If a substance has a bitter taste, feels slippery , conducts electricity, and has a high pH, it is a ?

The specific heat in J/(g·ºC) of an unknown substance if a 2. 50-g sample releases 12. 0 cal as its temperature changes from 25. 0ºC to 20. 0ºC. 2.02  J/(g·ºC).

The specific heat of the unknown substance can be calculated using the formula:
q = m x c x ΔT

where q is the heat released, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.

First, we need to convert the given heat release from calories to joules:
12.0 cal x 4.184 J/cal = 50.208 J

Next, we can plug in the given values and solve for c:
50.208 J = 2.50 g x c x (25.0°C - 20.0°C)
c = 2.02 J/(g·°C)


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The pH after the addition of 19.0 ml of 0.500 M HNO₃ to a 75.0 ml volume of 0.200 M NH₃ (Kb = 1.8 * 10⁻⁵) is 9.11.

1. Calculate moles of NH₃ and HNO₃: moles NH₃ = 75.0 ml * 0.200 mol/L = 15.0 mmol, moles HNO₃ = 19.0 ml * 0.500 mol/L = 9.5 mmol

2. Find moles of NH₃ remaining: 15.0 mmol - 9.5 mmol = 5.5 mmol

3. Calculate new concentrations: [NH₃] = 5.5 mmol / (75.0 ml + 19.0 ml) = 0.055 mol/L, [NH₄⁺] = 9.5 mmol / (75.0 ml + 19.0 ml) = 0.095 mol/L

4. Apply the Henderson-Hasselbalch equation: pH = pKa + log([NH₃]/[NH₄⁺])

5. Find pKa from Kb: pKa = 14 - log(Kb) = 14 - log(1.8 * 10⁻⁵) = 9.74

6. Calculate pH: pH = 9.74 + log(0.055/0.095) = 9.11

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To determine the molarity of the diluted solution, we need to use the equation:

M1V1 = M2V2

where M1 is the initial molarity of the solution, V1 is the initial volume of the solution, M2 is the final molarity of the solution, and V2 is the final volume of the solution.

In this case, the initial solution is a 0.75 M K2SO4 solution with a volume of 550 mL, and water is added to make a final volume of 450 mL. We can write:

M1 = 0.75 M

V1 = 550 mL

V2 = 450 mL

We can solve for M2:

M1V1 = M2V2

0.75 M × 550 mL = M2 × 450 mL

M2 = (0.75 M × 550 mL) / 450 mL

M2 = 0.92 M

Therefore, the molarity of the diluted solution is 0.92 M.

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1.35 liters of a 0.26 M solution of K2(MnO4) would contain 75 g of K2(MnO4).

To determine the volume of a 0.26 M solution of K2(MnO4) needed to contain 75 g of K2(MnO4), we need to use the formula:

Molarity (M) = moles of solute / volume of solution (L)

First, convert the mass of K2(MnO4) to moles using its molar mass:

Molar mass of K2(MnO4) = 2 * (39.1 g/mol for K) + (54.9 g/mol for Mn) + 4 * (16 g/mol for O) = 214.2 g/mol

Moles of K2(MnO4) = 75 g / 214.2 g/mol ≈ 0.35 moles

Now use the molarity formula to find the volume:

0.26 M = 0.35 moles / volume (L)

Volume (L) = 0.35 moles / 0.26 M ≈ 1.35 L

So, approximately 1.35 liters of a 0.26 M solution of K2(MnO4) would contain 75 g of K2(MnO4).

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Answer:

In Valence Bond Theory (VBT), the water molecule is formed by overlapping of two hydrogen 1s orbitals with two hybridized oxygen orbitals. The oxygen atom in the water molecule has two unpaired electrons in two 2p orbitals and two paired electrons in two 2s orbitals. It hybridizes the 2s and 2p orbitals to form four hybridized sp3 orbitals. These four sp3 hybridized orbitals point towards the corners of a tetrahedron.

The two hybridized orbitals of oxygen containing unpaired electrons overlap with the 1s orbitals of two hydrogen atoms. This overlapping results in the formation of two O-H sigma (σ) bonds. The two remaining hybridized orbitals containing the paired electrons do not participate in bond formation.

The bond angle in the water molecule is 104.5°, which is less than the tetrahedral angle (109.5°) because the two lone pairs of electrons on the oxygen atom exert greater repulsion than the two bonding pairs. This causes the bonding pairs to be pushed closer together, resulting in a smaller bond angle.

4.3 mL of 16.2 M [tex]NH3[/tex]is required to prepare 350.0 mL of 0.200 M [tex]NH3[/tex]

The molarity equation is:

Molarity (M) = moles of solute / liters of solution

We can rearrange this equation to solve for the number of moles of solute:

moles of solute = Molarity (M) x liters of solution

We can use this equation to determine the number of moles of [tex]NH3[/tex]required to prepare the 350.0 mL of 0.200 M [tex]NH3[/tex] solution:

moles of [tex]NH3[/tex] = (0.200 M) x (0.350 L) = 0.070 moles [tex]NH3[/tex]

Now, we need to determine the volume of 16.2 M [tex]NH3[/tex]required to obtain 0.070 moles of [tex]NH3[/tex]. We can use the following equation:

moles of solute = Molarity (M) x liters of solution

Rearranging the equation to solve for the volume of solution, we get:

liters of solution = moles of solute / Molarity (M)

Plugging in the values, we get:

liters of solution = 0.070 moles  / 16.2 M[tex]NH3[/tex] = 0.0043 L

Converting this to milliliters, we get:

volume of 16.2 M [tex]NH3[/tex] = 0.0043 L x 1000 mL/L = 4.3 mL

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This problem can be solved using Boyle's Law, which posits that the pressure of any given gas will be inversely proportional to its volume when temperature is kept as a constant.

Mathematically Boyle's Law can be expressed as:

P₁V₁ = P₂V₂

where P₁ and V₁ are the initial pressure and volume, respectively, and P₂ and V₂ are the final pressure and volume, respectively.

We are given that:

P₁ = 1.15 atm

V₁ = 640 mL

T = 23.9 °C (which is 297.05 K, using the Kelvin temperature scale)

We need to find V₂ when P₂ = 1.43 atm.

Using Boyle's Law, we can set up the following equation:

P₁V₁ = P₂V₂

(1.15 atm)(640 mL) = (1.43 atm)(V₂)

Solving for V₂:

V₂ = (1.15 atm)(640 mL) / (1.43 atm)

V₂ = 514.69 mL

Therefore, the final volume of the methane gas is 514.69 mL when compressed at constant temperature until its pressure is 1.43 atm.

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All else being equal, a reaction with a higher activation energy will have a slower reaction rate compared to one with a lower activation energy.

Activation energy refers to the minimum amount of energy required for a chemical reaction to occur. The higher the activation energy, the more energy is required to initiate the reaction, and thus the slower the reaction rate.

This is because a higher activation energy means that fewer reactant molecules will have enough energy to overcome the energy barrier and form products. Therefore, reactions with higher activation energies require more energy input to proceed and will typically have a slower reaction rate than those with lower activation energies.

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Answer:

0.79 L of concentrated nitric acid is needed to prepare 5.0 L of a 2.5 M solution.

Explanation:

We can use the formula:[tex]M_1V_1 = M_2V_2[/tex]where [tex]M_1[/tex] is the concentration of the concentrated nitric acid, [tex]V_1[/tex] is the volume of concentrated nitric acid needed, [tex]M_2[/tex] is the desired concentration of the final solution, and [tex]V_2[/tex] is the final volume of the solution.Plugging in the given values, we get:[tex](15.8 \text{ M})(V_1) = (2.5 \text{ M})(5.0 \text{ L})[/tex]Solving for [tex]V_1[/tex], we get:[tex]V_1 = \frac{(2.5 \text{ M})(5.0 \text{ L})}{15.8 \text{ M}} \approx 0.79 \text{ L}[/tex]Therefore, approximately 0.79 L of concentrated nitric acid is needed to prepare 5.0 L of a 2.5 M solution.

Answer:

0.79 L

I hope this helps! Cheers ^^

To find the density of ammonia (NH3) at 646 torr and 10°C, we need to use the Ideal gas law equation:

PV = nRT

Where R is the gas constant, n is the number of moles, P is pressure, V is volume, and T is temperature in Kelvin.

We must first change the pressure from torr to atm:

646 torr = 0.852 atm

The temperature is then changed from Celsius to Kelvin:

10°C + 273.15 = 283.15 K

Now, we can rearrange the ideal gas law equation to solve for density (d):

d = (PM) / (RT)

M is the ammonia's molar mass.

With the supplied values and constants, we obtain:

d = (0.852 atm)(17.04 g/mol) / ((0.0821 atm*L)/(mol*K))(283.15 K)

d = 0.736 g/L

Therefore, the density of ammonia at 646 torr and 10°C is 0.736 g/L.

What do you mean by density of ammonia?

The density of ammonia refers to the mass of ammonia gas per unit volume. The standard temperature and pressure (STP), which is defined as a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atmosphere (101.325 kilopascals or 760 millimeters of mercury), is used to measure the density of ammonia, a colorless gas that is lighter than air.

At STP, the density of ammonia gas is approximately 0.771 grams per liter (g/L) or 0.771 kilograms per cubic meter (kg/m3). However, the density of ammonia can vary depending on the temperature, pressure, and other factors such as the presence of impurities or moisture.

The density of ammonia is an important property in many applications, particularly in the chemical industry. It is used to calculate the amount of ammonia needed for a particular reaction or process, and can also be used to determine the mass or volume of ammonia gas in a storage tank or container.

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The specific heat capacity of iron is 0.449 J/g°C.

The quantity of heat energy needed to raise the temperature of one gram of a substance by one degree Celsius is the substance's specific heat capacity.

The specific heat capacity of iron can be calculated using the formula:

q = mcΔT

where q is the heat energy absorbed by the iron, m is the mass of the iron, c is the specific heat capacity of iron, and ΔT is the change in temperature of the iron.

Substituting the given values:

2056.5 J = (17.98 g) × c × (200°C - 25°C)

2056.5 J = (17.98 g) × c × (175°C)

Solving for c:

c = 0.449 J/g°C

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Student has a sample of copper (I) sulfate hydrate, then the percent of water in the sample is 40%.

In chemistry, a hydrate is a compound that contains water molecules that are chemically bound to the atoms or ions of the compound.

Mass of the anhydrous salt is given as 12 grams.

So, mass of water = total mass - mass of anhydrous salt

mass of water = 20 g - 12 g

mass of water = 8 g

Now, % water = (mass of water ÷ total mass) × 100

% water = (8 g ÷ 20 g) × 100

% water = 40%

Therefore, the percent of water in the sample is 40%.

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The pressure produced when the temperature is raised to 600 K is 1.77 atm.

The pressure of a gas is directly proportional to its temperature, according to the ideal gas law. This means that if the temperature of a gas increases, its pressure will increase proportionally, assuming that the volume and number of gas molecules remain constant.

In this problem, we are given the initial pressure of neon gas at 400 K, which is 1.18 atm. We need to find the pressure of the gas when the temperature is raised to 600 K.

To solve this problem, we can use the following formula:

where P₁ is the initial pressure, T₁ is the initial temperature, P₂ is the final pressure, and T₂ is the final temperature.

Substituting the given values, we get:

Therefore, the pressure produced when the temperature is raised to 600 K is 1.77 atm. This means that the pressure of the neon gas increases by a factor of 1.5 when the temperature is increased from 400 K to 600 K.

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75.83 grams of KNO3 are required to prepare a 0.50 M solution in 1.50 L of water.

To prepare a 0.50 M solution of KNO3 in 1.50 L of water, we can determine the amount of KNO3 required by using the formula:

Molarity (M) = moles of solute / liters of solution

Rearranging the formula, we can calculate the number of moles of KNO3:

moles of KNO3 = Molarity x liters of solution

Given the values, we find:

moles of KNO3 = 0.50 M x 1.50 L = 0.75 moles

To find the mass of KNO3 needed, we need to use its molar mass:

molar mass of KNO3 = 101.10 g/mol

Therefore, the mass of KNO3 required is:

mass of KNO3 = moles of KNO3 x molar mass of KNO3

Substituting the values, we obtain:

mass of KNO3 = 0.75 moles x 101.10 g/mol = 75.83 g

Hence, to prepare a 0.50 M solution in 1.50 L of water, you would need 75.83 grams of KNO3.

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If we had a 32-gram sample of C-14 today, there would be 4 grams of C-14 remaining in 10,470 years.

The half-life of C-14 is 5370 years, which means that in 5370 years, half of the original sample of C-14 would decay. After another 5370 years, half of what remains would decay, and so on.

This can be modeled by the equation:

[tex]N = N_0(1/2)^{(t/T)[/tex]

Where:

N is the amount of C-14 remaining after time t

N₀ is the initial amount of C-14

T is the half-life of C-14

Using the given information, we can substitute N₀ = 32 g, T = 5370 years, and t = 10,470 years into the equation to find N:

[tex]N = 32 g \cdot (1/2)^{(\frac{10,470 years}{5370 years})[/tex]

N = 4 g

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The volume of a sample of gas is 2.8 L when the pressure is 749.5 mm Hg and the temperature is 31. 2°C. (c) 480°C is the new temperature in degrees Celsius if the volume increases to 4. 3 L and the pressure increases to 776.2 mm Hg

Using the combined gas law:

(P1V1) / (T1) = (P2V2) / (T2)

Where:

P1 = 749.5 mm Hg

V1 = 2.8 L

T1 = 31.2 + 273.15 = 304.35 K (temperature converted to Kelvin)

P2 = 776.2 mm Hg

V2 = 4.3 L

T2 = ?

Solving for T2:

T2 = (P2V2T1) / (P1V1)

T2 = (776.2 mmHg * 4.3 L * 304.35 K) / (749.5 mmHg * 2.8 L)

T2 ≈ 758 K

Converting T2 back to Celsius:

T2 = 758 K - 273.15 = 484.85°C ≈ 480°C

Therefore, the new temperature is approximately 480°C.

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A total of 0.1102 moles of Ag will be produced from 3.50 g of Cu.

To determine the number of moles of Ag produced from 3.50 g of Cu, we need to use stoichiometry.

From the balanced chemical equation, we see that 1 mole of Cu reacts with 2 moles of Ag to produce 1 mole of Cu(NO₃)₂ and 2 moles of Ag.

First, we need to convert 3.50 g of Cu to moles by dividing by its molar mass, which is 63.55 g/mol.

3.50 g Cu / 63.55 g/mol = 0.0551 mol Cu

Next, we use the stoichiometry ratio to determine the number of moles of Ag produced:

0.0551 mol Cu x (2 mol Ag / 1 mol Cu) = 0.1102 mol Ag


In summary, we use stoichiometry to determine the number of moles of Ag produced from 3.50 g of Cu by first converting the mass of Cu to moles, and then using the stoichiometry ratio from the balanced chemical equation.

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The Earth is emitting the most longwave radiation at a wavelength of approximately 11.4 micrometers.

Longwave radiation emission, also known as infrared radiation, is the process by which the Earth releases heat into space. This radiation is absorbed by greenhouse gases in the atmosphere, which then trap the heat and prevent it from escaping back into space.

If the Earth had no atmosphere, this longwave radiation emission would be lost quickly to space, resulting in a much cooler planet.

To calculate the rate of radiation emitted (E) by the Earth at a temperature of 255 K, we can use the Stefan-Boltzmann Law, which states that E = σT⁴, where σ is the Stefan-Boltzmann constant (5.67 x 10⁻⁸ W/m²K⁴) and T is the temperature in Kelvin. Plugging in the values, we get:

E = 5.67 x 10⁻⁸ x (255)⁴
E = 3.8 x 10⁸ W/m²

This means that the Earth is emitting 3.8 x 10⁸ watts of longwave radiation per square meter at a temperature of 255 K.

The wavelength of maximum radiation emission can be determined using Wien's Law, which states that the wavelength of maximum emission (λmax) is equal to the constant of proportionality (b) divided by the temperature in Kelvin. The value of b is approximately equal to 2.898 x 10⁻³ mK.

Plugging in the values, we get:

λmax = b/T
λmax = 2.898 x 10⁻³ / 255
λmax = 1.14 x 10⁻⁵ meters

This means that the Earth is emitting the most longwave radiation at a wavelength of approximately 11.4 micrometers.

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The electrolysis of water involves the passage of an electric current through water, which leads to the splitting of water molecules into their constituent elements, hydrogen and oxygen. The correct answer is (a)

This process occurs through two simultaneous half-reactions at the cathode and anode of the electrolysis cell.

At the cathode, hydrogen ions (H+) are reduced to hydrogen gas (H2) as they gain electrons from the electrode: [tex]2H+ + 2e-[/tex]→ [tex]H2[/tex]

At the anode, water molecules (H2O) are oxidized to oxygen gas (O2) and positively charged hydrogen ions (H+):  [tex]2H2O[/tex]→ [tex]O2 + 4H+ + 4e-[/tex]

Therefore, the correct answer is (a) hydrogen is reduced; oxygen is oxidized. During the electrolysis of water, hydrogen is reduced at the cathode, while oxygen is oxidized at the anode.

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--The complete question is, what reactions take place during the electrolysis of water?

group of answer choices

a.  hydrogen is reduced; oxygen is oxidized

b. oxygen is reduced; hydrogen is oxidized

c. . oxygen gas is reduced; water is oxidized.

d.  water is reduced; oxygen gas is oxidized.

e. both oxygen and hydrogen are oxidized and reduced.--

The new pressure at 273 K, given that the initial pressure was 378 KPa, is 249.9 KPa

The following parameters were obtained from the question:

The new pressure of the gas at 273 K can be obtained as shown below:

P₁ / T₁ = P₂/ T₂

378 / 413 = P₂ / 273

Cross multiply

413 × P₂ = 378 × 273

413 × P₂ = 103194

Divide both sides by 413

P₂ = 103194 / 413

P₂ = 249.9 KPa

Thus, from the above calculation, we can conclude the new pressure at 273 K is 249.9 KPa

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A 4 L sample of gas at 30 degrees celcius and 1 atm is changed to 0 degrees celcius and 800torr. 4.51 L is its new volume.

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas.

[tex]P1V1/T1 = P2V2/T2[/tex]

where P1, V1, and T1 are the initial conditions, and P2, V2, and T2 are the final conditions.

Substituting the given values, we get:

[tex]\left(\frac{{1 , \text{atm} \cdot 4 , \text{L}}}{{303 , \text{K}}}\right) = \left(\frac{{0.8 , \text{atm} \cdot V2}}{{273 , \text{K}}}\right)[/tex]

Solving for V2, we get:

[tex]V2 = \frac{{1 , \text{atm} \cdot 4 , \text{L} \cdot 273 , \text{K}}}{{303 , \text{K} \cdot 0.8 , \text{atm}}} = 4.51 , \text{L}[/tex]

Therefore, the new volume of the gas is 4.51 L when the temperature is changed from 30 degrees Celsius to 0 degrees Celsius and the pressure is changed from 1 atm to 800 torr.

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Alanine is also a non-polar amino acid and would have similar Protein structure of leucine. Thus, enzyme activity should be maintained because the polar basic should not undergo major change.

Alanine is an amino acid this is used to make proteins. It is used to interrupt down tryptophan and nutrition B-6. It is a supply of power for muscular tissues and the principal frightened gadget. It strengthens the immune gadget and allows the frame use sugars. Alanine is a nonacidic α-amino acid. Its aspect chain (a methyl group) is neither acidic (it isn't greater acidic than water) nor basic (it does now no longer have a nitrogen atom lone pair that isn't delocalized via way of means of resonance).

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The set of coefficients that will balance the chemical equation are: B.) 1, 2, 2, 1

Chemical reactions are processes that cause one set of chemical elements to change into another set of chemical elements. During chemical reaction, atoms are rearranged, bonds between atoms are broken and formed and then new compounds or molecules are produced.

Chemical reactions can be represented using the chemical equations, that show reactants and products.

The balanced chemical equation for the given reaction is: H₂SO₄ (aq) + 2NH₄OH (aq) ---> 2H₂O (l) + (NH₄)2SO₄ (aq)

Therefore, the set of coefficients that will balance the chemical equation are: 1, 2, 2, 1.

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Electronic configuration of  Ba²⁺ is 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶; Ca²⁺ is 1s²2s²2p⁶3s²3p⁶; Li⁺ is 1s²; K⁺ is 1s²2s²2p⁶3s²3p⁶; Na⁺ is 1s²2s²2p⁶; Sr²⁺is  1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶.

Electronic configuration of the elements present in the periodic table is defined as the designation of atoms on the basis of the electrons present in their shells and subshells. The electrons entering in the same valence shell are grouped together which shows similarity in case of physical and chemical properties. Atoms tend to lose electron and attain stable positive charge so as to attain their nearest noble gas configuration.

Electronic configuration of

Ba²⁺ = 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶

Ca²⁺= 1s²2s²2p⁶3s²3p⁶

Li⁺=1s²

K⁺ = 1s²2s²2p⁶3s²3p⁶

Na⁺ = 1s²2s²2p⁶

Sr²⁺=  1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶

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A 100 gram sample of liquid water is heated from 20. 0°C to 50. 0°C, the mass of [tex]KClO_3[/tex](s) that must dissolve to make a saturated solution in 100 g of H2O at 50.0°C is 60 grams.

Table G must be consulted to establish the mass of [tex]KClO_3[/tex](s) that must dissolve to form a saturated solution in 100 g of H2O at 50.0°C.

According to the table, the solubility of  [tex]KClO_3[/tex] at 48°C is 60 grammes of  [tex]KClO_3[/tex] per 100 grammes of [tex]H_2O[/tex]. We must examine the solubility at 48°C since the water is heated from 20.0°C to 50.0°C.

Therefore, the mass of  [tex]KClO_3[/tex](s) that must dissolve to make a saturated solution in 100 g of  [tex]H_2O[/tex] at 50.0°C is 60 grams.

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Your question seems incomplete, the probable complete question is:

A 100. -gram sample of liquid water is heated from 20. 0°C to 50. 0°C. Enough KCIO3(s) is dissolved in the sample of water at 50. 0°C to form a saturated solution.

Based on Table G, determine the mass of KClO3(s) that must dissolve to make a saturated solution in 100. G of H20 at 50. 0°C

Answer: The answer is 2.50g.

I hope this helps and have a great day!

The effect of the stratosphere being colder at the bottom than at the top is: no vertical air movement.

The effect of the stratosphere being colder at the bottom than at the top is the separation of gases into layers. This temperature gradient creates a stable atmosphere, which prevents vertical air movement and sudden weather changes.

Additionally, the separation of gases can enhance radio reception, as radio waves are able to travel further and more easily through stable layers of air.

The effect of the stratosphere being colder at the bottom than at the top is: no vertical air movement. This temperature gradient results in a stable atmosphere with limited mixing, preventing significant vertical air movement within the stratosphere.

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The bonding you would expect to find in a coin made out of copper is metallic bonding.

Metallic bonding is the strong attraction between positively charged metal ions and the surrounding delocalized electrons in a metallic lattice. Copper is a metal and its atoms are closely packed together, forming a lattice structure. In the case of a copper coin, copper atoms lose their outermost electrons to form positively charged copper ions (Cu⁺), which are then held together by the delocalized electrons that move freely throughout the metal lattice. This type of bonding gives copper its characteristic properties such as electrical conductivity, malleability, and ductility making it an ideal material for coinage.

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Answer: 3. It appears as though nothing is happening, but there is constant change occurring.

Explanation:

equilibrium is the state when the changes cancel each other, and the net change is 0.

think of it like a stalemate in tug of war; both people are pulling, but you wont see anything change, because their forces are equal and in opposite direction :)

The reactions that will occur for each activity series include:

Reactive metals are metals that easily undergo chemical reactions with other substances, particularly with acids and water, to form new compounds. These metals are usually found in the lower part of the activity series, which means they have a high tendency to lose electrons and form cations.

Examples of reactive metals include alkali metals (such as lithium, sodium, and potassium) and alkaline earth metals (such as calcium and magnesium).

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Image trancribed:

Based on the activity series, which of the reactions will occur?

Least Reactive

Most Reactive Li Na K Mg Al Zn Fe Ni Sn Pb H Cu Ag Pt F₂ Cl₂ Br₂ I₂

Hint: Is the metal element more reactive than the metal ion in the compound?

A. Mg + NaNO3 →

B. AI+NISO4→

C. Zn + NaNO3 -

D. Sn+ Zn(NO3)2 →

Answer:A. Mg + NaNO₃ →

Explanation:

will occur since Mg is more reactive than Na.