclohexane compound is shown in the image attached. Two substituents in a compound are trans to each other if they appear on opposite sides of the structure. Cis substituents appear on the same side of the structure.

Considering the structure as shown in the image attached, the substituents that are trans to the bromo substituent are the substituents shown as letters C, E, G, H and K. They lie opposite to the bromo substituent.

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Answer: H , G , E , K, C

Explanation:

These are all opposite from Bromine, meaning they do not face the same way. Bromine faces upward (axial) while the rest of these face downward, (equatorial) to Bromine. We are looking for this opposite facing options. A is not considered an opposite facing option even though it at least looks opposite to Bromine only because it is connected directly to Bromine, making it neither the same or opposite to Bromine.

Answer: The options are not given, here are the options from another website.

A. They will attract each other

b. They will repel each other

C. Nothing will happen to them

d. They will get heated up

The answer is they will attract each other

Explanation:

This is because the charge from the rubbing of ebonite and acetate to a wool is called triboeletric effect.

When ebonite is rubbed with a wool it will produce negative charge, the electrons around it will be negative but when acetate is rubbed with it, it will produce positive charge. This two will attract each other because unlike charges(negative and positive) attract each other.

Option B; This radioisotope's half-life is 20.4 minutes. After 60.0 minutes, 13.03% of a sample is still present.

The decay of C-11 isotopes is known to follow first order kinetics.

The half-life time (t1/2) in a first order reaction is equal to 0.693/k, where k is the rate constant.

k = 0.693/(t1/2), 0.693/(20.4 min), or 0.03397 min1.

The following is the first order reaction integrated law:

ln[A0]/[A], or

where k is the rate constant (0.03397 seconds per second).

t (t = 60.0 min).

[A0] is C-11's starting concentration ([A0] = 100%).

[A] is the amount of C-11 that is still present ([A] =???%).

kt = ln[A0] / [A]

(60.0 min 0.03397 min) = ln(100%)/[A]

2.038 = ln(100%)/[A]

assuming e is on both sides:

7.677 = (100%)/[A]

[A] = (100%)/(7.677) = 13.03%.

After 60.0 minutes, 13.03% of a sample is still present.

Question:

Carbon-11 is used in medical imaging. The half-life of this radioisotope is 20.4 min. What percentage of a sample remains after 60.0min a)34.0 b)13.03 c)2871.2  d)2.94

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The peptide with a pi of 5.6 will bind to the cation-exchange column at pH = 3.8 and to the anion-exchange column at pH = 6.8 and pH = 10.0. The peptide with a pi of 8.7 will bind to the cation-exchange column at pH = 3.8 and pH = 6.8, and to the anion-exchange column at pH = 10.0.

The two peptides have isoelectric points (pi) of 5.6 and 8.7. When separating peptides using an ion exchange column, they can bind to a cation-exchange column when they are positively charged and to an anion-exchange column when they are negatively charged.

For the peptide with a pi of 5.6:
- It will bind to the cation-exchange column at pH values lower than 5.6. At pH = 3.8, it will be positively charged and bind to the cation-exchange column.
- It will bind to the anion-exchange column at pH values higher than 5.6. At pH = 6.8 and pH = 10.0, it will be negatively charged and bind to the anion-exchange column.

For the peptide with a pi of 8.7:
- It will bind to the cation-exchange column at pH values lower than 8.7. At pH = 3.8 and pH = 6.8, it will be positively charged and bind to the cation-exchange column.
- It will bind to the anion-exchange column at pH values higher than 8.7. At pH = 10.0, it will be negatively charged and bind to the anion-exchange column.

In summary, the peptide with a pi of 5.6 will bind to the cation-exchange column at pH = 3.8 and to the anion-exchange column at pH = 6.8 and pH = 10.0. The peptide with a pi of 8.7 will bind to the cation-exchange column at pH = 3.8 and pH = 6.8, and to the anion-exchange column at pH = 10.0.

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Answer:

1750

Explanation:

Greenhouse gases are those gases that are believed to contribute to the rise in the temperature of the earth by trapping infrared rays close to the surface of the earth.

Prominent among the greenhouse gases are; N2O, methane and CO2.

From the chart, we can see that the lowest sum of concentration of greenhouse gases measured in ppm was obtained in 1750. Thus this period  shows the lowest composition of greenhouse gases.

For an atom with an atomic number of 12, it belongs to the third group (Group 3 or Boron family) which means that it has three electrons in its outermost shell.In conclusion, an atom with an atomic number of 12 has 12 electrons and 3 electrons in the outermost shell.

The atomic number of an element represents the number of protons in the nucleus of an atom. Thus, if an atom's atomic number is 12, it means that there are 12 protons in its nucleus and therefore, the atom has 12 electrons as well. The number of electrons in the outermost shell of an atom is determined by its position in the periodic table.Elements in the first group of the periodic table (Group 1 or alkali metals) have one electron in their outermost shell, while elements in the second group (Group 2 or alkaline earth metals) have two electrons in their outermost shell. For an atom with an atomic number of 12, it belongs to the third group (Group 3 or Boron family) which means that it has three electrons in its outermost shell.In conclusion, an atom with an atomic number of 12 has 12 electrons and 3 electrons in the outermost shell.

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The concentration of a 20.00 mL of a H₂SO₄ solution with an unknown concentration was titrated to a phenolphthalein endpoint with 30.85 mL of a 0.1095 M NaOH solution is 84.39 M.

Phenolphthalein is an indicator which turns colourless in acidic medium and pink in basic medium. Therefore, when all the H₂SO₄ has reacted with NaOH to form H₂SO₄ and H₂O, the phenolphthalein will change colour from colourless to pink. This is the end point. We can write the balanced chemical equation as:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

To find out the number of moles of NaOH used.

Number of moles of NaOH used = Molarity × volume

= 0.1095 M × 30.85 × 10⁻³ L

= 0.003375675 mol

Now, according to the balanced chemical equation, one mole of H₂SO₄ reacts with two moles of NaOH. Therefore,

Number of moles of H₂SO₄ = 0.003375675 ÷ 2 = 0.0016878375 mol

Volume of H₂SO₄ = 20.00 mL = 20.00 × 10⁻³ L

Concentration of H₂SO₄ = Number of moles ÷ volume= 0.0016878375 mol ÷ 20.00 × 10⁻³ L= 84.39 M

Therefore, the concentration of the H₂SO₄ solution is 84.39 M.

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The concentration of the H2SO4 solution is 0.084432 M or 0.084 M

To find out the concentration of H2SO4, the balanced chemical equation of H2SO4 and NaOH will be used as shown below:

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

From the equation above, the stoichiometry of H2SO4 and NaOH is 1:2.

Moles of NaOH= Molarity × Volume in litres

= 0.1095 M × 0.03085 L

= 0.00337728 mol

Moles of H2SO4= 1/2 × Moles of NaOH

= 1/2 × 0.00337728

= 0.00168864 mol

Therefore, the concentration of the H2SO4 solution is given by:

C= Number of moles/Volume in litres

= 0.00168864 mol/0.02000 L

= 0.084432 M

Consequently, the concentration of the H2SO4 solution is 0.084432 M or 0.084 M

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Besides density, the physical properties that can you use to confirm that the gold is indeed real gold are the conduction of heat and electricity, malleability, ductility, lustrous nature, ductility, etc.

Gold is a type of chemical element which is classified into the category of precious metal. In the modern periodic table, it belongs to Group 11 and period 6. The chemical symbol which represents gold is Au.

The conduction of a test in order to confirm the identity of a given sample of gold is as follows:

Therefore, apart from density, several other physical properties are also used to confirm that the gold is indeed real gold.

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There are 3 hydrogen atoms present in 42 g of ammonium carbonate.

To find the number of hydrogen atoms in 42 g of ammonium carbonate, we first need to determine the number of moles of ammonium carbonate present. The molecular formula of ammonium carbonate is (NH4)2CO3, which contains a total of 8 hydrogen atoms. The molar mass of ammonium carbonate can be calculated by adding the atomic masses of all the atoms in the formula:
Molar mass of (NH4)2CO3 = (2 x 14.01 g/mol) + (4 x 1.01 g/mol) + 12.01 g/mol + (3 x 16.00 g/mol)
= 96.09 g/mol
Now, we can calculate the number of moles of ammonium carbonate in 42 g:
Number of moles = mass / molar mass
= 42 g / 96.09 g/mol
= 0.4374 mol
Finally, to find the number of hydrogen atoms in 0.4374 mol of ammonium carbonate, we multiply the number of moles by the number of hydrogen atoms per molecule:
Number of hydrogen atoms = 0.4374 mol x 8 hydrogen atoms/molecule
= 3.4992 hydrogen atoms
Since we cannot have a fraction of an atom, we can round down to get the final answer: Hence, there are 3 hydrogen atoms present in 42 g of ammonium carbonate.

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Answer:

phosphorus

Explanation:

Answer:

phosphorus

Explanation:

phosphorus has to lose electrons

Answer:

can u helP me?

Who was Agamemnon trying to persuade to come fight with him again?

Patroclus

Achilles

Odysseus

Hector

Explanation:

The rate of effusion of CO₂ under the same conditions is approximately 6.21 mole/min. Option B.

According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Therefore, we can use the ratio of molar masses to determine the rate of effusion of CO₂ (carbon dioxide) compared to NH₃ (ammonia).

The molar mass of NH₃ is approximately 17.03 g/mol, while the molar mass of CO₂ is approximately 44.01 g/mol.

Let's denote the rate of effusion of CO₂ as x mole/min. Using the ratio of molar masses, we can set up the following proportion:

(√molar mass of NH₃) / (√molar mass of CO₂) = rate of effusion of NH₃ / rate of effusion of CO₂

√17.03 / √44.01 = 2.40 mole/min / x mole/min

Solving for x, we find:

x = 2.40 mole/min * (√molar mass of CO₂) / (√molar mass of NH₃)

= 2.40 mole/min * (√44.01 g/mol) / (√17.03 g/mol)

≈ 6.21 mole/min

Therefore, the rate of effusion of CO₂ under the same conditions is approximately 6.21 mole/min.

The correct answer is (b) 6.21.

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Answer:

Transform boundaries because they're sliding past each other.

I hope this helped!

Answer:

129.50 N/kg

Explanation:

125 kg + 4.50 N

129.50 N/kg

Specific heat capacity is 139 J/kgC and molar heat capacity is 221 J/C/mol.

To calculate the specific heat capacity we will be using the equation
Q = mC∆T
Where Q = energy added or lost
m = the mass of the substancec = the specific heat of the substance
∆T is the difference in temperature usually written as T2-T1
Given Q=585Jm = 125.6 grams or 0.1256 kg

Where should find c =?
T2 =53.5°CT1 = 20 °C585 = 0.1256x C x(53.5–20)585 =4.2076xC585/4.2076
=139J/kg°C = C
Now,
Using the equation c_m = C/n

Where c_m = molar heat capacity
C = specific heat capacity
n = number of moles in kgc_m = 139/0.628
= 221 J/°C/mol

Therefore, specific heat capacity is 139 J/kgC and molar heat capacity is 221 J/C/mol.

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According to the given statement 56.31g mass of H₂O, in grams, must react to produce 50. 00 g of O₂.

The volume in grams of one mole of a compound is its molar mass. The molar mass of a substance can be found by adding the molality of its constituent atoms, as demonstrated in this video. The computed molar mass can then be used to convert both mass and the quantity of moles in the substance.

2H₂O ---> 2H₂ + O₂

Step 1: calculating moles O₂

⇒For each mole O₂ we have 2 moles of H₂

1) (50.00g O₂)/(32.00g/mol O₂) = 1.563mol O₂

⇒For each mole O₂ we have 2 moles of H2, so we multiply moles of O₂ by 2 to get moles of H₂O

Step 2: Calculating moles H₂O

2) (1.563mol O₂)(2mol H₂O/1mol O₂) = 3.125mol H₂O

Step 3: Calculating mass H₂O

3. (3.125mol H₂O) * (18.01g/mol H₂O) = 56.31g H₂O

We need 56.31g H₂O to make 50.00g O₂.

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Lead (Pb) has an atomic number of 82. Since it is a neutral atom, there are no excess electrons per lead atom.

Lead (Pb) has an atomic number of 82, indicating that it has 82 protons and 82 electrons in its neutral state. A neutral atom has an equal number of protons and electrons, resulting in a balanced charge. Therefore, there are no excess electrons per lead atom.

Each lead atom maintains electrical neutrality by having the same number of positively charged protons in its nucleus as negatively charged electrons in its electron cloud. This equilibrium of charges ensures that the lead atom is electrically neutral and does not possess any excess electrons or a net charge.

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Answer:

Ty for free points :)

Explanation:

Answer:

add a substace devider

Explanation:

i forgot what it was called but there is a chemical that allows the substances to seperste and than you can test the substaces ph and what it is

Answer:

no. of protons = atomic number of the elements

False. Electrostatic catalysis and covalent bonding interactions are two different types of chemical interactions that occur between atoms and molecules.

Electrostatic catalysis refers to a process in which a catalyst accelerates a chemical reaction by altering the charge distribution around the reactants, without participating in the reaction itself. This process relies on the electrostatic interactions between the catalyst and the reactants, which can help to stabilize the transition state of the reaction and lower the activation energy required for the reaction to proceed.

In contrast, covalent bonding interactions occur when atoms share electrons to form a chemical bond. These interactions are much stronger than electrostatic interactions and involve the sharing of electrons between atoms.

While both types of interactions can play important roles in chemical reactions, electrostatic catalysis does not typically involve covalent bonding interactions. Instead, it relies on the weaker electrostatic interactions between the catalyst and the reactants. These interactions can be enhanced by the geometric and electronic properties of the catalyst, as well as the nature of the reactants and the reaction conditions.

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The crash where the cone experience a stronger force is option D because: Crash 1: the force on the black hockey puck was stronger in this crash, so the force on the cone was also stronger.

The ball is drawn back to Earth by gravitational force. The ball returns to Earth as a result of friction. The ball is forced back toward Earth by magnetic force.

A puck's velocity changes when a player makes contact with it when it is still. He causes the puck to speed up, in other terms. The hockey stick's force, which causes the acceleration, is responsible. The velocity grows as long as this force is in motion.

Therefore, the force applied to an object must be larger than what is required for a progressive slowing down if the object must be slowed down quickly. For instance, a bicycle's brakes will slow or stop it more quickly the more force is given to it.

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See full question below

After hockey practice, Carissa and Keenan were playing a game where they were pushing some objects to get them to crash. They were using a cone and two different pucks—a black one with more mass for Crash 1 and a blue one with less mass for Crash 2. They want to know what happened to the cone.

Use the information from the diagram to answer.

In which crash did the cone experience a stronger force? How do you know?

answer choices

There was no force on the cone. In both crashes, only the hockey puck experienced a force.

The diagram doesn’t tell you anything about the force on the cone. It only gives information about the force on the pucks.

It was the same force in both crashes; the hockey puck changed speed by the same amount in each crash, so the force on the cone was the same each time.

Crash 1; the force on the black hockey puck was stronger in this crash, so the force on the cone was also stronger.

The empirical formula, which represents the simplest ratio of atoms in a compound, may be useful in the lab setting but not useful for predicting the properties and functions of materials.

In the lab, the empirical formula can be useful for identifying the composition of a compound, especially if the molecular formula is unknown. It can also help in determining the stoichiometry of a reaction, which can be important for conducting experiments. However, the empirical formula does not provide information about the actual number of atoms or the arrangement of atoms within a molecule, which can greatly affect the properties and functions of a material.

Therefore, while the empirical formula can be a useful tool in the lab setting, it may not be sufficient for predicting the properties and functions of materials. More detailed information about the molecular formula and structure is needed for accurate predictions.

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Answer: time

Explanation: hi

Answer:

1. (NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)

2. Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)

Explanation:

The dissociation of ammonium sulphide, (NH₄)₂S when dissolved in water is given in the equation below:

(NH₄)₂S(s) -----> NH₄+(aq) + S²-(aq)

However very little S²- ions are present in solution due to the very basic nature of the S²- ion (Kb = 1 x 105).

The ammonium ion being a better proton donor than water, donates a proton to sulphide ion to form hydrosulphide ion which exists in equilibrium with aqueous ammonia.

S²- (aq) + NH₄+ (aq) ⇌ SH- (aq) + NH₃ (aq)

Aqueous solutions of ammonium sulfide are smelly due to the release of hydrogen sulfide and ammonia, hence, their use in making stink bombs.

2. The reaction between aluminium nitrate and sodium phosphatein aqueous solution is a double decomposition reaction whish results in the precipitation of insoluble aluminium phosphate. The equation of the reaction is given below :

Al(NO₃)₃ (aq) + Na₃PO₄ (aq) ----> AlPO₄ (s) + 3 NaNO₃ (aq)

The net ionic equation is given below:

Al³+ (aq) + PO₄³+ (aq) ----> AlPO₄ (s)

Answer:

Methane + Bromine = (CH4+Br2)

Explanation:

When a mixture of methane and bromine is exposed to ultraviolet light - typically sunlight - a substitution reaction occurs and the organic product is bromomethane. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by bromine atoms. The reaction is a single-step reaction.

Answer:A square = b square

Explanation:

The correct option from the given statements concerning mixtures is (d) more than one correct response.

The statement (a) "The composition of a homogeneous mixture cannot vary" is incorrect as the composition of a homogeneous mixture can vary. For example, a mixture of salt and water is homogeneous and its composition can vary depending on the amount of salt and water mixed in it.

The statement (b) "A homogeneous mixture can have components present in two physical states" is correct. Homogeneous mixtures are mixtures that are uniform throughout their composition, meaning that there is no visible difference between the components of the mixture. For example, a mixture of ethanol and water is homogeneous and its components are present in two physical states (liquid and liquid).

The statement (c) "A heterogeneous mixture containing only one phase is an impossibility" is incorrect. A heterogeneous mixture is a mixture where the components are not evenly distributed and the mixture has different visible regions or phases. However, it is possible for a heterogeneous mixture to contain only one phase. For example, a mixture of oil and water is heterogeneous but can have only one phase.

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Answer:

you can do anything if you put your mind to it. Also, school is not everything in life there are amazinng things in life that you need school for. you got this.