Is -1 degree Celsius cold or hot?

The climatic change in Earth's history that has resulted in glaciers is the cold climate.

During the last 2.6 million years, the Earth has experienced a series of ice ages, or periods of colder global climate, which have led to the growth of glaciers in regions with sufficient snowfall.

These colder periods are associated with changes in the Earth's orbit, tilt, and precession, which affect the amount and distribution of solar radiation received by the Earth. These climatic changes have had significant impacts on the Earth's surface and have influenced the evolution of life on our planet.

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Answer:

Yes that is why you can even see how some chemicals might react but that is if you are not considering temperature, pressure, or if catalyst are involved.

Mass of Na₂SO₄= 14.346 g

Given

17 g NaHCO₃

0.4 mol H₂SO₄

Required

mass of Na₂SO₄

Solution

Reaction

H₂SO₄ (aq) + 2 NaHCO₃ (aq) ⇒ Na₂SO₄(aq) + 2 CO₂(g) + 2 H₂O (l)

mol NaHCO₃ :

= mass : MW

= 17 g : 84 g/mol

= 0.202

Find limiting reactant = mol : coefficient

H₂SO₄ = 0.4 :  1 = 0.4

NaHCO₃ = 0.202 : 2 = 0.101

NaHCO₃ as a limiting reactant (smaller ratio)

Mol Na₂SO₄ based on mol NaHCO₃

From the equation, mol ratio NaHCO₃ : Na₂SO₄ = 2 : 1, so mol Na₂SO₄ :

= 1/2 x mol NaHCO₃

= 1/2 x 0.202

= 0.101

Mass of Na₂SO₄ :

= mol x MW

= 0.101 x 142,04 g/mol

= 14.346 g

Answer:

504.57 K.

Explanation:

From the question given above, the following data were obtained:

Number of mole (n) = 6.81 moles

Pressure (P) = 2.99 atm

Volume (V) = 94.35 L

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) =.?

Using the ideal gas equation, the temperature of the ideal gas can be obtained as follow:

PV = nRT

2.99 × 94.35 = 6.81 × 0.0821 × T

282.1065 = 0.559101 × T

Divide both side by 0.559101

T = 282.1065 / 0.559101

T = 504.57 K.

Thus, the temperature of the ideal gas is 504.57 K.

Answer:

B) Group 1 and 17

Explanation:

alkali metals and the halogens are the most reactive because of valence electrons

LiAlH4 reagent can reduce aldehydes to primary alcohols, ketones to secondary alcohols, carboxylic acids and esters to primary alcohols, amides and nitriles to amines, epoxides to alcohols and lactones to diols.

LiAlH4 is a powerful reducing agent compared to sodium borohydride, NaBH4, since the Al-H bond is weaker and thus less stable than B-H bond.

Reduction of aldehydes or ketone to primary  and secondary alcohols:

Initially, a hydride ion is transferred onto the carbonyl carbon and the oxygen atom coordinates to the remaining aluminum hydride species to furnish an alkoxy tri hydro aluminate ion, which can reduce the next carbonyl molecule. Thus three of the hydride ions are used up in reduction. The reaction will be as follows :

    AlH4 + ROR   →  RHROH

Reduction of amides to amine:

Amides are converted to amines. The LAH reduction mechanism is slightly different from that depicted for esters. In iminium ion is formed during the reaction since nitrogen atom is relatively a good donor than oxygen atom. The reaction will be as follows:

RONR2 + AlH4 → RNR2

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Answer:

0.0559 mol

Explanation:

Step 1: Given and required data

Step 2: Convert "P" to atm

We will use the conversion factor 1 atm = 760 torr.

750 torr × 1 atm/760 torr = 0.987 atm

Step 3: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 17.0 °C + 273.15 = 290.2 K

Step 4: Calculate the number of moles of air

If we assume air behaves as an ideal gas, we can calculate the moles (n) of air using the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.987 atm × 1.35 L/(0.0821 atm.L/mol.K) × 290.2 K = 0.0559 mol

Answer:

The crushed tablet dissolves fastest among the three forms.

Explanation:

The experiment is aimed at determining the effect of surface area of reactants on reaction rate.

Sodium bicarbonate tablets are provided and divided into three groups of consisting of three trials each. The temperature of the water is kept constant at 20°C. The first group consists of three tablets of sodium bicarbonate dissolved in individual trials in water. The second group consists of three quarter tablets of sodium bicarbonate dissolved in individual trials in water. The third group consists of three crushed tablets of sodium bicarbonate dissolved in individual trials in water.

The results shows that the average time for the dissolution of the first group is 51 seconds; the second group is 42 seconds while the third group is 17 seconds.

Therefore, it can be concluded that increasing the surface area of reactants increases the rate of reaction. This is because more reactant molecules are exposed for dissolution by water with increase in surface area.

Answer:

cold water

Explanation:

Answer:

At a temperature of 300K, a gas has a volume of 3.0 L. If we double the temperature to 600K, the volume will increase to 6.0 L. However, the pressure will remain the same at 1 atm. Therefore, the new volume is 6.0 L x 1 atm = 6.0 L

The presence of electron donating methoxy group in  p-methoxybenzaldehyde , makes it less reactive towards  nucleophilic additions.

Nucleophilic addition reactions that involve of p-Nitrobenzaldehyde and p-methoxybenzaldehyde both occur at the C=O functional group. The electron density around the carbonyl is a key factor that controls the reactivity of each compound.

In  p-Nitrobenzaldehyde, the electron withdrawing nitro group decreases the electron density around the carbonyl making it more vulnerable to nucleophilic attack.

In p-methoxybenzaldehyde, the methoxy group donates electrons by resonance towards the carbonyl thereby increasing the electron density around it so that it is less reactive towards nucleophillic addition reaction.

The nucleophilic attack on p-Nitrobenzaldehyde leads to more resonance structures than nucleophilic attack on p-methoxybenzaldehyde.

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Answer:

See explanation

Explanation:

A popular experiment that describes the effect of heat on the position of equilibrum is the change of colour when copper II chloride is heated.

As the solution is heated, it's colour changes from blue to green, this implies the the colour change (blue to green) is an endothermic process (equilibrum position shifts to the right with increase in temperature)

The equilibrum is represented by the equation;

[Cu(H2O)6]^2+(aq) + 4Cl^-(aq)<------>[CuCl4]^2-(aq) + 6H2O(l) ∆H=positive

The equilibrium mixture undergoing cooling or heating have colour changes. The temperature affects the colour of the products formed and the forward reaction is endothermic.

In the reaction copper (II) chloride or \(\rm CuCl_{4}\) is the main species. The heat or the temperature affects the colour formation of copper (II) chloride as the equilibrium change affects the colouration of the product.

The heating of the solution affects the colour change from blue to the green of the reactant to products and the forward reaction shifts the equilibrium towards the right when the temperature is increased and is an endothermic reaction.

The reaction at the equilibrium can be shown as,

\(\rm [Cu(H_{2}O)_{6}]^{2+} (aq) + 4Cl^{-} (aq) \Leftrightarrow [CuCl_{4}]^{2-}(aq) + 6H_{2}O(l), \Delta H=positive\)

Therefore, temperature changes the colouration and the forward reaction is an endothermic reaction.

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Answer:

I believe that would be 100 mph.

Explanation:

Your distance (d) is 500 miles and your time (t) is 5 hours.

So when you replace the variables with numbers you bet 500/5 which is 100.

Therefore, you speed (S) would be 100mph.

The new volume assuming that the pressure and temperature remain constant is 0.46 L and the correct option is option 1.

The Ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behaviour of many gases under many conditions, although it has several limitations. The ideal gas equation can be written as-

                        PV = nRT

where,

P = Pressure

V = Volume

T = Temperature

n = number of moles

Given,

Initial volume = 1.5 L

Initial moles = 7.5 moles.

Moles remaining = 2.3 moles

\(\frac{n_{1} }{V_{1} } = \frac{n_{2} }{V_{2} }\)

\(\frac{7.5}{1.5 } = \frac{2.3}{V_{2} } }\)

V₂ = 0.46 L

Thus, the ideal selection is option 1.

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Answer:

1

 \(T_{d} = 6.9 ^oC\)  

2

 \(C =   0.7041 M\)

3

 \(x = 0.00355 \  mols \)

4

\(Z = 135.3 \ g/mol\)

Explanation:

Considering Question 1

From the question we are told that

The freezing point of Benzophenone is \(T_{pure} = 49.9^oC\)

The freezing point of Benzophenone + unknown is \(T_{im} = 43.0^o C\)

We are told that we can calculate the freezing point depression of the solution by evaluating the difference between the freezing points of the pure benzophenone and the unknown + benzophenone solution

This mathematically represented as

\(T_{d} = 49.9 - 43.0\)

=> \(T_{d} = 6.9 ^oC\)

Considering Question 2

From the question we are told that

The freezing point depression constant for benzophenone is \(K_{fp} = 9.80 \ ^oC /molal\)

Generally freezing point depression is also mathematically represented as

\(T_d = K_{fp} * C\)

Here C is the morality, now making C the subject we have

\(C = \frac{T_d}{K_{kp}}\)

=> \(C = \frac{6.9}{9.80 }\)

=> \(C = 0.7041 M\)

Considering Question 3

From the question we are told that

The mass of Benzophenone is \(m_b = 5.047\)

This morality obtained can be interpreted as

0.7041 moles of the solute(unknown) is in 1000 g of the solvent (benzophenone)

x moles of solute(unknown) is in 5.047 g of the solvent (benzophenone)

=> \(x = \frac{0.7041 * 5.047}{1000 }\)

=> \(x = 0.00355 \ mols \)

Considering Question 4

From the question we are told that

The mass of unknown solute is \(m = 0.480\)

Generally molar mass of the unknown solute is mathematically represented as

\(Z = \frac{m}{x}\)

=> \(Z = \frac{0.480}{0.00355}\)

=> \(Z = 135.3 \ g/mol\)

As per the details given, the molality of the solution of the unknown in benzophenone is approximately 0.704 m. 0.003550 moles of the unknown solute present in the solution. The molar mass of the unknown solute is approximately 135.21 g/mol.

1. Subtract the freezing point of the solution (43.0°C) from the freezing point of pure benzophenone (49.9°C) to compute the freezing point depression:

Freezing point depression = 49.9°C - 43.0°C = 6.9°C

2. By rearranging the equations, we can get the molality of the solution using the freezing point depression constant (Kfp) for benzophenone (9.80°C/molal):

Freezing point depression = Kfp * molality

Molality = Freezing point depression / Kfp = 6.9°C / 9.80°C/molal ≈ 0.704 molal

The molality of the solution of the unknown in benzophenone is approximately 0.704 m.

3. The number of moles:

Moles of solute = Molality * Mass of solvent (in kg)

Moles of solute = 0.704 molal * 0.005047 kg ≈ 0.003550 moles

There are 0.003550 moles of the unknown solute present in the solution.

4. Divide the mass of the solute (0.480 g) by the number of moles estimated in the previous step to get the approximate molar mass (gramme molecular weight) of the unknown solute:

Molar mass = Mass of solute / Moles of solute

= 0.480 g / 0.003550 moles

≈ 135.21 g/mol

Thus, the molar mass (gram molecular weight) of the unknown solute is approximately 135.21 g/mol.

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Answer:

the real answer is Iceland, Norway, Antarctica.

Explanation:

It is not Green land proof Glaciers are not an unusual sight in Greenland due to the gigantic ice sheet that covers most of the country. However, there are only a few places where the glaciers occur immediately adjacent to a town, and therefore can be appreciated on an excursion.  

proof for Antarctica These large-scale features of glacial erosion can be seen in Antarctica where rock protrudes above the surface of the ice sheet There are many glaciers in the Antarctic. ... The lists include outlet glaciers, valley glaciers, cirque glaciers, tidewater glaciers and ice streams. Ice streams are a type of glacier and many of them have "glacier" in their name, e.g. Pine Island Glacier

proof for Norway are the largest glaciers on mainland Norway. ... In total, Norway has around 1,600 glaciers - 900 of these are in North Norway, but 60% of the total glacier area is south of Trøndelag. 1% of mainland Norway is covered by glaciers.

proof for Iceland Glaciers in Iceland are among the best natural wonders in the world. In fact, the country's glaciers are so great that you can see them from space! With 11% of the total land area covered by ice caps, Iceland is a glacier paradise. Here are the five most amazing glaciers that you must see when visiting Iceland!

Answer:

T2+500k - 273 = 223 *C

Explanation:

Answer:

The answer is A: (more below)

Liquid A: Neutral

Liquid B: Acid

Liquid C: Base

Explanation:

Ph scale: 1 - 6 are acids, 7 is neutral, 8 - 14 are bases.  

14

13

12

11

10

9

8

7

6

5

4

3

2

1

Answer:

B. Molarity

Molarity is the number of moles of solute present in one litre of the solution.

Answer: Molarity

Explanation:

I took the test!

Answer:

C) one magnesium and one oxygen

\(K_{sp}\) of the compound is found to be  5.04 ×\(10^{-10}\).

Solubility :Solubility can be define as the amount of a substance that dissolves or mixes in a given amount of solvent at specific conditions.

Solubility equilibrium

Ksp = \([A^{+} ]^{a}\) \([B^{-} ]^{b}\)

Ksp = solubility product constant

A+ = cation in an aquious solution

B- = anion in an aqueous solution

a, b = relative concentrations of a and b

Given,

Solubility = s = 3.60 × \(10^{-2}\) g/L

molar mass = 288 g/ mol

∴ s= 3.60 × \(10^{-2}\) g/L ÷ 288 g/ mol = 1.25 ×\(10^{-4}\) mol/ L

Reaction:

MX3 ⇄ M + 3X

           s       3s

\(K_{sp}\) =[ \(M^{+3}\)] [ \(X^{-1}\)\(]^{3}\) = solubility product

∴ \(K_{sp}\) =\([s]^{} [3s]^{3}\)

∴ \(K_{sp}\) = 3 \(s^{4}\)

∴ \(K_{sp}\) = 3 × (3.60 × \(10^{-2}\) \()^{4}\)

∴ \(K_{sp}\) = 503.8848 ×\(10^{-8}\)  = 5.04 ×\(10^{-10}\)

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Answer:

A) –7.8944×10-13 J/atom

Explanation:

Mass defect of a radionuclide (m)

\(=-8.7839\)×\(10^{-30} kg\)

Formula for binding energy

\(E=mc^{2}\)

   \(=(-8.7839x10^{-30} kg)(3x10^{8} m/s)^{2}\)

\(E=-7.8944x10^{-13} J/atom\)

If the mass defect for a radionuclide is – 8.7839 × 10⁻³° kg, the binding energy per atom will be –7.8944 × 10⁻¹³  J/atom. The correct option is A.

A radionuclide is an unstable nuclide because it contains so much charge. The excess energy is used by the gamma radiation by the nucleus, the electron uses energy to move to another orbital.

Radionuclides are particles that are used to scanning or monitor the radioactive chemicals that are in the body due to swallowing or inhaling.

The binding energy per atom can be calculated by the formula

The mass of defect of a radionuclide (m) is – 8.7839 × 10⁻³° kg

E = mc²

E =  – 8.7839 × 10⁻³° x 3 x 10⁸ m/s

E = –7.8944 × 10⁻¹³  J/atom.

Thus, the correct option is A. –7.8944×10-13 J/atom.

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Answer:

Glass 1

Explanation:

Molarity is measured in moles of substance per liter.

For the sake of calculations, let's say that each glass contains 1 mole of juice and 1 liter after it is divided between the glasses. If you add an equal amount of water to glass 1 ( another liter), you now have:

Glass 1 = 1 mole / 2 liters = 0.5 M

Glass 1 = 1 mole/ 1 liter = 1 M

So glass 1 will have a lower molarity

The concentration of SO42- ions in the final solution is 0.01 mol L-1.

The neutralisation reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) requires the use of a chemical equation that is balanced:

H2SO4 + NaOH = Na2SO4 + 2H2O

We may deduce from the balanced equation that one mole of H2SO4 reacts with one mole of Na2SO4, meaning that the amount of SO42- ions produced is equal to the amount of H2SO4 utilised.

We may determine how many moles of H2SO4 were used by calculating the amount of 0.05 M H2SO4 solution needed to neutralise 25 mL of NaOH solution:

H2SO4 moles are equal to the product of the volume of H2SO4 (in L) and the concentration of H2SO4 (in mol L-1): 0.03214 L 0.05 mol L-1 = 0.001607 mol

The formation of SO42- ions results in the same amount of moles as the
The resulting solution contains the following amounts of SO42- ions per mole of H2SO4 used:

Moles of SO42- ions/volume of solution (in L) = 0.001607 mol/(0.025 L + 0.03214 L) = 0.01 mol L-1; concentration of SO42- ions (in mol L-1)

As a result, the final solution contains 0.01 mol L-1 of SO42- ions.

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the answer is 95 kg/mol

Using the data in the tables and Coulomb's law, the energy change for the given reaction is; ΔE = 95 KJ/mol

The question is a bit scattered but from online sources, the right order is observed and the question solved as below;

     We want to calculate the energy change (ΔE) for the given reaction. We are given their respective ionization energies and electron affinities.

Now, the formula for change in Energy here is;

ΔE = Ionization Energy + Electron Affinity

Our given reaction is;

K(g) + Br(g) → K⁺(g) + Br⁻(g)

Looking at the reaction, we can see that K transformed to positive from neutral while Br transformed to negative from neutral.

Thus, we can say that;

- K lost Electrons

- Br gained Electrons

      Energy required to lose electrons is called ionization energy and as such, we are given;

Ionization Energy of K = 0.696 aJ

From online conversion using the given ionic radius of 138 pm, we can convert 0.696 aJ to KJ/mol to give 419 KJ/mol

Thus; Ionization Energy of K = 419 KJ/mol

     Since Br gained electrons, the energy that is involved is Electron affinity.

We are given;

Electron affinity of Br = -0.54 aJ

Again, from online conversion using the given radius of 196 pm, we have -0.54 aJ = -324 KJ/mol

Thus, Electron affinity of Br = -324 KJ/mol

Plugging in the relevant values into the change in energy equation earlier stated gives us;

ΔE = 419 + (-324)

ΔE = 95 KJ/mol

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The density (in g/cm³) of a cube of metal that is 2 inches on a side and weighs 444.5 grams is 3.4 g/cm³ (1st option)

First, we shall obtain the volume of the metal. Details below:

Volume = L³

Volume = 5.08³

Volume = 131.1 cm³

Finally, we shall obtain the density of the metal. Details below:

Density = mass / volume

Density of metal = 444.5 / 131.1

Density of metal = 3.4 g/cm³

Thus, we can conclude from the above calculation that the density of the metal is 3.4 g/cm³ (1st option)

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Complete question:

What is the density (in g/cm³) of a cube of metal that is 2 inches on a side and weighs 444.5 grams?

3.4 g/cm³

8.89 g/cm³

55.56 g/cm³

0.180 g/cm³

The birth weight of cats at the 90th percentile would be approximately 3.7 ounces. b. The birth weight of cats at the 10th percentile would be approximately 2.3 ounces.

There are three main methods of calculating weight:

1. Balance Beam Scale: A balance beam scale is an example of a mechanical weighing system. It uses a set of calibrated weights to measure the weight of an object.

2. Digital Scale: A digital scale uses an electronic or digital readout to display the weight of an object.

3. Calipers: Calipers are devices used to measure the distance between two points, such as thickness or diameter. They come in various sizes and shapes, depending on the type of measurement the user wants to take.

Birth weight is the weight of a baby at the time of birth. It is usually measured soon after delivery, with a special baby scale, though sometimes the baby's weight is estimated. The normal range of birth weight is anywhere between 5 lbs 8 oz and 10 lbs, though preterm or premature babies may be significantly lighter. A baby's birth weight is important because it can provide a clue to the baby's overall health. High birth weights may indicate an underlying medical condition, and low birth weights can be a sign of premature or difficult delivery and health risks associated with such a delivery.

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Tthe initial temperature of the polystyrene sample is 39.4°C.

Given: Mass of polystyrene sample = 54.2 gSpecific heat of polystyrene = 1.880 J-g°CWater mass = 100.0 g Initial water temperature = 21.0°CWater final temperature = 34.3°CPressure remains constant at 1 atmFormula used:Heat gained by water = heat lost by polystyreneHence,Heat lost by polystyrene = Heat gained by water=> mcΔT = mcΔTwhere,m = mass of polystyrene or waterc = specific heat capacityΔT = change in temperatureThe temperature change is ΔT = 34.3°C - 21.0°C = 13.3°CNow we can use this temperature change to calculate the initial temperature of the polystyrene.Taking the water's specific heat capacity, c = 4.184 J/g°CHeat gained by water = (100.0 g)(4.184 J/g°C)(13.3°C) = 5574 JHeat lost by polystyrene = 5574 JTaking the polystyrene's specific heat capacity, c = 1.880 J/g° ) = 13.3°C Now let's calculate the mass of polystyrene using the specific heat capacity formula.5574 J = (54.2 g)(1.880 J/g°C)(13.3°C - Ti)Ti = 39.4°C

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There are approximately 6.022 × 10^23 atoms in 12 grams of Carbon-12 (12C).

The number of atoms in a given amount of a substance can be calculated using Avogadro's number, which represents the number of atoms or molecules in one mole of a substance. Avogadro's number is approximately 6.022 × 10^23.

Carbon-12 is a specific isotope of carbon, with an atomic mass of 12 atomic mass units (amu). One mole of Carbon-12 has a mass of 12 grams. Since one mole of any substance contains Avogadro's number of particles, in the case of Carbon-12, it contains 6.022 × 10^23 atoms.

Therefore, if we have 12 grams of Carbon-12, which is equal to one mole, we can conclude that there are approximately 6.022 × 10^23 atoms in this amount of Carbon-12.

In summary, 12 grams of Carbon-12 contains approximately 6.022 × 10^23 atoms. Avogadro's number allows us to relate the mass of a substance to the number of atoms or molecules it contains, providing a fundamental concept in chemistry and enabling us to quantify and understand the microscopic world of atoms and molecules.

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Answer:

Explanation:

The size of the orbital is determined by the principal quantum number (n), so the size of the orbital increases as this value increases. Therefore, an electron in a 1 orbital is closer to the nucleus than is an electron in a 2 orbital.

An atomic orbital can be thought  of as the wave function of an electron in an atom. The Principal Quantum Number (n)  is one of the quantum numbers that are required to describe the distribution of electrons in atoms. This quantum number has positive integral values such as 1, 2, 3 and so on. It also relates to the mean distance of the electron from the nucleus in a given orbital. The larger the quantum number, the greater the mean distance of an electron in that orbital from the nucleus, therefore the orbital will be larger.