if we are heated a metal with 28g mass then we add it in a coffee cup with 28g water. the water temperature raised from 19 degree to 23.8 oc. what was the initial temperature of the metal. the water specific heat of water is 4.184 j/g oc.

Answer:

carbon and oxygen

Explanation:

c stands for carbon and o stands for oxygen

Answer:

Covalent bonding

Explanation:

covalent bonding

Make suggestions for menu items free of the allergen. To make the order clear to the kitchen and service staff. To avoid cross-contact, deliver food in separate portions.

Handle any of the common allergens with caution, paying attention to what surfaces they have touched. To avoid coming into contact with other foods, wash your hands right away after handling allergen-containing foods and change gloves. Antihistamines. Since histamine causes many of the symptoms of an allergic reaction, antihistamines work by blocking its effects. You can buy a lot of antihistamines from your pharmacist without a prescription; stock up in case of emergency. It is best to use non-drowsy antihistamines. Food allergies have no known treatment, and the only way to prevent a reaction is to avoid the allergen. It helps to get rid of skin rashes, calms itching, and stops the skin from inflaming further.

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There is 1.0 mole of KNO₃ in 500.0 mL of a 2.0 M KNO₃ solution.

To determine the number of moles of KNO₃ in 500.0 mL of a 2.0 M KNO₃ solution, we need to use the equation:

moles = concentration (M) × volume (L)

Since 1 liter is equal to 1000 milliliters, we divide 500.0 mL by 1000 to get 0.5 L.

Next, we substitute the values into the equation:

moles = 2.0 M × 0.5 L

The concentration of 2.0 M indicates that there are 2.0 moles of KNO₃ in 1 liter of the solution. Therefore, multiplying the concentration (2.0 M) by the volume (0.5 L) gives us the number of moles of KNO₃:

moles = 2.0 M × 0.5 L = 1.0 mol

Hence, there is 1.0 mole of KNO3 in 500.0 mL of a 2.0 M KNO₃ solution.

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To solve this we must be knowing each and every concept related to cell. Therefore, the mass of DNA the scientist should obtain from a haploid human cell is  0.956.

In biology, a cell is a basic lattice unit that houses the essential components of life and is made up from which all living organisms are made. A single cell, for example a bacterium or yeast, is frequently a full organism in and of itself. As cells age, they develop specific roles.

These cells collaborate with some other specialized cells to form the foundation of huge multicellular creatures like humans and other mammals. Despite being far larger than atoms, cells are nonetheless incredibly tiny. A scientist obtained a mass of 0.0062 nanogram of dna from a diploid human cell. The mass of DNA the scientist should obtain from a haploid human cell is  0.956.

Therefore, the mass of DNA the scientist should obtain from a haploid human cell is 0.956.

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Answer:

Carbon

Explanation:

\(\frac{(98.9*12)+(1.1*13)}{100}\)

the answer gives you 12.011 which is closer to the amu of carbon

Answer:

See Explanation

Explanation:

For SF6;

Since;

1.25 g of S corresponds to 4.44g of F

1 g of sulphur corresponds to 1 * 4.44/1.25 = 3.55

For SF4;

Since;

1.88 g of S corresponds to 4.44g of F

1 g of sulphur corresponds to 1 * 4.44/ 1.88  = 2.36

Hence;

Mass of oxygen per gram of sulphur in SF6/Mass of oxygen per gram of sulphur in SF4

=

3.55/2.36 = 1.5

Hence the law of multiple proportion is obeyed here.

Answer:

Closed

Explanation:

Answer: chemical

Explanation:

Energy, potential energy, is stored in the covalent bonds holding atoms together in the form of molecules. This is often called chemical energy. Except at absolute zero (the coldest temperature it is possible to reach), all molecules move.

Answer:

For anyone that still needs the answer it's:

2.) Inspect the Bunsen burner and gas tubing for damage.

3.) Move her lab notebook and papers so they are away from the burner.

Explanation:

2 and 3 is right on edge

Bunsen burners are the laboratory equipment used in experiments. Maria should inspect the burner and move her papers and book away before burning. Thus, options b and c are correct.

Bunsen burners are the equipment used in laboratories for heating and sterilizing processes. It is a device used to produce flames using atmospheric oxygen gas and fuel in the burner.

While using the bunsen burner, laboratory safety procedures should be practiced to ensure the safety of the person as well as the working place. Before igniting the burner should be inspected and checked for any leakages.

The notebooks and papers must be placed away from eh burner to avoid any fire accidents as they are easily flammable and can catch fire.

Therefore, options b and c. gas tubing must be inspected before lighting the burner.

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The energy of an electron in a hydrogen atom with an orbit of n = 3 is -5.45 x 10⁻¹⁹ J.

To calculate the energy of an electron in a hydrogen atom with an orbit of n=3, we know that the value of k is given as k = 2.18 × 10⁻¹⁸ J. We can use the Rydberg formula to calculate the energy of an electron in the hydrogen atom. The Rydberg formula states that:

1/wavelength = R(1/n1² - 1/n2²)

where R is the Rydberg constant, which is equal to 1.097 x 10⁷ m⁻¹. We can use the formula E = hν to calculate the energy of a photon with frequency ν. Where h is the Planck constant, which is equal to 6.626 x 10⁻³⁴ J s.

The energy of an electron in the hydrogen atom can be calculated using the formula

E = -Rh/n²

where Rh is the Rydberg energy, which is equal to 2.18 x 10⁻¹⁸ J, and n is the principal quantum number. The negative sign indicates that the electron is bound to the nucleus.

Substituting n = 3 and Rh = 2.18 x 10⁻¹⁸ J into the formula gives:

E = - Rh/n²

= - 2.18 × 10⁻¹⁸ J / 3²

= - 5.45 x 10⁻¹⁹ J

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Answer:

Examples of longitudinal waves include: sound waves. ultrasound waves. seismic P-waves.

Explanation:

The balanced reaction equations for the described reactions are given below. The first reaction is a combination of a double displacement and a decomposition reaction, the second one represents single displacement reactions, and the third one is multiple double displacements.

The balanced reaction equation (including a double displacement and a decomposition) between aqueous sodium hydrogen carbonate and aqueous acetic acid (resulting in the formation of liquid water, gaseous carbon dioxide, and aqueous sodium acetate) looks like this:

NaHCO₃(aq) + AcOH(aq) → CO₂(g) + H₂O(l) + NaOAc(aq)

The balanced reaction equation between aluminum metal and iron(III) oxide (resulting in the formation of iron metal and aluminum oxide) looks like this:

2Al(s) + Fe₂O₃(s) → 2Fe(s) + Al₂O₃(s)

The balanced reaction equation between solid potassium oxide with aqueous magnesium chloride (resulting in the formation of solid magnesium hydroxide and aqueous potassium chloride) looks like this:

K₂O(s) + H₂O(l) + MgCl₂(aq) → Mg(OH)₂(s) + 2KCl(aq)

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Before any base is added, the pH of the naproxen solution is 2.60. At the half-equivalence point, the pH of the naproxen solution is 3.10. At the equivalence point, the pH of the naproxen solution is not provided in the given information.

Naproxen is a weak acid, and its dissociation reaction can be written as follows:

Naproxen (HA) ⇌ Naproxen⁻ (A⁻) + H⁺

The equilibrium constant expression for this reaction can be written as:

Ka = [A⁻][H⁺]/[HA]

where Ka is the acid dissociation constant, [A⁻] is the concentration of the conjugate base, [H⁺] is the concentration of hydrogen ions, and [HA] is the concentration of the weak acid.

The pKa of naproxen is given as 4.2, which means that:

pKa = -log Ka

4.2 = -log Ka

Ka = 10^(-4.2)

Ka = 6.31 x 10⁻⁵

(a) Before any base has been added, the concentration of H⁺ ions can be calculated using the expression:

Ka = [A⁻][H⁺]/[HA]

At the start of the titration, the concentration of the weak acid HA is 0.1 M, and the concentration of its conjugate base A^- is zero. Therefore, we can write:

Ka = [H⁺][A⁻]/[HA]

[H^+] = sqrt(Ka x [HA])

[H^+] = sqrt(6.31 x 10^(-5) x 0.1) = 2.52 x 10⁻³M

pH = -log[H⁺] = -log(2.52 x 10⁻³) = 2.60

Therefore, the pH before any base has been added is 2.60.

(b) At the half-equivalence point, half of the weak acid has been neutralized by the added base. At this point, the moles of weak acid and the moles of added base are equal. Therefore, the concentration of the weak acid and the conjugate base are equal.

At the half-equivalence point, the number of moles of NaOH added is:

0.5 L x 0.01 M = 0.005 moles

Since naproxen is a monoprotic acid, the number of moles of weak acid at the half-equivalence point is also 0.005 moles. Therefore, the concentration of weak acid is:

[HA] = 0.005 moles / 0.5 L = 0.01 M

At the half-equivalence point, the concentration of the conjugate base is also 0.01 M.

The equilibrium constant expression can be written as:

Ka = [A⁻][H⁺]/[HA]

At the half-equivalence point, [A⁻] = [HA] = 0.01 M. Therefore,

Ka = [H⁺]² / 0.01

[H^+] = sqrt(Ka x 0.01) = sqrt(6.31 x 10⁻⁵ x 0.01) = 7.94 x 10⁻⁴ M

pH = -log[H⁺] = -log(7.94 x 10⁻⁴) = 3.10

Therefore, the pH at the half-equivalence point is 3.10.

(c) At the equivalence point, all of the weak acid has been neutralized by the added base. Therefore, the concentration of the weak acid is zero, and the concentration of the conjugate base is equal to the initial concentration of the weak acid.

The number of moles of NaOH added at the equivalence point is:

0.5 L x 0.01 M = 0.005 moles

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In thin layer chromatography (TLC), the stationary phase is a thin, non-porous layer of a solid material and the mobile phase is a liquid. In column chromatography, the stationary phase is a solid material packed into a tube and the mobile phase is a liquid.

Thin layer chromatography (TLC) and column chromatography differ in their stationary and mobile phases. TLC and column chromatography differ in their stationary and mobile phases.

Both techniques can be used to identify compounds by comparing their retention times to those of known compounds. However, TLC is faster and more cost-effective than column chromatography, whereas column chromatography has higher resolution and can handle larger sample volumes.

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The equilibrium composition of CO₂ is determined to be 0.59 kmol.

To solve this problem, we can use the ideal gas law and the equilibrium constant expression for the reaction:

CO + 1/2O₂ ⇌ CO₂

Given the initial number of moles of CO and O₂, we can set up an ICE (Initial, Change, Equilibrium) table. Let's assume that x kmol of CO is consumed and converted to CO₂. Then, the change in the number of moles for each species is:

CO: -x kmol

O₂: -0.5x kmol

CO₂: +x kmol

At equilibrium, the number of moles of CO is (2 - x) kmol, O₂ is (3 - 0.5x) kmol, and CO₂ is x kmol. The equilibrium constant expression can be written as:

Kp = (P_CO₂) / (P_CO * P_O₂(1/2))

Given Kp = 16.461 and the pressure conditions, we can substitute the equilibrium partial pressures into the expression:

16.461 = x / ((2 - x) * (3 - 0.5x)(1/2))

Solving this equation yields x ≈ 0.59 kmol.

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The compound in which the octet is expanded to include 12 electrons is SF6 (sulphur hexafluoride). In SF6, the sulphur atom has six fluorine atoms surrounding it, and in order to bond with all six fluorine atoms, the sulphur atom must have an expanded octet, meaning it has 12 electrons in its outermost energy level.

The octet is expanded to include 12 electrons in compounds where the central atom can accommodate more than eight electrons. Such compounds typically involve elements from the 3rd period or below. A common example is sulfur hexafluoride (SF6), where sulfur has an expanded octet of 12 electrons.

The inorganic compound sulphur hexafluoride is a colourless, odourless, nonflammable, and nontoxic gas. With six fluorine atoms joined to a central sulphur atom, SF6 has an octahedral structure. As might be expected for a non-polar gas, SF6 dissolves poorly in water but readily in non-polar organic solvents. At sea level, it has a density of 6.12 g/L, which is significantly higher than the density of air (1.225 g/L). It is often carried as a compressed gas that has been liquefied.

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To determine the number of moles of the gaseous organic compound, we can 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.

First, we need to convert the density to mass per volume. The density of the gas is given as 340g/L. Therefore, the mass of 1 L of the gas is 340 g.

Next, we need to use the ideal gas law to calculate the number of moles. We know that the pressure is 1.7 atm, the temperature is 45°C (which is 318 K), and the volume can be calculated using the density and the molar mass of the compound. The molar mass can be determined from the molecular formula of the compound.

Assuming the compound is a hydrocarbon, we can use an average molar mass of 28. Thus, the volume of 1 mole of the gas can be calculated as follows:

V = (molar mass/density) × 1000 ml/L = (28/340) × 1000 = 82.35 ml/mol

Using the ideal gas law equation and plugging in the given values, we get:

n = (PV) / (RT) = (1.7 atm × 82.35 ml) / (0.0821 L atm/mol K ×318 K) = 0.839 mol

Therefore, the number of moles of the gaseous organic compound is 0.839 mol

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Answer:

Try doing this: 50.7/.9327 for the answer.

Group 1 metals, also known as alkali metals, are highly reactive and exhibit similar chemical behavior when reacting with water, oxygen, and chlorine.

Reaction with Water:

Observations:

The metal reacts vigorously with water, often producing effervescence or bubbling.

The metal may move rapidly on the water surface, releasing hydrogen gas.

The solution formed is alkaline and may turn phenolphthalein pink.

Word Equation:

Metal + Water → Metal Hydroxide + Hydrogen Gas

Symbol Equation (using sodium as an example):

2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

Reaction with Oxygen:

Observations:

The metal readily reacts with oxygen in the air.

The metal surface becomes tarnished or forms an oxide layer.

The reaction is often accompanied by the release of heat and sometimes light.

Word Equation:

Metal + Oxygen → Metal Oxide

Symbol Equation (using potassium as an example):

4K(s) + O₂(g) → 2K₂O(s)

Reaction with Chlorine:

Observations:

The metal vigorously reacts with chlorine gas.

A white or colored chloride salt is formed.

Word Equation:

Metal + Chlorine → Metal Chloride

Symbol Equation (using lithium as an example):

2Li(s) + Cl₂(g) → 2LiCl(s)

It's important to note that these equations are simplified and do not take into account the stoichiometric coefficients. The actual reaction equations would involve balanced coefficients to ensure the conservation of mass.

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In this example, the limiting reagent is the alkyl οr aryl halide (RX), as it is present in a lοwer quantity and will be cοmpletely cοnsumed during the reactiοn.

In οrganic chemistry, an alkyl grοup is an alkane missing οne hydrοgen. The term alkyl is intentiοnally unspecific tο include many pοssible substitutiοns. An acyclic alkyl has the general fοrmula οf −CnH₂n+1.

A cyclοalkyl grοup is derived frοm a cyclοalkane by remοval οf a hydrοgen atοm frοm a ring and has the general fοrmula −CnH₂n+1. Typically an alkyl is a part οf a larger mοlecule. In structural fοrmulae, the symbοl R is used tο designate a generic (unspecified) alkyl grοup. The smallest alkyl grοup is methyl, with the fοrmula −CH₃

In the preparatiοn οf Grignard reagents, the reactiοn typically invοlves the reactiοn between an alkyl οr aryl halide and magnesium metal in an ether sοlvent. The general equatiοn fοr the fοrmatiοn οf a Grignard reagent can be represented as:

RX + Mg -> RMgX

Tο determine the limiting reagent in this reactiοn, we need tο cοmpare the mοles οf the alkyl οr aryl halide (RX) and the mοles οf magnesium (Mg) and identify which reactant is present in the lοwer quantity. The reactant that is cοnsumed cοmpletely and limits the amοunt οf prοduct fοrmed is the limiting reagent.

Let's assume we have a specific example where we are preparing a Grignard reagent by reacting 2 mοles οf alkyl οr aryl halide (RX) with 3 mοles οf magnesium (Mg).

Mοles οf RX = 2 mοles

Mοles οf Mg = 3 mοles

Since the stοichiοmetric ratiο between RX and Mg is 1:1 (1 mοle οf RX reacts with 1 mοle οf Mg), we can see that we have an excess οf Mg in this example.

The stοichiοmetry indicates that 2 mοles οf RX wοuld require 2 mοles οf Mg fοr cοmplete reactiοn. Hοwever, we have 3 mοles οf Mg, which is mοre than enοugh tο react with the 2 mοles οf RX.

Therefοre, in this example, the limiting reagent is the alkyl οr aryl halide (RX), as it is present in a lοwer quantity and will be cοmpletely cοnsumed during the reactiοn.

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So, the mass gained by the copper electrode is also 3.12 grams.

According to the law of conservation of mass, the mass lost by the zinc electrode during the operation of the cell must be equal to the mass gained by the copper electrode. Therefore, if the zinc electrode loses 3.12 grams of mass, the copper electrode must gain exactly the same amount of mass. The copper ions in the solution receive electrons at this electrode to produce copper metal, which then deposits on the electrode.

As a result, the copper electrode's mass grows as the reaction progresses.  The cathode is where reduction happens. The cathode is the electrode whose mass rose as a result. On the anode, oxidation takes place. The anode is the electrode whose mass has dropped as a result. As the Cu electrode gains mass, the Pb electrode's mass falls. The oxidation-reduction process takes place between active electrodes. Metal atoms in the electrode would lose mass if they oxidised and entered solution because metals produce cations.

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Answer:

true

Explanation:

The chemical group which is most likely to be responsible for an orgainc molecule behaving as a base is Amino group.

Amines are the nitrogen containing organic bases. Simple organic molecules such as methylamine, H₃N-CH₃, and ethylamine, H₃N-CH₂CH₃ are the bronsted bases.

Functional group is a group of atoms in molecule which have the different and separate chemical properties.  It provide certain characteristics to the molecule. The functional group gets attached to the carbon atom.

Functional groups can behave as a base or an acid. It relies in their structure's ability to accept or to donate the H⁺ ions.

Amino group is represented as :

Amino group has ability to accept the H⁺ ions from the solution, it forms NH₃⁺.

                  --NH₂ + H⁺ →  NH₃⁺

Therefore, Amino group is most likely to be responsible for an organic molecule behaving as a base.

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Answer:

The second choice i think

Explanation:

Hope it helps

Answer:

What question?

Explanation:

Answer:

how u doing good?

<(^-^)>

helloooo

The alcohol that could be prepared by the greatest number of different combinations of Grignard reagents and carbonyl compounds is the primary alcohol. This is because primary alcohols can be prepared by the reaction of Grignard reagents with aldehydes, ketones, and esters. Secondary alcohols can also be prepared by the reaction of Grignard reagents with ketones, but they cannot be prepared from aldehydes or esters. Tertiary alcohols, on the other hand, cannot be prepared by the reaction of Grignard reagents with carbonyl compounds at all. Therefore, the primary alcohol has the greatest number of possible combinations of Grignard reagents and carbonyl compounds for its synthesis.
The alcohol that can be prepared by the greatest number of different combinations of Grignard reagents and carbonyl compounds is a secondary alcohol. This is because secondary alcohols can be synthesized from both aldehydes and ketones through the reaction with Grignard reagents, providing a wide range of possibilities for varying the reactants.Grignard reagent is an organometallic compound that is commonly used in organic chemistry as a nucleophile. It is named after its discoverer, French chemist Victor Grignard.

Grignard reagents are formed by the reaction of an alkyl halide or an aryl halide with magnesium metal in the presence of anhydrous ether. The resulting compound is a highly reactive species that can react with a wide range of electrophiles, such as carbonyl compounds, to form a new carbon-carbon bond.

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Answer: it think the answer is c.) The cell gets help from white blood cells or b.) The cell gets help from its neighbor cells.

Explanation: