How much heat is created by 79.2 g O2?

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Answer:

it belongs to alakli metals

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Answer:An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.

Explanation:Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.

This causes them to be soft and highly reactive because:

The single valance electron leads to weak bonds amongst the element's atoms which makes them soft

The elements want to lose the single valance electron so as to become stable so they will react with other elements to give away the electron.

Examples of alkali electrons include:

Lithium

Sodium

Potassium etc

In conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.

Answer:

1. hydrogen bonds form between neighboring hydrogen and oxygen antonyms of adjacent water molecules 2. the increasing in temperature results in expansion of the gas and for the decrease they cool the molecules down they will slow and the pressure 3. I don't really get what you're saying this one but if I'm reading it correctly d is density m is math and v is volume so the three variables in the density equation or something is density mass and volume

Explanation:

1.) Cohesion: Hydrogen Bonds Make Water Sticky

Oxygen’s lone pair and electronegativity create a pseudobond (hydrogen bond) with hydrogen.

In the case of water, hydrogen bonds form between neighboring hydrogen and oxygen atoms of adjacent water molecules. The attraction between individual water molecules creates a bond known as a hydrogen bond.

2.)This law states that the volume and temperature of gas have a direct relationship: As temperature increases, volume increases when pressure is held constant. Heating a gas increases the kinetic energy of the particles, causing the gas to expand.

3.) Scientific research identifies variables that are dependent, independent, and controlled (constants).

The dependent variable is usually the

My field is psychology, so the dependent variable is always some measure of

Most studies are merely correlational in that they attempt to identify some other variable(s) that might be used to predict the dependent variable. These predictors could be background factors (e.g., age, sex, ethnicity, social class), current situations, or even previous behaviors, performance, choices or attitudes.

What makes an experiment different is that it identifies an independent variable which will be tested out as a potential cause of the dependent variable. The experiment must then design some way to manipulate (intentionally vary) that independent variable. The best way to do this is by a randomized control trial: assign the participants of the sample into two groups, and treat those groups differently. The treatment is the independent variable.

Here is an example of psychiatric research. The dependent variable is depression, measured by scores on a standardized depression scale. The patients will be randomly assigned to one of two groups. One group gets a new psychiatric medication while the other gets a placebo (a fake pill that the patient might mistake for the real medication). The placebo is necessary to control the expectation factor. So, if there is a difference between the two groups in terms of resulting depression scores, we can attribute that difference to the impact of the independent variable (the medication) and not to the patient’s expectation that the medication will work.

Answer:

B). Electrons are more attracted to the oxygen atom, so there is a partial negative charge on the oxygen atom.

Explanation:

The above statement most aptly describes the polarity or uneven distribution of the density of electrons. In water, two hydrogen atoms and one atom of oxygen leading to the formation of polar covalent bonds. Thus, there is the presence of a partial negative charge close to the oxygen atom and partial positive charge near the hydrogen atom as the former has unshared pair of electrons. This polarity of water allows it to dissolve sugar or other ionic substances in it. Thus, option B is the correct answer.

Answer: Argon, neon, and xenon are examples of noble gases.

The noble gases (Group 18), which are found at the far right of the periodic table,Argon, neon, and xenon are examples of  noble or inert gases.

A gas is said to be inert if it does not easily undergo chemical reactions with other chemicals and does not easily produce chemical compounds as a result. The noble gases, formerly known as the inert gases, frequently do not react with a wide variety of substances.

Any one of the seven chemical elements that compose Group 18 (VIII a) of the periodic table is referred to as a noble gas. Helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson are the constituent elements (Og).

Due to their filled valence shells (octets), which render them incredibly nonreactive, the noble gases (Group 18), which are found at the far right of the periodic table, were formerly known as the "inert gases."

Thus, Argon, neon, and xenon are examples of  noble or inert gases.

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This problem is providing the initial volume and pressure of a gas as 675 mL and 1.5 atm and then mentions the final volume as 345 mL and asks for the new pressure of the gas.

Thus, we start off this problem by recalling the concept of Boyle's law as an inversely proportional relationship between pressure and volume:

\(P_1V_1=P_2V_2\)

In which we solve for the final pressure, \(P_2\) as shown below:

\(P_2=\frac{P_1V_1}{V_2} \\\\\\P_2=\frac{675mL*1.5atm}{345mL} \\\\\\P_2=2.93atm\)

Considering we had to cancel out the milliliters to obtain atm as units of pressure.

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Answer:

Yes, water molecules remain the same despite the phase

Explanation:

When water is in the form of ice, it molecules remain the same only the distance between the molecule. This distance is higher than that of the inter molecular distance between liquid water molecules. Due to this reason ice is lighter than water.  

Now in gaseous phase, the intermolecular distance increases thereby making it lighter than solid ice and liquid water.

Answer:

b

Explanation:

Answer:

Na+

Cl-

H+

OH-

Explanation:

NaCl + H2O -----> Na+. Cl- + H+ OH-

: PM2.5 is defined as the mass concentration of particles in the air less than or equal to 2.5 micrometers in diameter.Question 15: False, carbon dioxide (CO2) is not considered a criteria air pollutant.

Question 16: The closest answer is 50%, but the exact percentage is not provided in the question.Question 17: The term "photochemical smog" is most synonymous with ozone (O3), which is a criteria air pollutant.Question 18: True, attainment of ambient air quality standards requires that measured concentrations at all monitoring stations within an air district are below ambient air standards.

Question 14 asks about the definition of PM2.5. PM2.5 refers to particulate matter with a diameter less than or equal to 2.5 micrometers. It represents the mass concentration of particles suspended in the air, which are small enough to be inhaled into the respiratory system and can have adverse health effects.

Question 15 states whether carbon dioxide (CO2) is a criteria air pollutant. Criteria air pollutants are a set of pollutants regulated by environmental agencies due to their detrimental impact on air quality and human health. However, carbon dioxide is not considered a criteria air pollutant because it does not directly cause harm to human health or the environment in the same way as pollutants like ozone or particulate matter.

Question 16 asks about the percentage of carbon monoxide (CO) emissions from mobile sources in Santa Clara County. While the exact percentage is not provided in the question, the closest answer option is 50%. However, it is important to note that the precise percentage may vary depending on specific local conditions and emissions sources.

Question 17 inquires about the criteria air pollutant most synonymous with the term "photochemical smog." Photochemical smog is primarily associated with high levels of ground-level ozone (O3). Ozone is formed when nitrogen oxides (NOx) and volatile organic compounds (VOCs) react in the presence of sunlight, creating a hazy and polluted atmospheric condition.

Question 18 addresses the concept of "attainment" of ambient air quality standards. To achieve attainment, measured concentrations of pollutants at all monitoring stations within an air district must be below the established ambient air quality standards. This ensures that the air quality in the given area meets the required standards for protecting human health and the environment.

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The pH of the solution after 25.0 mL of KOH has been added to the acid is  10.37.

HCOOH is a weak acid that reacts with KOH (a strong base) to form the HCOO⁻ ion and water:

HCOOH + KOH → HCOO⁻ + H₂O

The balanced chemical equation shows that the stoichiometric ratio of HCOOH to KOH is 1:1, so 25.0 mL of 0.20 M KOH corresponds to the same amount of moles of HCOOH. This means that 25.0 mL of the original 0.35 M HCOOH solution has reacted with the 25.0 mL of 0.20 M KOH solution.

moles of HCOOH remaining = moles of HCOOH initially - moles of KOH added

moles of HCOOH initially = 0.35 mol/L × 0.0250 L = 0.00875 mol

moles of KOH added = 0.20 mol/L × 0.0250 L = 0.00500 mol

moles of HCOOH remaining = 0.00875 mol - 0.00500 mol = 0.00375 mol

The concentration of the remaining HCOOH is:

[ HCOOH ] = moles of HCOOH remaining / volume of solution remaining

= 0.00375 mol / (25.0 mL + 25.0 mL)

= 0.075 M

Now we can use the expression for the dissociation constant of HCOOH to calculate the pH of the solution:

Ka = [ H⁺ ][ HCOO⁻ ] / [ HCOOH ]

We can assume that the HCOO⁻ ion behaves as a weak base and calculate its concentration using the equation:

[ HCOO⁻ ] = Ka / [ HCOOH ]

[ HCOO⁻ ] = (1.77 × 10⁻⁴) / 0.075 ≈ 2.36 × 10⁻³ M

Now we can use the equation for the ionization of water to calculate [ H⁺ ]:

Kw = [ H⁺ ][ OH⁻ ]

1.00 × 10⁻¹⁴ = [ H⁺ ][ 2.36 × 10⁻³ ]

[ H⁺ ] = 4.24 × 10⁻¹¹ M

Therefore, the pH of the solution is:

pH = -log[H⁺] ≈ 10.37

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The equilibrium cοncentratiοn οf Zn²⁺ is measured as 0.065M  when   equal vοlumes οf 0.140 m AgNO₃  and 0.140 m ZnCl₂  sοlutiοn are mixed.

               2AgNO₃ + ZnCl₂ → 2AgCl + Zn(NO₃)₂

2 mοles οf AgNO₃ respοnd with 1 mοle οf ZnCl₂ tο deliver 2 mοles οf AgCl and 1 mοle οf Zn(NO₃)₂, as determined by the reasοnable cοnditiοn.

Mοlarity οf AgNO₃ is 0.14M

Mοlarity οf ZnCl₂ is 0.2M

ZnCl₂ is in excess and AgNO₃ is the limiting reactant because ZnCl₂ is mοre cοncentrated than that. Hοwever, ZnCl₂ at 0.07 M wοuld cοmpletely react with AgNO₃ at 0.14 M.

Disregarding the Ksp οf AgCl: There is nο Ag+ cοncentratiοn because the precipitate cοntains all οf the Ag.

(0.2 M ZnCl2 οriginally) - (0.07 M ZnCl2 reacted)

           = 0.13 M ZnCl₂ = 0.13 M Zn²⁺

Hοwever, tο dilute the initial sοlutiοn, an equal vοlume οf AgNO₃ has been added, resulting in a cοncentratiοn half that οf the initial sοlutiοn:

                             0.13 M Zn²⁺ / 2

                               = 0.065 M Zn²⁺

In additiοn, AgCl(s) is delivered quickly tο separate the AgCl particles frοm their arrangement.

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Stoichiometric equations for the combustion reactions ΔHf° (C2H2) = (2 x (-393.5)) + (-285.8) - (-1299.5) = +226.7 kJ mol-1(c) Acetylene hydrogenation reaction

Acetylene combustion reaction:C2H2 (g) + (5/2) O2 (g) → 2 CO2 (g) + H2O (l) ΔHc° = -1299.5 kJ mol-1 Hydrogen combustion reaction:2H2 (g) + O2 (g) → 2 H2O (l) ΔHc° = - 483.7 kJ mol-1Ethane combustion reaction:C2H6 (g) + (7/2) O2 (g) → 2 CO2 (g) + 3 H2O (l) ΔHc° = - 1560 kJ mol-1(b) Heat of formation method for verifying the standard heat of combustion of acetylene: The standard heat of combustion of acetylene from the heat of formation method is:ΔHc° (C2H2) = 2 ΔHf° (CO2) + ΔHf° (H2O) - 2 ΔHf° (C2H2) = -1299.5 kJ mol-1ΔHf° (CO2) = -393.5 kJ mol-1ΔHf° (H2O) = -285.8 kJ mol-1.

For verifying the answer in Q4(e)(1) using Hess's Law, we need to convert acetylene hydrogenation reaction into a combination of other reactions:Reaction 1:C2H2 (g) + (2.5) O2 (g) → 2 CO2 (g) + H2O (l) ΔH1 = -1299.5 kJ mol-1Reaction 2:2 CO2 (g) + 2.5 H2 (g) → C2H6 (g) + 5 O2 (g) ΔH2 = +1560 kJ mol-1After multiplying and adding the above equations, we get the required reaction as:C2H2 (g) + 2 H2 (g) → C2H6 (g) ΔH = -396.1 kJ mol-1.

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In architectural drawings, architects use precise dimensions or measurements to convey exact details about various parts of a drawing without leaving any scope for misinterpretation.

These dimensions are typically represented using numerical values, units of measurement, and specific symbols or annotations.

By including accurate dimensions in their drawings, architects ensure that the size, shape, and position of architectural elements are clearly communicated. This allows builders, contractors, and other professionals involved in the construction process to accurately interpret the design and implement it accordingly.

Architects also utilize symbols, notations, and labels to provide additional clarity and information about specific elements or features. For example, they may use symbols to represent different materials, finishes, or construction techniques. They may also include text annotations or callouts to provide further explanations or specifications.

Overall, these precise markings, dimensions, symbols, and annotations in architectural drawings play a crucial role in communicating design intent and facilitating accurate construction, reducing the potential for errors or misunderstandings during the building process.

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Answer:

A. the path for the movement of charge.

Explanation:

A circuit is the path for the movement of charge.

By determining whether each of the following equations is balanced or not balanced, we get : Part A: Not balanced, Part B: Balanced, Part C: Not balanced, Part D: Balanced

Let's analyze each equation to determine if it is balanced or not:

Part A: S(s) + O₂(g) ----> SO₃(g)

This equation is not balanced. On the left side, there is one sulfur atom, but on the right side, there are three sulfur atoms in the SO₃ molecule. Therefore, the equation is unbalanced.

Part B: Zn(s) + H₂SO₄(aq) ----> ZnSO₄(aq) + H₂(g)

This equation is balanced. On both sides of the equation, there is an equal number of atoms for each element: zinc (Zn), hydrogen (H), sulfur (S), and oxygen (O).

Part C: H₂(g) + O₂(g) ----> H₂O(g)

This equation is not balanced. On the left side, there are two hydrogen atoms, but on the right side, there is only one hydrogen atom in the H₂O molecule. Therefore, the equation is unbalanced.

Part D: C₃H₈(g) + 5O₂(g) ----> 3CO₂(g) + 4H₂O(g)

This equation is balanced. On both sides of the equation, there is an equal number of atoms for each element: carbon (C), hydrogen (H), and oxygen (O).

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Answer:

#Ar atoms = 1.47 x 10²²

Explanation:

PV = nRT => n = PV/RT

P = 8.80 X 10⁵ mmHg =  8.80 X 10⁵ mmHg/760mmHg·atm⁻¹ = 11.58atm

V = 137ml = 0.137 Liter

n = ? moles Ar = ? moles Ar x 6.02 x 10²³ Atoms Ar/mole Ar

R = 0.08206 L·atm/mol·K

T = 794K

PV = nRT => n = PV/RT = 11.58atm·0.137L / 0.08206L·atm/mol·K·794K

=> n = 0.0243mole Ar = 0.0244mole Ar x 6.02 x 10²³ atoms Ar/mole Ar

= 1.47 x 10²² atoms Ar

Answer:

A split-plot design lets us replicate our treatments twice using only 24 runs in a single time period, hence it is the most efficient use of resource.

Explanation:

hope this helps

Answer: (A) a cobalt atom contains 27 protons, and most cobalt atoms contain 32 neutrons.

Explanation: just answered the question and that was the correct answer

Answer:     The answer is :     A cobalt atom contains 27 protons and MOST  neutral cobalt atoms contain 32 electrons

Explanation:   I took the test

Who evolved the expression for the broad spectrum of hydrogen to encompass lines within the ultraviolet and infrared portions of the electromagnetic spectrum : Johannes Rydberg

Electromagnetic Spectrum: The electromagnetic (EM) spectrum is the range of all varieties of EM radiation. Radiation is the power that travels and spreads out because it is going – the visible light that comes from a lamp in your own home and the radio waves that come from a radio station are styles of electromagnetic radiation. Scientists name them all electromagnetic radiation. The waves of energy are known as electromagnetic (EM) due to the fact they've oscillated in electric-powered and magnetic fields. Scientists classify them through their frequency or wavelength, going from high to low frequency (quick to long wavelength).

The electromagnetic spectrum observed over a century ago is the idea behind our universe(Opens in a new tab). Without it we would not be able to see, the stars would not shine, and existence would not exist. It is one of the most important concepts that govern everything around us. Electromagnetic radiation occurs whenever a charged particle changes speed with an electron, i.e., increases or slows significantly. The power of the generated electromagnetic radiation comes from charged particles and is therefore lost with use.

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The energy of combustion per gram of vanillin is 158.61 kJ/g and per mole of vanillin is 1.044 kJ/mol.

Given data:

The energy released by combustion of benzoic acid is 26.42 kJ/g. The combustion of 0.1584 g benzoic acid increases the temperature of a bomb calorimeter by 2.54 °C. The heat capacity of the calorimeter needs to be calculated. A 0.2130 g sample of vanillin (C8H8O3) is then burned in the same calorimeter, and the temperature increases by 3.25 °C.

To calculate the heat capacity of the calorimeter, we will use the following formula:

Q = CΔT where, Q = Heat absorbed by the calorimeter ,C = Specific heat capacity of the calorimeter ,

ΔT = Change in temperature

We are given: Q = Heat absorbed by the calorimeter = Energy released during combustion of benzoic acid = 26.42 kJ/gΔT = Change in temperature = 2.54 °C = 2.54 K

And we need to find C.

Substituting the values in the formula, we get: 26.42 kJ/g = C × 2.54 KC = 26.42/2.54 = 10.39 kJ/K

Now, we need to calculate the energy of combustion per gram of vanillin.To calculate the energy of combustion per gram of vanillin, we will use the following formula:

Energy of combustion per gram of vanillin = Q/m

where, Q = Heat absorbed by the calorimeter  , m = Mass of the sample of vanillin burned

3.25 °C = 3.25 K

Mass of the sample of vanillin burned = 0.2130 g

Energy released during combustion of vanillin:ΔT = 3.25 K = Change in temperature

C = 10.39 kJ/K = Specific heat capacity of the calorimeter

Q = CΔT = 10.39 kJ/K × 3.25 K = 33.76 kJ

The energy released during combustion of vanillin:

Energy of combustion per gram of vanillin = Q/m= 33.76 kJ/0.2130 g= 158.61 kJ/g

Now, we need to calculate the energy of combustion per mole of vanillin.

To calculate the energy of combustion per mole of vanillin, we will use the following formula:

Energy of combustion per mole of vanillin = Energy of combustion per gram of vanillin/Molecular weight of vanillin

We are given:

Energy of combustion per gram of vanillin = 158.61 kJ/g

Molecular weight of vanillin = 8 × 12 + 8 × 1 + 3 × 16 = 152 g/mol

Energy of combustion per mole of vanillin:

Energy of combustion per mole of vanillin = Energy of combustion per gram of vanillin/Molecular weight of vanillin= 158.61 kJ/g / 152 g/mol= 1.044 kJ/mol

Hence, the energy of combustion per gram of vanillin is 158.61 kJ/g and per mole of vanillin is 1.044 kJ/mol.

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Answer:

Explanation:

Ca = +2

P = -3

Se = 0

NH3=

N = -3 , H = +1 , H3 = 3 x +1 = +3

-3 + (+3) = 0

NH3 = 0

NF3

N = -3, F = -1 , F3 = 3 x (-1) = -3

-3 + -3 = -6

Al2(CO3)3

CO3

C= +4, O = -2, O3 = 3 x (-2) = -6

CO3 = +4 + (-6) = +4 - 6 = -2

(CO3)3 = 3 x (-2) = -6

Al = +3, Al2 = 2 x (+3) = +6

Al2(CO3)3 = +6 + (-6)= 0

FALSE. PM2.5 is a term used for "fine" particles that are smaller than or equal to 2.5 µm (micrometers) in diameter, not larger.

Fine particles are categorized as part of the particulate matter pollution, which refers to a mixture of solid particles and liquid droplets found in the air. PM2.5 specifically denotes particles with a diameter of 2.5 micrometers or smaller.

These fine particles are of particular concern due to their small size. Because of their microscopic nature, PM2.5 particles can remain suspended in the air for longer periods and have the ability to penetrate deep into the respiratory system when inhaled. This makes them more likely to have adverse health effects on humans compared to larger particles that are filtered out by the nose and throat.

PM2.5 particles can originate from various sources, including combustion processes, industrial emissions, vehicle exhaust, and natural sources like dust and pollen. Their presence in the air can contribute to respiratory issues, cardiovascular problems, and other health complications. Monitoring and controlling PM2.5 levels are important for assessing and mitigating air pollution's impact on public health.

In conclusion, the statement that PM2.5 refers to "fine" particles larger than or equal to 2.5 µm in diameter is false. PM2.5 refers to fine particles that are smaller than or equal to 2.5 µm in diameter.

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Answer:

e.g converting 6.55×10^24 molecules of Sodium to moles.

We use Avogadro's number NA= 6.02×10^23 molecules.

Explanation:

\(6.02 \times {10 }^{23} molecules \: are \: in \: 1 \: mole \: of \: sodium \\ 6.55 \times {10}^{24} molecules \: will \: be \: in \: ( \frac{6.55 \times {10}^{24} \times 1}{6.02 \times {10}^{23} } ) \\ = 10.88 \: moles \: of \: sodium\)

Answer:

It does follow the law of conservation of mass.

Explanation:

You start with 4 phosphorous atoms and 6 oxygen atoms, you end with 4 phosphorous atoms and 6 oxygen atoms.

Answer:

The mass number of an isotope is the sum of neutrons and protons.

Explanation:

In any elemental isotope, the only things that will affect molar mass and mass number is the number of protons and neutrons. Electrons are not counted because we usually assume they are equal to the amount of protons and have no weight.

Protons are what gives the element its atomic number and the neutrons determine the type of isotope it is within the element.

For instance:

There can be a regular Carbon - 12

But there are isotopes like Carbon - 13 and Carbon - 14.

*The number of protons stays the same but the number of neutrons are different

Assume that every gas in the list will act perfectly. Using the mole fraction of O2 and the specified pressure, the partial pressure of O2 may be computed (P).

What is specified?

The following diagram illustrates the mathematical link between partial pressure of oxygen (P1) and oxygen mole fraction (X1):

P₁=X₁P

By dividing the total number of moles of gases (0.76 + 0.18 + 0.031 + 0.026) by the number of moles of O2 (0.18 mol), it is possible to determine the mole fraction of O2 as follows:

X₁= 0.18 / (0.76+0.18+0.031+0.026)= 0.1805

In this manner, the partial pressure of O2 (P1) is determined:

P1=0.1805x739mmHg, or 133.4mmHg

According to the estimate above, the partial pressure of oxygen is almost equal to 130 mm Hg. As a result, option C is the best one.

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Answer:

thermal energy is heat energy it is used for man kitchen appliances and also in nature from the sun for photosynthesys

Explanation:

Answer:

The answer is

Explanation:

The density of a substance can be found by using the formula

From the question

mass of block = 36 g

volume = 9 cm³

The density is

\(density = \frac{36}{9} \)

We have the final answer as

4.0 g/cm³

Hope this helps you

The balanced chemical equation for epsom salt being heated to produce magnesium sulfate solid and water vapor is:

\(MgSO4*7H2O (s) -- > MgSO4 (s) + 7H2O (g)\)

Magnesium sulphate heptahydrate, often known as epsomite, is a mineral with the chemical formula MgSO47H2O. Anhydrous salt plus water vapor plus hydrated salt. The molar ratio of salt to water can be calculated since many hydrates contain water in a stoichiometric amount. Magnesium sulphate heptahydrate crystals lose seven water molecules when they are gently heated, turning them into anhydrous magnesium sulphate. A single replacement occurs in this reaction. This is a single-replacement reaction because the general formula is met. The crystals of magnesium sulphate heptahydrate are typically needle-like and white, crystalline, or dazzling. It is easily soluble in water, more easily soluble in boiling water, and almost insoluble in alcohol.

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