Which of the following is NOT one of the common Science and Engineering Practices?
a) Planning and Carrying Out Investigations
b) Constructing Explanations and Designing Solutions
c) Communicating information
d) Forming opinions based on personal beliefs

Answer:

idrk im sorry and iknow this didnt help bt im sorry

Explanation:

you would lose $3,900 in total annual income over one year

If your monthly benefit at full retirement age (67) is $1300, then your annual benefit would be $15,600 ($1300 x 12).
If you choose to retire at age 62, your benefits will be reduced by 25%, which means your monthly benefit will be $975 ($1300 x 0.75).
Therefore, your annual benefit at age 62 will be $11,700 ($975 x 12).

To calculate how much you would lose in total annual income over one year, you can subtract your annual benefit at age 62 ($11,700) from your annual benefit at full retirement age ($15,600).

$15,600 - $11,700 = $3,900

So, if you retire at age 62, you would lose $3,900 in total annual income over one year compared to if you had waited until full retirement age to retire.

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Answer:

where is the evidance?

Explanation:

A student mixes two solutions, planning to produce a chemical reaction. The evidence that no chemical reaction is happening is a change in appearance only.

A chemical reaction is in which there is reactant and product and also involve chemical substance. When two or more reactant combines, the substance changed, and the result is called product.

The options are attached:

(a.) a color change

(b.) a temperature decrease

(c.) a new smell

(d.) a change in appearance only

Thus, the correct option is (d.) a change in appearance only.

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The statement "Certified organic farms must avoid the use of synthetic fertilizers and toxic pesticides" is generally true. Organic farming practices prioritize the use of natural methods and substances for fertilization and pest control, avoiding synthetic fertilizers and toxic pesticides commonly used in conventional agriculture.

Certified organic farms are subject to strict regulations and standards that govern their farming practices. One of the core principles of organic farming is the promotion of ecological balance and environmental sustainability. To achieve this, organic farms are required to minimize their reliance on synthetic inputs, including fertilizers and pesticides.

Organic farmers primarily rely on natural fertilizers such as compost, manure, and cover crops to enrich the soil and provide essential nutrients to plants. These methods enhance soil health, promote biodiversity, and reduce the risk of harmful chemical runoff into water sources.

Similarly, organic farmers employ various strategies to manage pests and diseases without the use of toxic synthetic pesticides. They utilize techniques such as crop rotation, natural predators, beneficial insects, and cultural practices to control pests and maintain plant health.

By avoiding the use of synthetic fertilizers and toxic pesticides, certified organic farms aim to produce food in a more environmentally friendly and sustainable manner. This approach supports the principles of organic agriculture, which prioritize soil health, biodiversity, and the reduction of potential harm to human health and the environment.

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Buckminsterfullerene, also known as C60 or buckyball, is a unique and fascinating molecule in the field of chemistry.

It was first discovered in 1985 by a team of scientists led by Richard Smalley, Robert Curl, and Harold Kroto, who were awarded the Nobel Prize in Chemistry in 1996 for their discovery.

Buckminsterfullerene is a carbon allotrope composed of 60 carbon atoms arranged in a hollow sphere resembling a soccer ball. Its name is derived from its resemblance to the geodesic dome designs created by architect Buckminster Fuller.

One of the remarkable aspects of buckminsterfullerene is its symmetrical structure, which confers extraordinary stability. Its structure allows for the distribution of strain throughout the molecule, making it highly resistant to chemical reactions and providing exceptional thermal and mechanical stability.

Buckminsterfullerene exhibits a range of unique properties that have attracted significant scientific interest. It is an excellent electron acceptor and can undergo various chemical reactions due to its high reactivity. Its electronic properties have applications in organic electronics, photovoltaics, and molecular electronics.

Moreover, buckminsterfullerene has shown potential in various fields, including medicine, material science, and nanotechnology. Its hollow structure can encapsulate other atoms or molecules, making it useful for drug delivery systems.

In summary, buckminsterfullerene is a fascinating carbon molecule with a distinctive structure and exceptional properties. Its discovery has opened up new avenues for research and applications in chemistry, physics, materials science, and other interdisciplinary fields.

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Answer:

The oxygen atom in water is partially negative and the hydrogen atoms are partially positive. In ethanol, the oxygen atom is partially negative and the carbon and hydrogen atoms are partially positive.

Answer:

D

Explanation:

cracks from temperature changes is a type of physical weathering often known as freezing and thawing

The element X forms the chloride XCl₄ ( 4 being a sub script of Cl ) containing 75% Cl by mass. The  X element in the compound is titanium.

The molecular weight of chlorine is 35.5 g/mol.

Four chlorine atoms, or molecules, make up the chemical XCl₄and account for 0.75 of its total weight.

All four chlorine atoms weigh a total of 4×35 = 140 g/mol

The compound's total weight of all chlorine atoms is now 140/0.75.

The compound's overall weight is now equal to 186.66≈ 187

The mass of element X is equal to 187 - 140 = 47 g/mol.

If we match any element from the periodic table, we can discover that the molecular weight of titanium is almost equivalent to 47.

TiCl₄ is the substance, and its molecular weight is 140 g/mol.

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The reaction at the given concentration proceed towards left to form reactants. Thus, option C is correct.

The given balanced equation has been:

\(\rm 2\;NOCl\;\leftrightharpoons 2\;NO\;+\;Cl_2\)

The equilibrium constant, Keq for the reaction, is given as:

\(Keq=\rm \dfrac{[NO]^2\;[Cl_2]}{[NOCl]^2}\)

The given concentration of NOCl is 1 M

The given concentration of NO is  0. 5 M

The given concentration of Chlorine is  0. 5 M.

The value of equilibrium constant for given concentrations are:

\(Keq=\dfrac{[ 0. 5]^2\;[ 0. 5]}{[1]^2}\\ Keq=0.125\)

The given value of equilibrium constant is \(\rm 4.4\;\times\;10^-^4\). Since the given value of constant is less than the reaction constant, the reaction proceeds towards the reactant  to establish equilibrium.

Thus, option C is correct.

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Answer:

ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation. Example 4.8. Moles of Reactant Required in a Reaction.

Explanation:

The density of a metal crystallizing into a face-centered cubic (FCC) unit cell can be calculated using the given atomic radius. In this case, the atomic radius is 174 picometers (pm).

The density of a material is defined as its mass per unit volume. To determine the density, we need to find the mass and volume of the unit cell. In an FCC structure, there are four atoms at the corners of the unit cell and one atom at the center of each face. Each of these atoms contributes to the overall mass of the unit cell.

The mass of the unit cell can be calculated by multiplying the atomic mass of the metal by the number of atoms in the unit cell. The atomic mass can be obtained from the periodic table.

The volume of the unit cell can be determined by considering the arrangement of atoms in the FCC structure. Each atom at the corner contributes 1/8th of its volume to the unit cell, while each atom at the face contributes 1/2 of its volume.

Once the mass and volume of the unit cell are determined, the density can be calculated by dividing the mass by the volume.

In conclusion, the density of the metal can be calculated by dividing the mass of the unit cell (determined by multiplying the atomic mass by the number of atoms in the unit cell) by the volume of the unit cell (determined by considering the arrangement of atoms in the FCC structure). This calculation allows us to obtain the density of the metal based on the given atomic radius of 174 pm.

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Due to their improved charge transfer and great environmental stability, 2D Dion-Jacobson (DJ) perovskites have recently received a lot of attention.

Unfortunately, due to the scarcity of high-quality single crystals for precise measurements, their fundamental optoelectronic capabilities are mainly unknown. Here, a reactive, low-temperature-gradient crystallization method is created using 1,4-butanediammonium as a short-chain insulating spacer to generate high-quality 2D perovskite single crystal (BDAPbI4). It is discovered that the BDAPbI4 single crystal exhibits a direct bandgap with effective charge collection (μτ = 1.45 × 10−3 cm2 V−1). The BDAPbI4 single crystal in particular exhibits the expected high ion migration activation energy (0.88 eV).

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While Bromine water can be used to test for the presence of unsaturation in organic compounds, it may not be the best approach for detecting triple bonds. Other tests should be considered for a more accurate analysis.

Bromine water, also known as Br2 or Bromine, is commonly used in organic chemistry to test for the presence of unsaturation in organic compounds. Unsaturated compounds, such as alkenes and alkynes, are capable of reacting with Bromine water to produce characteristic color changes. This reaction occurs due to the electrophilic addition of Bromine to the double or triple bond, resulting in the formation of a Di halogenated compound.  
However, the use of Bromine water to test for the level of unsaturation in an organic compound consisting of triple bonds may not be the best approach. This is because triple bonds are much stronger than double bonds and are less reactive towards electrophilic additions. Therefore, the reaction with Bromine water may not occur as readily, resulting in false negative results.
Other tests, such as the Baeyer's test or the ozone test, may be more suitable for detecting the presence of triple bonds in organic compounds. These tests involve the oxidative cleavage of the triple bond and the subsequent formation of characteristic products that can be identified using various techniques.

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Answer:

temperate

Explanation:

The reaction used in reverse of glycolysis for gluconeogenesis. The three irreversible reactions are catalyzed by : hexokinase, phosphofructokinase-1, pyruvate kinase.

Option E is correct .

Most of the time, this is because gluconeogenesis needs to avoid the energy-saving and irreversible steps of glycolysis. In gluconeogenesis, these three irreversible steps cannot be reversed directly due to their exergonic nature.

Pyruvate is produced by the reactions of glycolysis on glucose 6-phosphate. The whole interaction is cytosolic. Fructose 6-phosphate is produced by the reversible isomerization of glucose 6-phosphate. The physiologically irreversible phosphorylation of fructose 6-phosphate to shape fructose 1,6-bisphosphate is catalyzed by phosphofructokinase.

Incomplete question :

Seven of the ten reactions of glycolysis are reversible (DG near zero) and can be used in reverse of glycolysis for gluconeogenesis. The three irreversible reactions are catalyzed by:


A. hexokinase, phosphoglycerate kinase, pyruvate kinase.


B.  triose phosphate isomerase, phosphoglycerate mutase, pyruvate kinase.


C.  enolase, phosphoglycerate kinase, phosphofructokinase-1.


D.  hexokinase, phosphoglucoisomerase, glyceraldehyde-3-phosphate dehydrogenase.


E.  hexokinase, phosphofructokinase-1, pyruvate kinase.

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Answer:

B.) Multicellular organisms are made of many cells and have specialized shapes that allow them to perform specific functions

Explanation:

Multicellular means that there is more than one cell.

Hope this helps!

Yes for instance.

once given the physical properties of the alkali metal you can be able to indicate the group and where you can find them in the periodic table. properties like it being soft and having relatively low melting point (Li, Na, K, RB, CS, Fr)

or once been said that it reacts with group 7 Elements that means we are quick to know it out of those elements.

now to answer your question specifically, if the information given upon says an element burns in air with a yellow flame. Then we are quick to say it's Sodium.

so yeah.

pCO2 is an important factor that affects various aspects of water chemistry and the impacts of ocean acidification. When pCO2 increases, the concentration of total CO2 dissolved in water also increases. This leads to changes in pH, which decreases due to increasing atmospheric CO2 concentration.

When pCO2 rises, the concentration of only CO2 dissolved in water increases. The dissolved CO2 forms carbonic acid, which contributes to the acidification of the ocean. This increase in CO2 affects the equilibrium between CO2, HCO3-, and CO3^2-, shifting it towards higher levels of dissolved CO2 and H+ ions, resulting in a lower pH.

In terms of the impacts of ocean acidification on different organisms, the effects can vary depending on their specific needs. An organism that requires CO2 is likely to fare better under ocean acidification since the increase in dissolved CO2 can provide them with a favorable environment. However, organisms that rely on HCO3- or CO3^2- may fare worse under ocean acidification, as the lower pH interferes with the availability of these carbonate ions, which are essential for shell formation and calcification in some marine organisms.

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Answer:

A- Before the mechanic left, the molecules were moving away from each other. When she returned, they were moving around each other.

Explanation:

Brainliest?

Answer:

3. a. I⁻

Explanation:

The charge of the ion is determined by which group (vertical on the periodic table) the element is in. Iodine is in the halogens and therefore is most likely to have a  1- charge.

Answer:

person and earth since they are closer to the core ? maybe

Answer:

the charge of the aluminum ion is +3

Answer: A FILTERING

Explanation:

WHEN YOU FILTER YOU SEPARATE (sorry for caps) btw can you check out my recent question plz

The correct response is b) N - H. Because nitrogen is one of the most electronegative elements in the periodic table, its bond with hydrogen will be the most polar of all.

A polar covalent bond is a covalent association in which the atoms' electron attraction is uneven, resulting in unequal electron sharing. In a polar covalent bond, also known as a polar bond, the distribution of electrons surrounding the molecule is no longer symmetrical. Polar covalent bonds have an unequal distribution of electrons. The difference in electronegativity between the two elements would be between 0.4 and 1.7. The electronegativity gap between polar and nonpolar bonds is less than 0.4.

In this case, the right answer is b) N - H. Because nitrogen is one of the periodic table's most electronegative elements, its bond with hydrogen will be the most polar of all.

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Answer:

19.05 g have been oxidized

Explanation:

This is the unballanced equation:

H₂SO₄ + Cu → CuSO₄ + SO₂ + H₂O

Let's balance as a redox one:

Cu° oxidizes to Cu²⁺

Cu → Cu²⁺ + 2e⁻

SO₄⁻² reduces to SO₂

We add 2 water in the left side in order to ballance the O, so we add 4 protons to the right side, to ballance the H.

SO₄⁻² + 4H⁺ + 2e⁻ → SO₂ + 2H₂O

In the 2 half reactions, we have the same electrons, we sum both equations, so we have the main equation ballanced (we cancel the e⁻):

2H₂SO₄ + Cu → CuSO₄ + SO₂ + 2H₂O

We determine the moles of sulfuric acid:

M = moles /L → 0.5 mol/L . 1.2L = 0.6 moles

As ratio is 2:1, we propose

2 moles of sulfuric react to 1 mol of Cu

0.6 moles of sulfuric will react to (0.6 . 1) /2 = 0.3 moles of Cu

We determine the mass: 0.3 mol . 63.5 g/mol = 19.05 g

The physical mechanism causing the change in strength when using brine as the pore fluid is the presence of salt ions that weaken the interparticle bonds. Salt can reduce the shear strength of clays by increasing the repulsive forces between clay particles.

The strength of clay types at 50% water content can vary depending on whether DI water or brine is used as the pore fluid. Generally, there is a significant difference in strength between the two.

The presence of salt in brine can have an effect on the shear strength of clays. When salt is dissolved in water, it creates ions that can interact with the clay particles. These interactions can lead to the formation of electrical double layers around the clay particles, which can increase the interparticle repulsion and decrease the shear strength of the clay.

On the other hand, when DI water is used as the pore fluid, there is no presence of salt ions to affect the interparticle interactions. As a result, the clay particles can have stronger bonds and higher shear strength compared to when brine is present.

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Kc=28.7  is the equilibrium constant

[CO]=26.7g(28.01g/mol)(5.25L)=0.182M

[H2]=2.33g(2.016g/mol)(5.25L)=0.220M

[CH3OH]=8.65g(32.04g/mol)(5.25L)=0.0514M

Kc= x/(0.182−x)(0.220−2x)2

   = 0.0514/(0.182−0.0514)(0.220−2(0.0514))2

   = 28.7

a number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

So, this shows the equilibirum constant is 28.7

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Answer:

54.26 × 10²³ representative particles

Explanation:

Given data:

Number of moles of water = 9.01 mol

Number of representative particle = ?

Solution:

one mole of any substance contain 6.022 ×10²³ representative particles.

9.01 mol × 6.022 ×10²³ representative particles

54.26  ×10²³ representative particles

The number 6.022 × 10²³ is called Avogadro number.

"It is the number of atoms , ions , molecules or representative particles  in one gram atom of element, one gram molecules of compound and one gram ions of a substance "

Answer:

he distance to a star can be found by comparing its luminosity to its apparent brightness.

Explanation: A star that is far away will be very dim compared to its luminosity. So it proves that  the ratio of its luminosity to apparent brightness will be greater than for a nearby star.